Integrand size = 31, antiderivative size = 272 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {\left (4 a A b-6 a^2 B-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{2 b^4 d}+\frac {2 a^2 \left (2 a^2 A b-3 A b^3-3 a^3 B+4 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}+\frac {\left (2 a^2 A b-A b^3-3 a^3 B+2 a b^2 B\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}+\frac {a (A b-a B) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:
-1/2*(4*A*a*b-6*B*a^2-B*b^2)*arctanh(sin(d*x+c))/b^4/d+2*a^2*(2*A*a^2*b-3* A*b^3-3*B*a^3+4*B*a*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2 ))/(a-b)^(3/2)/b^4/(a+b)^(3/2)/d+(2*A*a^2*b-A*b^3-3*B*a^3+2*B*a*b^2)*tan(d *x+c)/b^3/(a^2-b^2)/d-1/2*(2*A*a*b-3*B*a^2+B*b^2)*sec(d*x+c)*tan(d*x+c)/b^ 2/(a^2-b^2)/d+a*(A*b-B*a)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d *x+c))
Time = 6.96 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.61 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {2 a^2 \left (-2 a^2 A b+3 A b^3+3 a^3 B-4 a b^2 B\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} \left (-a^2+b^2\right ) d}+\frac {\left (4 a A b-6 a^2 B-b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac {\left (-4 a A b+6 a^2 B+b^2 B\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 b^4 d}+\frac {B}{4 b^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {B}{4 b^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A b \sin \left (\frac {1}{2} (c+d x)\right )-2 a B \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {A b \sin \left (\frac {1}{2} (c+d x)\right )-2 a B \sin \left (\frac {1}{2} (c+d x)\right )}{b^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {-a^3 A b \sin (c+d x)+a^4 B \sin (c+d x)}{b^3 (-a+b) (a+b) d (b+a \cos (c+d x))} \] Input:
Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^2,x]
Output:
(-2*a^2*(-2*a^2*A*b + 3*A*b^3 + 3*a^3*B - 4*a*b^2*B)*ArcTanh[((-a + b)*Tan [(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*(-a^2 + b^2)*d) + (( 4*a*A*b - 6*a^2*B - b^2*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2*b^ 4*d) + ((-4*a*A*b + 6*a^2*B + b^2*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/ 2]])/(2*b^4*d) + B/(4*b^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - B/( 4*b^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (A*b*Sin[(c + d*x)/2] - 2*a*B*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + ( A*b*Sin[(c + d*x)/2] - 2*a*B*Sin[(c + d*x)/2])/(b^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (-(a^3*A*b*Sin[c + d*x]) + a^4*B*Sin[c + d*x])/(b^3*( -a + b)*(a + b)*d*(b + a*Cos[c + d*x]))
Time = 1.83 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4517, 3042, 4580, 25, 3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4517 |
| \(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (-3 B a^2+2 A b a+b^2 B\right ) \sec ^2(c+d x)\right )-b (A b-a B) \sec (c+d x)+2 a (A b-a B)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (3 B a^2-2 A b a-b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a (A b-a B)\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4580 |
| \(\displaystyle \frac {\frac {\int -\frac {\sec (c+d x) \left (-2 \left (-3 B a^3+2 A b a^2+2 b^2 B a-A b^3\right ) \sec ^2(c+d x)-b \left (-B a^2+2 A b a-b^2 B\right ) \sec (c+d x)+a \left (-3 B a^2+2 A b a+b^2 B\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (-2 \left (-3 B a^3+2 A b a^2+2 b^2 B a-A b^3\right ) \sec ^2(c+d x)-b \left (-B a^2+2 A b a-b^2 B\right ) \sec (c+d x)+a \left (-3 B a^2+2 A b a+b^2 B\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-2 \left (-3 B a^3+2 A b a^2+2 b^2 B a-A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-B a^2+2 A b a-b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (-3 B a^2+2 A b a+b^2 B\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (-3 B a^2+2 A b a+b^2 B\right )+\left (a^2-b^2\right ) \left (-6 B a^2+4 A b a-b^2 B\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (-3 B a^2+2 A b a+b^2 B\right )+\left (a^2-b^2\right ) \left (-6 B a^2+4 A b a-b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \int \sec (c+d x)dx}{b}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}+\frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {a (A b-a B) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {-\frac {\left (-3 a^2 B+2 a A b+b^2 B\right ) \tan (c+d x) \sec (c+d x)}{2 b d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (-6 a^2 B+4 a A b-b^2 B\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^2 \left (-3 a^3 B+2 a^2 A b+4 a b^2 B-3 A b^3\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 \left (-3 a^3 B+2 a^2 A b+2 a b^2 B-A b^3\right ) \tan (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}\) |
Input:
Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^2,x]
Output:
(a*(A*b - a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])) + (-1/2*((2*a*A*b - 3*a^2*B + b^2*B)*Sec[c + d*x]*Tan[c + d*x])/( b*d) - ((((a^2 - b^2)*(4*a*A*b - 6*a^2*B - b^2*B)*ArcTanh[Sin[c + d*x]])/( b*d) - (4*a^2*(2*a^2*A*b - 3*A*b^3 - 3*a^3*B + 4*a*b^2*B)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b - (2 *(2*a^2*A*b - A*b^3 - 3*a^3*B + 2*a*b^2*B)*Tan[c + d*x])/(b*d))/(2*b))/(b* (a^2 - b^2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Time = 0.83 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {\frac {B}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A b -4 B a -B b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (4 A a b -6 B \,a^{2}-b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {B}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A b -4 B a -B b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-4 A a b +6 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {2 a^{2} \left (\frac {a b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 A \,a^{2} b -3 A \,b^{3}-3 B \,a^{3}+4 B a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\) | \(330\) |
| default | \(\frac {\frac {B}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A b -4 B a -B b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (4 A a b -6 B \,a^{2}-b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {B}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A b -4 B a -B b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-4 A a b +6 B \,a^{2}+b^{2} B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}-\frac {2 a^{2} \left (\frac {a b \left (A b -B a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 A \,a^{2} b -3 A \,b^{3}-3 B \,a^{3}+4 B a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\) | \(330\) |
| risch | \(\text {Expression too large to display}\) | \(1276\) |
Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBO SE)
Output:
1/d*(1/2*B/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(2*A*b-4*B*a-B*b)/b^3/(tan(1/2 *d*x+1/2*c)-1)+1/2*(4*A*a*b-6*B*a^2-B*b^2)/b^4*ln(tan(1/2*d*x+1/2*c)-1)-1/ 2*B/b^2/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*A*b-4*B*a-B*b)/b^3/(tan(1/2*d*x+1/ 2*c)+1)+1/2/b^4*(-4*A*a*b+6*B*a^2+B*b^2)*ln(tan(1/2*d*x+1/2*c)+1)-2*a^2/b^ 4*(a*b*(A*b-B*a)/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan( 1/2*d*x+1/2*c)^2*b-a-b)-(2*A*a^2*b-3*A*b^3-3*B*a^3+4*B*a*b^2)/(a+b)/(a-b)/ ((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))) )
Leaf count of result is larger than twice the leaf count of optimal. 643 vs. \(2 (260) = 520\).
Time = 10.79 (sec) , antiderivative size = 1343, normalized size of antiderivative = 4.94 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="f ricas")
Output:
[-1/4*(2*((3*B*a^6 - 2*A*a^5*b - 4*B*a^4*b^2 + 3*A*a^3*b^3)*cos(d*x + c)^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3*b^3 + 3*A*a^2*b^4)*cos(d*x + c)^2)*s qrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2* sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos( d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - ((6*B*a^7 - 4*A*a^6*b - 11*B*a^5 *b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*cos(d*x + c)^3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B*a^2*b^5 - 4*A *a*b^6 + B*b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((6*B*a^7 - 4*A*a^ 6*b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*co s(d*x + c)^3 + (6*B*a^6*b - 4*A*a^5*b^2 - 11*B*a^4*b^3 + 8*A*a^3*b^4 + 4*B *a^2*b^5 - 4*A*a*b^6 + B*b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*( B*a^4*b^3 - 2*B*a^2*b^5 + B*b^7 - 2*(3*B*a^6*b - 2*A*a^5*b^2 - 5*B*a^4*b^3 + 3*A*a^3*b^4 + 2*B*a^2*b^5 - A*a*b^6)*cos(d*x + c)^2 - (3*B*a^5*b^2 - 2* A*a^4*b^3 - 6*B*a^3*b^4 + 4*A*a^2*b^5 + 3*B*a*b^6 - 2*A*b^7)*cos(d*x + c)) *sin(d*x + c))/((a^5*b^4 - 2*a^3*b^6 + a*b^8)*d*cos(d*x + c)^3 + (a^4*b^5 - 2*a^2*b^7 + b^9)*d*cos(d*x + c)^2), -1/4*(4*((3*B*a^6 - 2*A*a^5*b - 4*B* a^4*b^2 + 3*A*a^3*b^3)*cos(d*x + c)^3 + (3*B*a^5*b - 2*A*a^4*b^2 - 4*B*a^3 *b^3 + 3*A*a^2*b^4)*cos(d*x + c)^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b ^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - ((6*B*a^7 - 4*A*a^6 *b - 11*B*a^5*b^2 + 8*A*a^4*b^3 + 4*B*a^3*b^4 - 4*A*a^2*b^5 + B*a*b^6)*...
\[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**2,x)
Output:
Integral((A + B*sec(c + d*x))*sec(c + d*x)**4/(a + b*sec(c + d*x))**2, x)
Exception generated. \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="m axima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.23 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.41 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{5} - 2 \, A a^{4} b - 4 \, B a^{3} b^{2} + 3 \, A a^{2} b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {4 \, {\left (B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {{\left (6 \, B a^{2} - 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} + \frac {{\left (6 \, B a^{2} - 4 \, A a b + B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac {2 \, {\left (4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x, algorithm="g iac")
Output:
-1/2*(4*(3*B*a^5 - 2*A*a^4*b - 4*B*a^3*b^2 + 3*A*a^2*b^3)*(pi*floor(1/2*(d *x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*ta n(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^2)) - 4*(B*a^4*tan(1/2*d*x + 1/2*c) - A*a^3*b*tan(1/2*d*x + 1/2*c))/((a^2*b^3 - b^5)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)) - (6 *B*a^2 - 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 + (6*B*a^ 2 - 4*A*a*b + B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 - 2*(4*B*a*tan (1/2*d*x + 1/2*c)^3 - 2*A*b*tan(1/2*d*x + 1/2*c)^3 + B*b*tan(1/2*d*x + 1/2 *c)^3 - 4*B*a*tan(1/2*d*x + 1/2*c) + 2*A*b*tan(1/2*d*x + 1/2*c) + B*b*tan( 1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3))/d
Time = 21.13 (sec) , antiderivative size = 6678, normalized size of antiderivative = 24.55 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int((A + B/cos(c + d*x))/(cos(c + d*x)^4*(a + b/cos(c + d*x))^2),x)
Output:
(atan(-((((8*tan(c/2 + (d*x)/2)*(72*B^2*a^10 + B^2*b^10 - 2*B^2*a*b^9 - 72 *B^2*a^9*b + 16*A^2*a^2*b^8 - 32*A^2*a^3*b^7 + 20*A^2*a^4*b^6 + 64*A^2*a^5 *b^5 - 64*A^2*a^6*b^4 - 32*A^2*a^7*b^3 + 32*A^2*a^8*b^2 + 11*B^2*a^2*b^8 - 20*B^2*a^3*b^7 + 23*B^2*a^4*b^6 - 26*B^2*a^5*b^5 + 17*B^2*a^6*b^4 + 120*B ^2*a^7*b^3 - 120*B^2*a^8*b^2 - 8*A*B*a*b^9 - 96*A*B*a^9*b + 16*A*B*a^2*b^8 - 40*A*B*a^3*b^7 + 64*A*B*a^4*b^6 - 40*A*B*a^5*b^5 - 176*A*B*a^6*b^4 + 17 6*A*B*a^7*b^3 + 96*A*B*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - (((8* (2*B*b^15 + 12*A*a^2*b^13 + 12*A*a^3*b^12 - 20*A*a^4*b^11 - 4*A*a^5*b^10 + 8*A*a^6*b^9 + 6*B*a^2*b^13 - 16*B*a^3*b^12 - 14*B*a^4*b^11 + 28*B*a^5*b^1 0 + 6*B*a^6*b^9 - 12*B*a^7*b^8 - 8*A*a*b^14))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - (4*tan(c/2 + (d*x)/2)*(6*B*a^2 + B*b^2 - 4*A*a*b)*(8*a*b^13 - 8 *a^2*b^12 - 16*a^3*b^11 + 16*a^4*b^10 + 8*a^5*b^9 - 8*a^6*b^8))/(b^4*(a*b^ 8 + b^9 - a^2*b^7 - a^3*b^6)))*(6*B*a^2 + B*b^2 - 4*A*a*b))/(2*b^4))*(6*B* a^2 + B*b^2 - 4*A*a*b)*1i)/(2*b^4) + (((8*tan(c/2 + (d*x)/2)*(72*B^2*a^10 + B^2*b^10 - 2*B^2*a*b^9 - 72*B^2*a^9*b + 16*A^2*a^2*b^8 - 32*A^2*a^3*b^7 + 20*A^2*a^4*b^6 + 64*A^2*a^5*b^5 - 64*A^2*a^6*b^4 - 32*A^2*a^7*b^3 + 32*A ^2*a^8*b^2 + 11*B^2*a^2*b^8 - 20*B^2*a^3*b^7 + 23*B^2*a^4*b^6 - 26*B^2*a^5 *b^5 + 17*B^2*a^6*b^4 + 120*B^2*a^7*b^3 - 120*B^2*a^8*b^2 - 8*A*B*a*b^9 - 96*A*B*a^9*b + 16*A*B*a^2*b^8 - 40*A*B*a^3*b^7 + 64*A*B*a^4*b^6 - 40*A*B*a ^5*b^5 - 176*A*B*a^6*b^4 + 176*A*B*a^7*b^3 + 96*A*B*a^8*b^2))/(a*b^8 + ...
Time = 0.17 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^2} \, dx=\frac {-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (d x +c \right )^{2} a^{3}+4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a^{3}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{4}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{4}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{4}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{4}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{4}-\sin \left (d x +c \right ) a^{2} b^{2}+\sin \left (d x +c \right ) b^{4}}{2 b^{3} d \left (\sin \left (d x +c \right )^{2} a^{2}-\sin \left (d x +c \right )^{2} b^{2}-a^{2}+b^{2}\right )} \] Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^2,x)
Output:
( - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**3 + 4*sqrt( - a**2 + b**2)*atan(( tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**3 + 2*co s(c + d*x)*sin(c + d*x)*a**3*b - 2*cos(c + d*x)*sin(c + d*x)*a*b**3 - 2*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 + log(tan((c + d*x)/2) - 1)*s in(c + d*x)**2*a**2*b**2 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**4 + 2*log(tan((c + d*x)/2) - 1)*a**4 - log(tan((c + d*x)/2) - 1)*a**2*b**2 - log(tan((c + d*x)/2) - 1)*b**4 + 2*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **2*a**4 - log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**2*b**4 - 2*log(tan((c + d*x)/2) + 1)*a**4 + log(tan((c + d*x)/2) + 1)*a**2*b**2 + log(tan((c + d*x)/2) + 1)*b**4 - si n(c + d*x)*a**2*b**2 + sin(c + d*x)*b**4)/(2*b**3*d*(sin(c + d*x)**2*a**2 - sin(c + d*x)**2*b**2 - a**2 + b**2))