Integrand size = 45, antiderivative size = 291 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \left (8 A b^3-5 a^3 B-10 a b^2 B+a^2 b (7 A+15 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right ) \sqrt {\sec (c+d x)}}{15 a^3 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 A \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (4 A b-5 a B) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d \sqrt {\sec (c+d x)}} \]
-2/15*(8*A*b^3-5*B*a^3-10*B*a*b^2+a^2*b*(7*A+15*C))*(cos(1/2*d*x+1/2*c)^2) ^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^( 1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/a^3/d/(a+b*sec(d*x+c ))^(1/2)+2/5*A*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d/sec(d*x+c)^(3/2)-2/15 *(4*A*b-5*B*a)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/d/sec(d*x+c)^(1/2)+2/ 15*(8*A*b^2-10*B*a*b+3*a^2*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2 *d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*(a+b*sec (d*x+c))^(1/2)/a^3/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 12.13 (sec) , antiderivative size = 3039, normalized size of antiderivative = 10.44 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*Sqrt [a + b*Sec[c + d*x]]),x]
((b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-4*(9*a^2* A + 8*A*b^2 - 10*a*b*B + 15*a^2*C)*Cot[c])/(15*a^3*d) + (4*(-4*A*b + 5*a*B )*Cos[d*x]*Sin[c])/(15*a^2*d) + (2*A*Cos[2*d*x]*Sin[2*c])/(5*a*d) + (4*(-4 *A*b + 5*a*B)*Cos[c]*Sin[d*x])/(15*a^2*d) + (2*A*Cos[2*c]*Sin[2*d*x])/(5*a *d)))/((A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(3/2 )*Sqrt[a + b*Sec[c + d*x]]) - (8*A*b*AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]* (b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + C ot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2])))]*Sqrt[b + a*Cos[c + d*x]]*Csc[c]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[(a*Sqrt[ 1 + Cot[c]^2] - a*Sqrt[1 + Cot[c]^2]*Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt[1 + Cot[c]^2] - b*Csc[c])]*Sqrt[(a*Sqrt[1 + Cot[c]^2] + a*Sqrt[1 + Cot[c]^2] *Sin[d*x - ArcTan[Cot[c]]])/(a*Sqrt[1 + Cot[c]^2] + b*Csc[c])]*Sqrt[b - a* Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]])/(15*a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*Sec[c + d*x]^(3 /2)*Sqrt[a + b*Sec[c + d*x]]) - (4*B*AppellF1[1/2, 1/2, 1/2, 3/2, (Csc[c]* (b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + C ot[c]^2]*(1 + (b*Csc[c])/(a*Sqrt[1 + Cot[c]^2]))), (Csc[c]*(b - a*Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]))/(a*Sqrt[1 + Cot[c]^2]*(-1...
Time = 2.15 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.04, number of steps used = 19, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.422, Rules used = {3042, 4592, 27, 3042, 4592, 27, 3042, 4523, 3042, 4343, 3042, 3134, 3042, 3132, 4345, 3042, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {-2 A b \sec ^2(c+d x)-a (3 A+5 C) \sec (c+d x)+4 A b-5 a B}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 A b \sec ^2(c+d x)-a (3 A+5 C) \sec (c+d x)+4 A b-5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b-5 a B}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4523 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4343 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4345 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\) |
(2*A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - ( -1/3*((-2*(8*A*b^3 - 5*a^3*B - 10*a*b^2*B + a^2*b*(7*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[Sec[c + d*x]])/(a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(8*A*b^2 - 10*a*b*B + 3*a^2*( 3*A + 5*C))*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]] )/(a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]))/a + (2*(4*A *b - 5*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x ]]))/(5*a)
3.11.56.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[Sqrt[a + b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*S qrt[b + a*Sin[e + f*x]]) Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; FreeQ[{a , b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[Sqrt[d*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/S qrt[a + b*Csc[e + f*x]]) Int[1/Sqrt[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_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d _.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Simp[A/a I nt[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Simp[(A*b - a*B) /(a*d) Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ [{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[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 + A*n + C*n)* 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, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(4323\) vs. \(2(321)=642\).
Time = 12.40 (sec) , antiderivative size = 4324, normalized size of antiderivative = 14.86
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(4324\) |
| default | \(\text {Expression too large to display}\) | \(4514\) |
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2 ),x,method=_RETURNVERBOSE)
2/15*A/d/a^3/((a-b)/(a+b))^(1/2)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/s ec(d*x+c)^(5/2)/(1+cos(d*x+c))*(2*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)) )^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c )+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b-8*(1/(a+b)*(b+a*cos(d*x+c))/(1+c os(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*( -cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2-9*(1/(a+b)*(b+a*cos(d* x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b ))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b+8*(1/(a+b)*( b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(( (a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2-9* (1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*E llipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2) )*a^3*sec(d*x+c)^2+9*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1 +cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)) ,(-(a+b)/(a-b))^(1/2))*a^3*sec(d*x+c)^2-8*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos (d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-c ot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^3*sec(d*x+c)^2-18*(1/(a+b)*( b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF(( (a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*sec( d*x+c)+18*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(1/(1+cos(d*x...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.14 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-15 i \, B a^{3} + 6 i \, {\left (2 \, A + 5 \, C\right )} a^{2} b - 20 i \, B a b^{2} + 16 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) + \sqrt {2} {\left (15 i \, B a^{3} - 6 i \, {\left (2 \, A + 5 \, C\right )} a^{2} b + 20 i \, B a b^{2} - 16 i \, A b^{3}\right )} \sqrt {a} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right ) - 3 \, \sqrt {2} {\left (-3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} + 10 i \, B a^{2} b - 8 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) + 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) - 3 \, \sqrt {2} {\left (3 i \, {\left (3 \, A + 5 \, C\right )} a^{3} - 10 i \, B a^{2} b + 8 i \, A a b^{2}\right )} \sqrt {a} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (3 \, a^{2} - 4 \, b^{2}\right )}}{3 \, a^{2}}, \frac {8 \, {\left (9 \, a^{2} b - 8 \, b^{3}\right )}}{27 \, a^{3}}, \frac {3 \, a \cos \left (d x + c\right ) - 3 i \, a \sin \left (d x + c\right ) + 2 \, b}{3 \, a}\right )\right ) + \frac {6 \, {\left (3 \, A a^{3} \cos \left (d x + c\right )^{2} + {\left (5 \, B a^{3} - 4 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{45 \, a^{4} d} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) )^(1/2),x, algorithm="fricas")
1/45*(sqrt(2)*(-15*I*B*a^3 + 6*I*(2*A + 5*C)*a^2*b - 20*I*B*a*b^2 + 16*I*A *b^3)*sqrt(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a) + sqrt( 2)*(15*I*B*a^3 - 6*I*(2*A + 5*C)*a^2*b + 20*I*B*a*b^2 - 16*I*A*b^3)*sqrt(a )*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3 , 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) - 3*sqrt(2)*(-3*I*( 3*A + 5*C)*a^3 + 10*I*B*a^2*b - 8*I*A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3* (3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3* (3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3 *I*a*sin(d*x + c) + 2*b)/a)) - 3*sqrt(2)*(3*I*(3*A + 5*C)*a^3 - 10*I*B*a^2 *b + 8*I*A*a*b^2)*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*( 9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*( 9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) + 2*b)/a) ) + 6*(3*A*a^3*cos(d*x + c)^2 + (5*B*a^3 - 4*A*a^2*b)*cos(d*x + c))*sqrt(( a*cos(d*x + c) + b)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^4*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}} \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)/(sqrt(a + b*sec(c + d*x) )*sec(c + d*x)**(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) )^(1/2),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(sqrt(b*sec(d*x + c) + a )*sec(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{\sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) )^(1/2),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/(sqrt(b*sec(d*x + c) + a )*sec(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))^(1/2)*(1 /cos(c + d*x))^(5/2)),x)