Integrand size = 45, antiderivative size = 260 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {(8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(5 A-9 B+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(2 A-6 B+7 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(A-B+2 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}} \]
1/4*(8*A-12*B+19*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^( 3/2)/d-1/2*(A-B+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/ 4*(5*A-9*B+13*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/( a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/4*(2*A-6*B+7*C)*sec(d*x+c)^(3/2 )*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/2*(A-B+2*C)*sec(d*x+c)^(5/2)*sin (d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(1034\) vs. \(2(260)=520\).
Time = 7.89 (sec) , antiderivative size = 1034, normalized size of antiderivative = 3.98 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {B \sqrt {1+\sec (c+d x)} \left (-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{2 d (1+\sec (c+d x))^{3/2}}+\frac {1}{2} \left (\frac {3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}+\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}+\frac {3 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {9 \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {9 \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{\sqrt {2} d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{a \sqrt {a (1+\sec (c+d x))}}+\frac {A \sqrt {1+\sec (c+d x)} \left (-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{2 d (1+\sec (c+d x))^{3/2}}+\frac {1}{2} \left (-\frac {\sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {\arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {5 \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {5 \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{\sqrt {2} d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{a \sqrt {a (1+\sec (c+d x))}}+\frac {C \sqrt {1+\sec (c+d x)} \left (-\frac {\sec ^{\frac {11}{2}}(c+d x) \sin (c+d x)}{2 d (1+\sec (c+d x))^{3/2}}+\frac {1}{2} \left (-\frac {7 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{2 d \sqrt {1+\sec (c+d x)}}+\frac {2 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}+\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{d \sqrt {1+\sec (c+d x)}}-\frac {7 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{2 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {13 \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {13 \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{\sqrt {2} d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )\right )}{a \sqrt {a (1+\sec (c+d x))}} \]
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
(B*Sqrt[1 + Sec[c + d*x]]*(-1/2*(Sec[c + d*x]^(9/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + ((3*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[1 + Se c[c + d*x]]) - (Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]] ) + (Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) + (3*ArcS in[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) + (9*ArcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) - (9*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d *x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(Sqrt[2]*d*Sqrt[1 - Sec[c + d* x]]*Sqrt[1 + Sec[c + d*x]]))/2))/(a*Sqrt[a*(1 + Sec[c + d*x])]) + (A*Sqrt[ 1 + Sec[c + d*x]]*(-1/2*(Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + (-((Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d* x]])) + (Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d*x]]) - (Ar cSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[ 1 + Sec[c + d*x]]) - (5*ArcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) + (5*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(Sqrt[2]*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]))/2))/(a*Sqrt[a*(1 + Sec[c + d*x])]) + (C*Sqr t[1 + Sec[c + d*x]]*(-1/2*(Sec[c + d*x]^(11/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(3/2)) + ((-7*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(2*d*Sqrt[1 + Sec[ c + d*x]]) + (2*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*Sqrt[1 + Sec[c + d*...
Time = 1.65 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.05, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.356, Rules used = {3042, 4572, 27, 3042, 4509, 27, 3042, 4509, 27, 3042, 4511, 3042, 4288, 222, 4295, 219}
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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) (a (A-5 B+5 C)-4 a (A-B+2 C) \sec (c+d x))}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (A-5 B+5 C)-4 a (A-B+2 C) \sec (c+d x))}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (A-5 B+5 C)-4 a (A-B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle -\frac {\frac {\int -\frac {2 \sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a^2 (A-B+2 C)-a^2 (2 A-6 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4509 |
| \(\displaystyle -\frac {-\frac {\frac {\int -\frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x) a+a}}dx}{a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {\sqrt {\sec (c+d x)} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x) a+a}}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^3 (2 A-6 B+7 C)-a^3 (8 A-12 B+19 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4511 |
| \(\displaystyle -\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {\sec (c+d x) a+a}}dx-a^2 (8 A-12 B+19 C) \int \sqrt {\sec (c+d x)} \sqrt {\sec (c+d x) a+a}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-a^2 (8 A-12 B+19 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4288 |
| \(\displaystyle -\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {2 a^2 (8 A-12 B+19 C) \int \frac {1}{\sqrt {\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 222 |
| \(\displaystyle -\frac {-\frac {-\frac {2 a^3 (5 A-9 B+13 C) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 4295 |
| \(\displaystyle -\frac {-\frac {-\frac {-\frac {4 a^3 (5 A-9 B+13 C) \int \frac {1}{2 a-\frac {a^2 \sin (c+d x) \tan (c+d x)}{\sec (c+d x) a+a}}d\left (-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {-\frac {\frac {2 \sqrt {2} a^{5/2} (5 A-9 B+13 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {2 a^{5/2} (8 A-12 B+19 C) \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{2 a}-\frac {a^2 (2 A-6 B+7 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{a}-\frac {2 a (A-B+2 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d \sqrt {a \sec (c+d x)+a}}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}\) |
Int[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se c[c + d*x])^(3/2),x]
-1/2*((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) ^(3/2)) - ((-2*a*(A - B + 2*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]) - (-1/2*((-2*a^(5/2)*(8*A - 12*B + 19*C)*ArcSinh[(Sqrt[ a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (2*Sqrt[2]*a^(5/2)*(5*A - 9*B + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqr t[a + a*Sec[c + d*x]])])/d)/a - (a^2*(2*A - 6*B + 7*C)*Sec[c + d*x]^(3/2)* Sin[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]))/a)/(4*a^2)
3.7.13.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt [a])]/Rt[b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(a/(b*f))*Sqrt[a*(d/b)] Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; FreeQ[{a , b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*(d/(a*f)) Subst[Int[1/(2*b - d*x^2), x], x, b*(Cot[e + f*x]/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]]))], x] /; FreeQ[{a, b, d, e, f}, x] && 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[(-B)*d* Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[d/(b*(m + n)) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m + n) + a*B*m)*Csc[e + f*x], x], x], x] /; Fr eeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[n, 1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)/b Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Simp[B/b Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b , d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0]
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*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(1173\) vs. \(2(221)=442\).
Time = 1.92 (sec) , antiderivative size = 1174, normalized size of antiderivative = 4.52
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(1174\) |
| parts | \(\text {Expression too large to display}\) | \(1233\) |
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2 ),x,method=_RETURNVERBOSE)
1/8/a^2/d*sec(d*x+c)^(5/2)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)^2/(-1/( cos(d*x+c)+1))^(1/2)*(10*A*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(- 1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^4-18*B*arctan(1/2*sin(d*x+c)*2 ^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*2^(1/2)*cos(d*x+c)^4+26*C *arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))*2 ^(1/2)*cos(d*x+c)^4+8*A*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/ (cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-8*A*cos(d*x+c)^4*arctan(1/2*(cos (d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+10*A*2^(1/ 2)*cos(d*x+c)^3*arctan(1/2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+ c)+1))^(1/2))-4*A*sin(d*x+c)*cos(d*x+c)^3*(-1/(cos(d*x+c)+1))^(1/2)-12*B*c os(d*x+c)^4*arctan(1/2*(cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d *x+c)+1))^(1/2))+12*B*cos(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(c os(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-18*B*cos(d*x+c)^3*2^(1/2)*arctan(1 /2*sin(d*x+c)*2^(1/2)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))+12*B*cos(d *x+c)^3*sin(d*x+c)*(-1/(cos(d*x+c)+1))^(1/2)+19*C*cos(d*x+c)^4*arctan(1/2* (cos(d*x+c)-sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-19*C*c os(d*x+c)^4*arctan(1/2*(cos(d*x+c)+sin(d*x+c)+1)/(cos(d*x+c)+1)/(-1/(cos(d *x+c)+1))^(1/2))+26*C*2^(1/2)*cos(d*x+c)^3*arctan(1/2*sin(d*x+c)*2^(1/2)/( cos(d*x+c)+1)/(-1/(cos(d*x+c)+1))^(1/2))-14*C*sin(d*x+c)*cos(d*x+c)^3*(-1/ (cos(d*x+c)+1))^(1/2)+8*A*cos(d*x+c)^3*arctan(1/2*(cos(d*x+c)-sin(d*x+c...
Time = 0.65 (sec) , antiderivative size = 805, normalized size of antiderivative = 3.10 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="fricas")
[1/16*(2*sqrt(2)*((5*A - 9*B + 13*C)*cos(d*x + c)^3 + 2*(5*A - 9*B + 13*C) *cos(d*x + c)^2 + (5*A - 9*B + 13*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(c os(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*co s(d*x + c) + 1)) + ((8*A - 12*B + 19*C)*cos(d*x + c)^3 + 2*(8*A - 12*B + 1 9*C)*cos(d*x + c)^2 + (8*A - 12*B + 19*C)*cos(d*x + c))*sqrt(a)*log((a*cos (d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqr t(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c )) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) - 4*((2*A - 6*B + 7*C)*cos(d* x + c)^2 - (4*B - 3*C)*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d *x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d* cos(d*x + c)^2 + a^2*d*cos(d*x + c)), 1/8*(2*sqrt(2)*((5*A - 9*B + 13*C)*c os(d*x + c)^3 + 2*(5*A - 9*B + 13*C)*cos(d*x + c)^2 + (5*A - 9*B + 13*C)*c os(d*x + c))*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/co s(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + ((8*A - 12*B + 19*C)*co s(d*x + c)^3 + 2*(8*A - 12*B + 19*C)*cos(d*x + c)^2 + (8*A - 12*B + 19*C)* cos(d*x + c))*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*((2*A - 6*B + 7*C)*cos(d*x + c)^2 - (4*B - 3*C)*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x...
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 15483 vs. \(2 (221) = 442\).
Time = 1.45 (sec) , antiderivative size = 15483, normalized size of antiderivative = 59.55 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Too large to display} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="maxima")
1/16*(4*(4*(sin(2*d*x + 2*c) + 2*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d *x + 2*c))))*cos(3/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*(sqr t(2)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c) *sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2* c) + sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr t(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), c os(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) ))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2* sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) - 2*(sqr t(2)*cos(2*d*x + 2*c)^2 + 4*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos( 2*d*x + 2*c)))^2 + sqrt(2)*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*sin(2*d*x + 2*c) *sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 4*sqrt(2)*sin(1/2* arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c)))^2 + 4*(sqrt(2)*cos(2*d*x + 2* c) + sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2*sqr t(2)*cos(2*d*x + 2*c) + sqrt(2))*log(2*cos(1/4*arctan2(sin(2*d*x + 2*c), c os(2*d*x + 2*c)))^2 + 2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) ))^2 + 2*sqrt(2)*cos(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 2* sqrt(2)*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + 2) + 2*(...
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^(3/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*se c(d*x + c) + a)^(3/2), x)
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
int(((1/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(3/2),x)