Integrand size = 33, antiderivative size = 217 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^3 d}+\frac {2 (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d}+\frac {2 a^2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (A b-a B) \sin (c+d x)}{3 b^2 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (5 a A b-5 a^2 B-3 b^2 B\right ) \sin (c+d x)}{5 b^3 d \sqrt {\cos (c+d x)}} \] Output:
2/5*(5*A*a*b-5*B*a^2-3*B*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^3/d+ 2/3*(A*b-B*a)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^2/d+2*a^2*(A*b-B*a) *EllipticPi(sin(1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))/b^3/(a+b)/d+2/5*B*sin(d* x+c)/b/d/cos(d*x+c)^(5/2)+2/3*(A*b-B*a)*sin(d*x+c)/b^2/d/cos(d*x+c)^(3/2)- 2/5*(5*A*a*b-5*B*a^2-3*B*b^2)*sin(d*x+c)/b^3/d/cos(d*x+c)^(1/2)
Time = 3.03 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.50 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {\frac {b \left (45 a^2 A b+10 A b^3-45 a^3 B-19 a b^2 B\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}-\frac {b^2 \left (-20 a A b+20 a^2 B+9 b^2 B\right ) \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 b \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{a}+\frac {6 b^3 B \sin (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}+\frac {10 b^2 (A b-a B) \sin (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}+\frac {6 b \left (-5 a A b+5 a^2 B+3 b^2 B\right ) \sin (c+d x)}{\sqrt {\cos (c+d x)}}-\frac {3 \left (-5 a A b+5 a^2 B+3 b^2 B\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 b (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (a^2-2 b^2\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a \sqrt {\sin ^2(c+d x)}}}{15 b^4 d} \] Input:
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])),x ]
Output:
((b*(45*a^2*A*b + 10*A*b^3 - 45*a^3*B - 19*a*b^2*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/(a + b) - (b^2*(-20*a*A*b + 20*a^2*B + 9*b^2*B)*(2*El lipticF[(c + d*x)/2, 2] - (2*b*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/ (a + b)))/a + (6*b^3*B*Sin[c + d*x])/Cos[c + d*x]^(5/2) + (10*b^2*(A*b - a *B)*Sin[c + d*x])/Cos[c + d*x]^(3/2) + (6*b*(-5*a*A*b + 5*a^2*B + 3*b^2*B) *Sin[c + d*x])/Sqrt[Cos[c + d*x]] - (3*(-5*a*A*b + 5*a^2*B + 3*b^2*B)*(-2* a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*b*(a + b)*EllipticF[ArcS in[Sqrt[Cos[c + d*x]]], -1] + (a^2 - 2*b^2)*EllipticPi[-(a/b), ArcSin[Sqrt [Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]))/(15*b^4*d)
Time = 2.09 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.07, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.606, Rules used = {3042, 3433, 3042, 3479, 27, 3042, 3534, 27, 3042, 3534, 27, 3042, 3538, 27, 3042, 3119, 3481, 3042, 3120, 3284}
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.
\(\displaystyle \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3433 |
\(\displaystyle \int \frac {A \cos (c+d x)+B}{\cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+b)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+b\right )}dx\) |
\(\Big \downarrow \) 3479 |
\(\displaystyle \frac {2 \int \frac {3 a B \cos ^2(c+d x)+3 b B \cos (c+d x)+5 (A b-a B)}{2 \cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))}dx}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {3 a B \cos ^2(c+d x)+3 b B \cos (c+d x)+5 (A b-a B)}{\cos ^{\frac {5}{2}}(c+d x) (b+a \cos (c+d x))}dx}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a B \sin \left (c+d x+\frac {\pi }{2}\right )^2+3 b B \sin \left (c+d x+\frac {\pi }{2}\right )+5 (A b-a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {2 \int -\frac {-5 a (A b-a B) \cos ^2(c+d x)-b (5 A b+4 a B) \cos (c+d x)+3 \left (-5 B a^2+5 A b a-3 b^2 B\right )}{2 \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}+\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 a (A b-a B) \cos ^2(c+d x)-b (5 A b+4 a B) \cos (c+d x)+3 \left (-5 B a^2+5 A b a-3 b^2 B\right )}{\cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))}dx}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-5 a (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2-b (5 A b+4 a B) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (-5 B a^2+5 A b a-3 b^2 B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3534 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 \int -\frac {3 a \left (-5 B a^2+5 A b a-3 b^2 B\right ) \cos ^2(c+d x)+b \left (-20 B a^2+20 A b a-9 b^2 B\right ) \cos (c+d x)+5 \left (3 a^2+b^2\right ) (A b-a B)}{2 \sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 a \left (-5 B a^2+5 A b a-3 b^2 B\right ) \cos ^2(c+d x)+b \left (-20 B a^2+20 A b a-9 b^2 B\right ) \cos (c+d x)+5 \left (3 a^2+b^2\right ) (A b-a B)}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\int \frac {3 a \left (-5 B a^2+5 A b a-3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+b \left (-20 B a^2+20 A b a-9 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+5 \left (3 a^2+b^2\right ) (A b-a B)}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3538 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {3 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx-\frac {\int -\frac {5 \left (b (A b-a B) \cos (c+d x) a^2+\left (3 a^2+b^2\right ) (A b-a B) a\right )}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {3 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \int \sqrt {\cos (c+d x)}dx+\frac {5 \int \frac {b (A b-a B) \cos (c+d x) a^2+\left (3 a^2+b^2\right ) (A b-a B) a}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {3 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {5 \int \frac {b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (3 a^2+b^2\right ) (A b-a B) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \int \frac {b (A b-a B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2+\left (3 a^2+b^2\right ) (A b-a B) a}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3481 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 a^3 (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx+a b (A b-a B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )}{a}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 a^3 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+a b (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (3 a^3 (A b-a B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{a}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {10 (A b-a B) \sin (c+d x)}{3 b d \cos ^{\frac {3}{2}}(c+d x)}-\frac {\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) \sin (c+d x)}{b d \sqrt {\cos (c+d x)}}-\frac {\frac {5 \left (\frac {6 a^3 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 a b (A b-a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}\right )}{a}+\frac {6 \left (-5 a^2 B+5 a A b-3 b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{b}}{3 b}}{5 b}+\frac {2 B \sin (c+d x)}{5 b d \cos ^{\frac {5}{2}}(c+d x)}\) |
Input:
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])),x]
Output:
(2*B*Sin[c + d*x])/(5*b*d*Cos[c + d*x]^(5/2)) + ((10*(A*b - a*B)*Sin[c + d *x])/(3*b*d*Cos[c + d*x]^(3/2)) - (-(((6*(5*a*A*b - 5*a^2*B - 3*b^2*B)*Ell ipticE[(c + d*x)/2, 2])/d + (5*((2*a*b*(A*b - a*B)*EllipticF[(c + d*x)/2, 2])/d + (6*a^3*(A*b - a*B)*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2])/((a + b)*d)))/a)/b) + (6*(5*a*A*b - 5*a^2*B - 3*b^2*B)*Sin[c + d*x])/(b*d*Sqrt [Cos[c + d*x]]))/(3*b))/(5*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(A*b^2 - a*b*B))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin [e + f*x])^(1 + n)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(a*A - b*B)*(b*c - a*d)*(m + 1) + b*d*(A*b - a*B)*(m + n + 2) + (A*b - a*B)*(a*d*(m + 1) - b*c*(m + 2))*Sin[e + f*x] - b*d*(A*b - a*B) *(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n }, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Rat ionalQ[m] && m < -1 && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(I ntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0]) ))
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b* c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int [(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b*c - a* d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A *b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ [n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) | | EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(757\) vs. \(2(206)=412\).
Time = 5.79 (sec) , antiderivative size = 758, normalized size of antiderivative = 3.49
Input:
int((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c)),x,method=_RETURNVER BOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/5*B/b/(8*sin (1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/ 2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*EllipticE(co s(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c )^4+12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)* (2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c) *sin(1/2*d*x+1/2*c)^2-3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x +1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2))*(-2*sin(1/2*d*x+1/2*c)^ 4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(A*b-B*a)/b^2*(-1/6*cos(1/2*d*x+1/2*c)*(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/ 2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(- 2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1 /2*c),2^(1/2)))-2*(A*b-B*a)*a^3/b^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2) *(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/ 2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*(A*b-B*a) /b^3*a/sin(1/2*d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2 *c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c) ^2-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s in(1/2*d*x+1/2*c)^2-1)^(1/2)))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)...
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c)),x, algorithm= "fricas")
Output:
Timed out
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)**(7/2)/(a+b*sec(d*x+c)),x)
Output:
Timed out
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c)),x, algorithm= "maxima")
Output:
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c)),x, algorithm= "giac")
Output:
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*cos(d*x + c)^(7/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{7/2}\,\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )} \,d x \] Input:
int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))),x)
Output:
int((A + B/cos(c + d*x))/(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))), x)
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right )}}{\cos \left (d x +c \right )^{4}}d x \] Input:
int((A+B*sec(d*x+c))/cos(d*x+c)^(7/2)/(a+b*sec(d*x+c)),x)
Output:
int(sqrt(cos(c + d*x))/cos(c + d*x)**4,x)