Integrand size = 33, antiderivative size = 402 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\frac {\left (9 a^2 A b-3 A b^3-5 a^3 B-a b^2 B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{4 a^2 \left (a^2-b^2\right )^2 d}+\frac {\left (8 a^4 A-5 a^2 A b^2+3 A b^4-7 a^3 b B+a b^3 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 \left (a^2-b^2\right )^2 d}-\frac {\left (15 a^4 A b-6 a^2 A b^3+3 A b^5-3 a^5 B-10 a^3 b^2 B+a b^4 B\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{4 a^3 (a-b)^2 (a+b)^3 d}-\frac {(A b-a B) \sqrt {\sec (c+d x)} \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}-\frac {\left (7 a^2 A b-A b^3-3 a^3 B-3 a b^2 B\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 a \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:
1/4*(9*A*a^2*b-3*A*b^3-5*B*a^3-B*a*b^2)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2 *d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/a^2/(a^2-b^2)^2/d+1/4*(8*A*a^4-5*A*a ^2*b^2+3*A*b^4-7*B*a^3*b+B*a*b^3)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x +1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^3/(a^2-b^2)^2/d-1/4*(15*A*a^4*b-6*A*a^2 *b^3+3*A*b^5-3*B*a^5-10*B*a^3*b^2+B*a*b^4)*cos(d*x+c)^(1/2)*EllipticPi(sin (1/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/a^3/(a-b)^2/(a+b)^3/d- 1/2*(A*b-B*a)*sec(d*x+c)^(1/2)*sin(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^2-1 /4*(7*A*a^2*b-A*b^3-3*B*a^3-3*B*a*b^2)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/(a^2- b^2)^2/d/(a+b*sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(885\) vs. \(2(402)=804\).
Time = 7.21 (sec) , antiderivative size = 885, normalized size of antiderivative = 2.20 \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
Integrate[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3 ,x]
Output:
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((2*(-5*a^2*A* b - A*b^3 + a^3*B + 5*a*b^2*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d *x])*(1 - Cos[c + d*x]^2)) + (2*(16*a^3*A + 8*a*A*b^2 - 24*a^2*b*B)*Cos[c + d*x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*Cos[2*(c + d*x)]*(a + b*Sec[c + d*x])*(-4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*Elliptic E[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x] ^2] - 2*a*(a - 2*b)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec [c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*Ellip ticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d* x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2))))/(16*a*(a - b)^2*(a + b)^2 *d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^3) + ((b + a*Cos[c + d*x])^3* Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(((-9*a^2*A*b + 3*A*b^3 + 5*a^3*B + a*b^2*B)*Sin[c + d*x])/(4*a^2*(-a^2 + b^2)^2) - (A*b^3*Sin[c + d*x] - a* b^2*B*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (11*a^...
Time = 2.53 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4515, 27, 3042, 4588, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 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.
\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 4515 |
\(\displaystyle -\frac {\int -\frac {-3 (A b-a B) \sec ^2(c+d x)+4 (a A-b B) \sec (c+d x)+A b-a B}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}dx}{2 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {-3 (A b-a B) \sec ^2(c+d x)+4 (a A-b B) \sec (c+d x)+A b-a B}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^2}dx}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {-3 (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 (a A-b B) \csc \left (c+d x+\frac {\pi }{2}\right )+A b-a B}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 4588 |
\(\displaystyle \frac {-\frac {\int -\frac {-5 B a^3+9 A b a^2-b^2 B a+4 \left (2 A a^2-3 b B a+A b^2\right ) \sec (c+d x) a-3 A b^3-\left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \sec ^2(c+d x)}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {-5 B a^3+9 A b a^2-b^2 B a+4 \left (2 A a^2-3 b B a+A b^2\right ) \sec (c+d x) a-3 A b^3-\left (-3 B a^3+7 A b a^2-3 b^2 B a-A b^3\right ) \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {-5 B a^3+9 A b a^2-b^2 B a+4 \left (2 A a^2-3 b B a+A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right ) a-3 A b^3+\left (3 B a^3-7 A b a^2+3 b^2 B a+A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 4594 |
\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (-5 B a^3+9 A b a^2-b^2 B a-3 A b^3\right )+\left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (-5 B a^3+9 A b a^2-b^2 B a-3 A b^3\right )+\left (8 A a^4-7 b B a^3-5 A b^2 a^2+b^3 B a+3 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {\frac {\frac {a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {\frac {a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {\left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {\frac {2 a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 4336 |
\(\displaystyle \frac {\frac {\frac {\frac {2 a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\frac {2 a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {\left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
\(\Big \downarrow \) 3284 |
\(\displaystyle \frac {\frac {\frac {\frac {2 a \left (-5 a^3 B+9 a^2 A b-a b^2 B-3 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {2 \left (8 a^4 A-7 a^3 b B-5 a^2 A b^2+a b^3 B+3 A b^4\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {2 \left (-3 a^5 B+15 a^4 A b-10 a^3 b^2 B-6 a^2 A b^3+a b^4 B+3 A b^5\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}}{2 a \left (a^2-b^2\right )}-\frac {\left (-3 a^3 B+7 a^2 A b-3 a b^2 B-A b^3\right ) \sin (c+d x) \sqrt {\sec (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{4 \left (a^2-b^2\right )}-\frac {(A b-a B) \sin (c+d x) \sqrt {\sec (c+d x)}}{2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\) |
Input:
Int[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^3,x]
Output:
-1/2*((A*b - a*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*d*(a + b*S ec[c + d*x])^2) + ((((2*a*(9*a^2*A*b - 3*A*b^3 - 5*a^3*B - a*b^2*B)*Sqrt[C os[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*(8*a^4*A - 5*a^2*A*b^2 + 3*A*b^4 - 7*a^3*b*B + a*b^3*B)*Sqrt[Cos[c + d*x]]*Ellipti cF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 - (2*(15*a^4*A*b - 6*a^2*A*b ^3 + 3*A*b^5 - 3*a^5*B - 10*a^3*b^2*B + a*b^4*B)*Sqrt[Cos[c + d*x]]*Ellipt icPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a + b)*d))/( 2*a*(a^2 - b^2)) - ((7*a^2*A*b - A*b^3 - 3*a^3*B - 3*a*b^2*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(4*(a^2 - b^2 ))
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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]] Int[ 1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[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[(-d)*(A *b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b *Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[d*(n - 1)*(A*b - a*B) + d*(a*A - b*B)*(m + 1)*Csc[e + f*x] - d*(A*b - a*B)*(m + n + 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] && LtQ[0, n, 1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc [e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Sim p[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f *x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x ] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] && ILtQ[n, 0])
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2) Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2 Int[(a*A - (A*b - a *B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(1959\) vs. \(2(381)=762\).
Time = 6.96 (sec) , antiderivative size = 1960, normalized size of antiderivative = 4.88
Input:
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x,method=_RETURNV ERBOSE)
Output:
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^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^2*(3*A*b-B*a)/(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)*Elli pticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^3*b*(3*A*b-2*B*a)*(a^2/b/ (a^2-b^2)*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)/(2*a*cos(1/2*d*x+1/2*c)^2-a+b)-1/2/(a+b)/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(a^2-b^2) *(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))-1/2*a/b/(a^2-b^2)*(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)*Ellipti cE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/ 2))+3/2*b/(a^2-b^2)/(a^2-a*b)*a*(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*b^2*(A*b-B*a)/a^3*(...
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorith m="fricas")
Output:
Timed out
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)**(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**3,x)
Output:
Timed out
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorith m="maxima")
Output:
Timed out
\[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(b*sec(d*x + c) + a)^3, x)
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \] Input:
int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(1/2))/(a + b/cos(c + d*x))^3,x )
Output:
int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(1/2))/(a + b/cos(c + d*x))^3, x)
\[ \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2} b^{2}+2 \sec \left (d x +c \right ) a b +a^{2}}d x \] Input:
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^3,x)
Output:
int(sqrt(sec(c + d*x))/(sec(c + d*x)**2*b**2 + 2*sec(c + d*x)*a*b + a**2), x)