Integrand size = 21, antiderivative size = 222 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{2 b^4 d}-\frac {2 a^3 \left (3 a^2-4 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^4 (a+b)^{3/2} d}-\frac {a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b^3 \left (a^2-b^2\right ) d}+\frac {\left (3 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d}-\frac {a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:
1/2*(6*a^2+b^2)*arctanh(sin(d*x+c))/b^4/d-2*a^3*(3*a^2-4*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-a*(3 *a^2-2*b^2)*tan(d*x+c)/b^3/(a^2-b^2)/d+1/2*(3*a^2-b^2)*sec(d*x+c)*tan(d*x+ c)/b^2/(a^2-b^2)/d-a^2*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+ c))
Time = 5.32 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.28 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {8 a^3 \left (3 a^2-4 b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-12 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {b^2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 a^4 b \sin (c+d x)}{(-a+b) (a+b) (b+a \cos (c+d x))}-8 a b \tan (c+d x)}{4 b^4 d} \] Input:
Integrate[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]
Output:
((8*a^3*(3*a^2 - 4*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2 ]])/(a^2 - b^2)^(3/2) - 12*a^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*b^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*a^2*Log[Cos[(c + d*x)/ 2] + Sin[(c + d*x)/2]] + 2*b^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - b^2/(Cos[(c + d*x)/2] + Sin[ (c + d*x)/2])^2 + (4*a^4*b*Sin[c + d*x])/((-a + b)*(a + b)*(b + a*Cos[c + d*x])) - 8*a*b*Tan[c + d*x])/(4*b^4*d)
Time = 1.66 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.05, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4332, 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.
\(\displaystyle \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4332 |
\(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) \left (2 a^2-b \sec (c+d x) a-\left (3 a^2-b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a^2-b \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (b^2-3 a^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4580 |
\(\displaystyle -\frac {\frac {\int -\frac {\sec (c+d x) \left (-2 a \left (3 a^2-2 b^2\right ) \sec ^2(c+d x)-b \left (a^2+b^2\right ) \sec (c+d x)+a \left (3 a^2-b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {-\frac {\int \frac {\sec (c+d x) \left (-2 a \left (3 a^2-2 b^2\right ) \sec ^2(c+d x)-b \left (a^2+b^2\right ) \sec (c+d x)+a \left (3 a^2-b^2\right )\right )}{a+b \sec (c+d x)}dx}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (-2 a \left (3 a^2-2 b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+a \left (3 a^2-b^2\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (a b \left (3 a^2-b^2\right )+\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)}dx}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a b \left (3 a^2-b^2\right )+\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4486 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \int \sec (c+d x)dx}{b}-2 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-2 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-2 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle -\frac {-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b d}}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}}{b \left (a^2-b^2\right )}-\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {-\frac {\left (3 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{2 b d}-\frac {\frac {\frac {\left (a^2-b^2\right ) \left (6 a^2+b^2\right ) \text {arctanh}(\sin (c+d x))}{b d}-\frac {4 a^3 \left (\frac {3 a^2}{b}-4 b\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}}{b}-\frac {2 a \left (3 a^2-2 b^2\right ) \tan (c+d x)}{b d}}{2 b}}{b \left (a^2-b^2\right )}\) |
Input:
Int[Sec[c + d*x]^5/(a + b*Sec[c + d*x])^2,x]
Output:
-((a^2*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])) ) - (-1/2*((3*a^2 - b^2)*Sec[c + d*x]*Tan[c + d*x])/(b*d) - ((((a^2 - b^2) *(6*a^2 + b^2)*ArcTanh[Sin[c + d*x]])/(b*d) - (4*a^3*((3*a^2)/b - 4*b)*Arc Tanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b] *d))/b - (2*a*(3*a^2 - 2*b^2)*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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a^2)*d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^( m + 1)*((d*Csc[e + f*x])^(n - 3)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[d^3/ (b*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x]) ^(n - 3)*Simp[a^2*(n - 3) + a*b*(m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*( m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n , 2]))
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_)]*((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 = 0.95 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {\frac {1}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-4 a -b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-6 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {1}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-4 a -b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (6 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (3 a^{2}-4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\) | \(277\) |
default | \(\frac {\frac {1}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-4 a -b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-6 a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 b^{4}}-\frac {1}{2 b^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-4 a -b}{2 b^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (6 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 b^{4}}+\frac {2 a^{3} \left (\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (3 a^{2}-4 b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4}}}{d}\) | \(277\) |
risch | \(-\frac {i \left (-3 a^{3} b \,{\mathrm e}^{5 i \left (d x +c \right )}+a \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-6 a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+2 a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-12 b \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+8 a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-12 a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+10 a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+7 b^{3} a \,{\mathrm e}^{i \left (d x +c \right )}-6 a^{4}+4 a^{2} b^{2}\right )}{d \,b^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2} \left (-a^{2}+b^{2}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{b^{4} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 b^{2} d}+\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}-\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {3 a^{5} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{4}}+\frac {4 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 b^{2} d}\) | \(693\) |
Input:
int(sec(d*x+c)^5/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/2/b^2/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(-4*a-b)/b^3/(tan(1/2*d*x+1/2*c) -1)+1/2/b^4*(-6*a^2-b^2)*ln(tan(1/2*d*x+1/2*c)-1)-1/2/b^2/(tan(1/2*d*x+1/2 *c)+1)^2-1/2*(-4*a-b)/b^3/(tan(1/2*d*x+1/2*c)+1)+1/2*(6*a^2+b^2)/b^4*ln(ta n(1/2*d*x+1/2*c)+1)+2*a^3/b^4*(a*b/(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)-(3*a^2-4*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. 426 vs. \(2 (209) = 418\).
Time = 0.51 (sec) , antiderivative size = 909, normalized size of antiderivative = 4.09 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="fricas")
Output:
[1/4*(2*((3*a^6 - 4*a^4*b^2)*cos(d*x + c)^3 + (3*a^5*b - 4*a^3*b^3)*cos(d* x + c)^2)*sqrt(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*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - ((6*a^7 - 11*a^5*b^2 + 4*a^3* b^4 + a*b^6)*cos(d*x + c)^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*cos (d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(a^4*b^3 - 2*a^2*b^5 + b^7 - 2*(3* a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2 - 3*(a^5*b^2 - 2*a^3*b^4 + a *b^6)*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*a^6 - 4*a^4*b^2)*cos(d*x + c)^3 + (3*a^5*b - 4*a^3*b^3)*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*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (6* a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*cos(d*x + c)^2)*log(sin(d*x + c) + 1 ) + ((6*a^7 - 11*a^5*b^2 + 4*a^3*b^4 + a*b^6)*cos(d*x + c)^3 + (6*a^6*b - 11*a^4*b^3 + 4*a^2*b^5 + b^7)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*( a^4*b^3 - 2*a^2*b^5 + b^7 - 2*(3*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2 - 3*(a^5*b^2 - 2*a^3*b^4 + a*b^6)*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)*...
\[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**5/(a+b*sec(d*x+c))**2,x)
Output:
Integral(sec(c + d*x)**5/(a + b*sec(c + d*x))**2, x)
Exception generated. \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="maxima")
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 = 0.16 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.35 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} - \frac {4 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{4} - b^{6}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {{\left (6 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac {{\left (6 \, a^{2} + b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} + \frac {2 \, {\left (4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} b^{3}}}{2 \, d} \] Input:
integrate(sec(d*x+c)^5/(a+b*sec(d*x+c))^2,x, algorithm="giac")
Output:
1/2*(4*a^4*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)) - 4*(3*a^5 - 4*a^3*b^2)*(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*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^2*b^4 - b^6)*sqrt(-a^2 + b^ 2)) + (6*a^2 + b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^4 - (6*a^2 + b^2) *log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^4 + 2*(4*a*tan(1/2*d*x + 1/2*c)^3 + b*tan(1/2*d*x + 1/2*c)^3 - 4*a*tan(1/2*d*x + 1/2*c) + b*tan(1/2*d*x + 1/2* c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*b^3))/d
Time = 16.21 (sec) , antiderivative size = 3685, normalized size of antiderivative = 16.60 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^5*(a + b/cos(c + d*x))^2),x)
Output:
- ((tan(c/2 + (d*x)/2)^5*(3*a*b^3 - 3*a^3*b + 6*a^4 + b^4 - 5*a^2*b^2))/(( a*b^3 - b^4)*(a + b)) + (2*tan(c/2 + (d*x)/2)^3*(b^4 - 6*a^4 + 3*a^2*b^2)) /(b*(a*b^2 - b^3)*(a + b)) + (tan(c/2 + (d*x)/2)*(3*a^3*b - 3*a*b^3 + 6*a^ 4 + b^4 - 5*a^2*b^2))/(b^3*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2 )^2*(3*a + b) - tan(c/2 + (d*x)/2)^6*(a - b) + tan(c/2 + (d*x)/2)^4*(3*a - b))) - (atan((((6*a^2 + b^2)*((8*tan(c/2 + (d*x)/2)*(72*a^10 - 72*a^9*b - 2*a*b^9 + b^10 + 11*a^2*b^8 - 20*a^3*b^7 + 23*a^4*b^6 - 26*a^5*b^5 + 17*a ^6*b^4 + 120*a^7*b^3 - 120*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) - ( (6*a^2 + b^2)*((8*(2*b^15 + 6*a^2*b^13 - 16*a^3*b^12 - 14*a^4*b^11 + 28*a^ 5*b^10 + 6*a^6*b^9 - 12*a^7*b^8))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) - ( 4*tan(c/2 + (d*x)/2)*(6*a^2 + b^2)*(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))))/(2*b^4))*1i)/(2*b^4) + ((6*a^2 + b^2)*((8*tan(c/2 + (d*x)/2)*(72*a^1 0 - 72*a^9*b - 2*a*b^9 + b^10 + 11*a^2*b^8 - 20*a^3*b^7 + 23*a^4*b^6 - 26* a^5*b^5 + 17*a^6*b^4 + 120*a^7*b^3 - 120*a^8*b^2))/(a*b^8 + b^9 - a^2*b^7 - a^3*b^6) + ((6*a^2 + b^2)*((8*(2*b^15 + 6*a^2*b^13 - 16*a^3*b^12 - 14*a^ 4*b^11 + 28*a^5*b^10 + 6*a^6*b^9 - 12*a^7*b^8))/(a*b^11 + b^12 - a^2*b^10 - a^3*b^9) + (4*tan(c/2 + (d*x)/2)*(6*a^2 + b^2)*(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))))/(2*b^4))*1i)/(2*b^4))/((16*(108*a^11 - 54*a^10*b + ...
Time = 0.19 (sec) , antiderivative size = 1637, normalized size of antiderivative = 7.37 \[ \int \frac {\sec ^5(c+d x)}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^5/(a+b*sec(d*x+c))^2,x)
Output:
( - 12*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b) /sqrt( - a**2 + b**2))*cos(c + d*x)*sin(c + d*x)**2*a**6 + 16*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b** 2))*cos(c + d*x)*sin(c + d*x)**2*a**4*b**2 + 12*sqrt( - a**2 + b**2)*atan( (tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d* x)*a**6 - 16*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x) /2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**4*b**2 - 12*sqrt( - a**2 + b* *2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*s in(c + d*x)**2*a**5*b + 16*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*b**3 + 12* sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**5*b - 16*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*b**3 - 6*cos(c + d*x)*log (tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**7 + 11*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5*b**2 - 4*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2*a**3*b**4 - cos(c + d*x)*log(tan((c + d*x)/2) - 1 )*sin(c + d*x)**2*a*b**6 + 6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**7 - 11*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**5*b**2 + 4*cos(c + d*x)*log( tan((c + d*x)/2) - 1)*a**3*b**4 + cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a *b**6 + 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**7 -...