Integrand size = 31, antiderivative size = 427 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {\left (a^2 A-3 A b^2+2 a b B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 b \sqrt {a+b} d}+\frac {(3 A b+a (A-2 B)) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d}+\frac {\sqrt {a+b} (3 A b-2 a B) \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}+\frac {b \left (a^2 A-3 A b^2+2 a b B\right ) \tan (c+d x)}{a^2 \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]
(A*a^2-3*A*b^2+2*B*a*b)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^ (1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c ))/(a-b))^(1/2)/a^2/b/d/(a+b)^(1/2)+(3*A*b+a*(A-2*B))*cot(d*x+c)*EllipticF ((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c)) /(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/d/(a+b)^(1/2)+(3*A*b-2*B *a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b )/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ c))/(a-b))^(1/2)/a^3/d+A*sin(d*x+c)/a/d/(a+b*sec(d*x+c))^(1/2)+b*(A*a^2-3* A*b^2+2*B*a*b)*tan(d*x+c)/a^2/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1597\) vs. \(2(427)=854\).
Time = 21.33 (sec) , antiderivative size = 1597, normalized size of antiderivative = 3.74 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((-2*b*(A*b - a*B)*Sin[c + d*x])/(a ^2*(-a^2 + b^2)) + (2*(-(A*b^3*Sin[c + d*x]) + a*b^2*B*Sin[c + d*x]))/(a^2 *(a^2 - b^2)*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(3/2)) - ((b + a*Cos[c + d*x])^(3/2)*Sec[c + d*x]^(3/2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^( -1)]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[( c + d*x)/2]^2)]*(a^3*A*Tan[(c + d*x)/2] + a^2*A*b*Tan[(c + d*x)/2] - 3*a*A *b^2*Tan[(c + d*x)/2] - 3*A*b^3*Tan[(c + d*x)/2] + 2*a^2*b*B*Tan[(c + d*x) /2] + 2*a*b^2*B*Tan[(c + d*x)/2] - 2*a^3*A*Tan[(c + d*x)/2]^3 + 6*a*A*b^2* Tan[(c + d*x)/2]^3 - 4*a^2*b*B*Tan[(c + d*x)/2]^3 + a^3*A*Tan[(c + d*x)/2] ^5 - a^2*A*b*Tan[(c + d*x)/2]^5 - 3*a*A*b^2*Tan[(c + d*x)/2]^5 + 3*A*b^3*T an[(c + d*x)/2]^5 + 2*a^2*b*B*Tan[(c + d*x)/2]^5 - 2*a*b^2*B*Tan[(c + d*x) /2]^5 - 6*a^2*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b) ]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[ (c + d*x)/2]^2)/(a + b)] + 6*A*b^3*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]] , (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d *x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 4*a^3*B*EllipticPi[-1, ArcSin[ Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 4*a*b^2*B*Ell ipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d *x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a...
Time = 1.90 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4522, 27, 3042, 4549, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
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.
\(\displaystyle \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4522 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-A b \sec ^2(c+d x)+3 A b-2 a B}{2 (a+b \sec (c+d x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-A b \sec ^2(c+d x)+3 A b-2 a B}{(a+b \sec (c+d x))^{3/2}}dx}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\int \frac {-A b \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 A b-2 a B}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{2 a}\) |
\(\Big \downarrow \) 4549 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {-\frac {2 \int -\frac {b \left (A a^2+2 b B a-3 A b^2\right ) \sec ^2(c+d x)-2 a b (A b-a B) \sec (c+d x)+\left (a^2-b^2\right ) (3 A b-2 a B)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (A a^2+2 b B a-3 A b^2\right ) \sec ^2(c+d x)-2 a b (A b-a B) \sec (c+d x)+\left (a^2-b^2\right ) (3 A b-2 a B)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\int \frac {b \left (A a^2+2 b B a-3 A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-2 a b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+\left (a^2-b^2\right ) (3 A b-2 a B)}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 4546 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {\left (a^2-b^2\right ) (3 A b-2 a B)+\left (-2 a b (A b-a B)-b \left (A a^2+2 b B a-3 A b^2\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {\left (a^2-b^2\right ) (3 A b-2 a B)+\left (-2 a b (A b-a B)-b \left (A a^2+2 b B a-3 A b^2\right )\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 4409 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^2-b^2\right ) (3 A b-2 a B) \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx-b (a-b) (a (A-2 B)+3 A b) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {\left (a^2-b^2\right ) (3 A b-2 a B) \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (a (A-2 B)+3 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 4271 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b (a-b) (a (A-2 B)+3 A b) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {b \left (a^2 A+2 a b B-3 A b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (a (A-2 B)+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {A \sin (c+d x)}{a d \sqrt {a+b \sec (c+d x)}}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (a^2 A+2 a b B-3 A b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b d}-\frac {2 \sqrt {a+b} \left (a^2-b^2\right ) (3 A b-2 a B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}-\frac {2 (a-b) \sqrt {a+b} (a (A-2 B)+3 A b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}}{a \left (a^2-b^2\right )}-\frac {2 b \left (a^2 A+2 a b B-3 A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{2 a}\) |
(A*Sin[c + d*x])/(a*d*Sqrt[a + b*Sec[c + d*x]]) - (((-2*(a - b)*Sqrt[a + b ]*(a^2*A - 3*A*b^2 + 2*a*b*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec [c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b*d) - (2*(a - b)*Sqrt[a + b]*(3*A*b + a*(A - 2*B))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]* Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/d - (2*Sqrt[a + b]*(a^2 - b^2)*(3 *A*b - 2*a*B)*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)] *Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a*(a^2 - b^2)) - (2*b*(a ^2*A - 3*A*b^2 + 2*a*b*B)*Tan[c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(2*a)
3.4.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim p[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 + a^2*C)*Cot[e + f*x]*((a + b*Csc[ e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b* (A + C)*(m + 1)*Csc[e + f*x] + (A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m ] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(3864\) vs. \(2(396)=792\).
Time = 10.37 (sec) , antiderivative size = 3865, normalized size of antiderivative = 9.05
-1/d/a^2/(a+b)/(a-b)*(-6*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) *(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a -b)/(a+b))^(1/2))*a^2*b*cos(d*x+c)^2-4*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d* x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticPi(cot(d*x+c)-csc (d*x+c),-1,((a-b)/(a+b))^(1/2))*a*b^2*cos(d*x+c)^2+4*A*EllipticF(cot(d*x+c )-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1) )^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)+2*A*EllipticF(c ot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d *x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*cos(d*x+c)^2-8*B*( 1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1)) ^(1/2)*EllipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*a*b^2*cos( d*x+c)+A*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+ a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3- 3*A*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos (d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3+6*A*E llipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos( d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^3+4*B*El lipticPi(cot(d*x+c)-csc(d*x+c),-1,((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d *x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3-2*B*Ell ipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(1/(a+b)*(b+a*cos(d*x...
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((B*cos(d*x + c)*sec(d*x + c) + A*cos(d*x + c))*sqrt(b*sec(d*x + c ) + a)/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\cos \left (c+d\,x\right )\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]