Integrand size = 45, antiderivative size = 291 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \left (8 A b^3-5 a^3 B-10 a b^2 B+a^2 b (7 A+15 C)\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d} \]
-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)/a^3/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c ))^(1/2)+2/5*A*cos(d*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d-2/15 *(4*A*b-5*B*a)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^2/d+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))*cos(d*x+ c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^3/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
Result contains complex when optimal does not.
Time = 36.08 (sec) , antiderivative size = 2764, normalized size of antiderivative = 9.50 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Result too large to show} \]
Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqr t[a + b*Sec[c + d*x]],x]
((b + a*Cos[c + d*x])*((2*(-4*A*b + 5*a*B)*Sin[c + d*x])/(15*a^2) + (A*Sin [2*(c + d*x)])/(5*a)))/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - ( 2*Cos[c + d*x]^(3/2)*((3*A*Sqrt[Cos[c + d*x]])/(5*Sqrt[b + a*Cos[c + d*x]] *Sqrt[Sec[c + d*x]]) + (8*A*b^2*Sqrt[Cos[c + d*x]])/(15*a^2*Sqrt[b + a*Cos [c + d*x]]*Sqrt[Sec[c + d*x]]) - (2*b*B*Sqrt[Cos[c + d*x]])/(3*a*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (C*Sqrt[Cos[c + d*x]])/(Sqrt[b + a*C os[c + d*x]]*Sqrt[Sec[c + d*x]]) + (2*A*b*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(15*a*Sqrt[b + a*Cos[c + d*x]]) + (B*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(3*Sqrt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*(Cos[(c + d*x)/2 ]^2*Sec[c + d*x])^(3/2)*((-I)*(a + b)*(8*A*b^2 - 10*a*b*B + 3*a^2*(3*A + 5 *C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x )/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*a*(8*A* b^2 + 2*a*b*(A - 5*B) + a^2*(9*A + 5*(B + 3*C)))*EllipticF[I*ArcSinh[Tan[( c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d* x])*Sec[(c + d*x)/2]^2)/(a + b)] - (8*A*b^2 - 10*a*b*B + 3*a^2*(3*A + 5*C) )*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(15*a ^3*d*Sqrt[a + b*Sec[c + d*x]]*(-1/15*(Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2] ^2*Sec[c + d*x])^(3/2)*Sin[c + d*x]*((-I)*(a + b)*(8*A*b^2 - 10*a*b*B + 3* a^2*(3*A + 5*C))*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]* Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + ...
Time = 2.37 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.11, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {3042, 4753, 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.
\(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^{5/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )}{\sqrt {a+b \sec (c+d x)}}dx\) |
\(\Big \downarrow \) 4753 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sec ^2(c+d x)+B \sec (c+d x)+A}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \csc \left (c+d x+\frac {\pi }{2}\right )^2+B \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {-2 A b \sec ^2(c+d x)-a (3 A+5 C) \sec (c+d x)+4 A b-5 a B}{2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 A b \sec ^2(c+d x)-a (3 A+5 C) \sec (c+d x)+4 A b-5 a B}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}dx}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-2 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (3 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b-5 a B}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 a}\right )\) |
\(\Big \downarrow \) 4592 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{2 \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \sec (c+d x) a+8 A b^2}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}}dx}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {3 (3 A+5 C) a^2-10 b B a+(2 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right ) a+8 A b^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 4523 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \int \frac {\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 4343 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \cos (c+d x)}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a \sqrt {\sec (c+d x)} \sqrt {a \cos (c+d x)+b}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3134 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {\left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} \int \sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{a \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3132 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 4345 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \cos (c+d x)}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {a \cos (c+d x)+b} \int \frac {1}{\sqrt {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3142 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {\sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{a \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\) |
\(\Big \downarrow \) 3140 |
\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 A \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {2 \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{a d \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \sqrt {\sec (c+d x)} \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{a d \sqrt {a+b \sec (c+d x)}}}{3 a}}{5 a}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((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))
3.14.57.3.1 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]
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m Int[ActivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Leaf count of result is larger than twice the leaf count of optimal. \(3029\) vs. \(2(321)=642\).
Time = 11.76 (sec) , antiderivative size = 3030, normalized size of antiderivative = 10.41
int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2 ),x,method=_RETURNVERBOSE)
2/15/d*(3*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos(d*x+ c)))^(1/2)+15*C*(1/(a+b)*(b+a*cos(d*x+c))/(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*co s(d*x+c)+5*B*(1/(1+cos(d*x+c)))^(1/2)*((a-b)/(a+b))^(1/2)*a^3*cos(d*x+c)^2 *sin(d*x+c)+3*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos( d*x+c)))^(1/2)+9*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*(1/(1+cos (d*x+c)))^(1/2)+15*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c))) ^(1/2)*sin(d*x+c)+9*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c )))^(1/2)-4*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^2*(1/(1+cos(d*x+c)))^(1/2 )+5*B*sin(d*x+c)*cos(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*a ^3-8*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*Elliptic E(((a-b)/(a+b))^(1/2)*(cot(d*x+c)-csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^2- 2*A*cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(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*A *cos(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(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-A*cos( d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)+9*A*c os(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(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+5*B*((a- b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b*(1/(1+cos(d*x+c)))^(1/2)-10*B*((a-b)/(...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.13 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.84 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {6 \, {\left (3 \, A a^{3} \cos \left (d x + c\right ) + 5 \, B a^{3} - 4 \, A a^{2} b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\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 ) + \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 )}{45 \, a^{4} d} \]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^(1/2),x, algorithm="fricas")
1/45*(6*(3*A*a^3*cos(d*x + c) + 5*B*a^3 - 4*A*a^2*b)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + 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)*weierstrassPI nverse(-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)*w eierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weier strassPInverse(-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)))/(a^4*d)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*cos(d*x + c)^(5/2)/sqrt( b*sec(d*x + c) + a), x)
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}} \,d x } \]
integrate(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)*cos(d*x + c)^(5/2)/sqrt( b*sec(d*x + c) + a), x)
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]
int((cos(c + d*x)^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/co s(c + d*x))^(1/2),x)