Integrand size = 25, antiderivative size = 310 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {e^{5/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {4 e^3}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^3 \cos (c+d x)}{a^2 d \sqrt {e \tan (c+d x)}}+\frac {4 e^2 \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{a^2 d \sqrt {\sin (2 c+2 d x)}} \]
1/2*e^(5/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1 /2*e^(5/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)-1/ 4*e^(5/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a^2/ d*2^(1/2)+1/4*e^(5/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan( d*x+c))/a^2/d*2^(1/2)-4*e^3/a^2/d/(e*tan(d*x+c))^(1/2)+4*e^3*cos(d*x+c)/a^ 2/d/(e*tan(d*x+c))^(1/2)-4*e^2*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin( c+1/4*Pi+d*x)*EllipticE(cos(c+1/4*Pi+d*x),2^(1/2))*(e*tan(d*x+c))^(1/2)/a^ 2/d/sin(2*d*x+2*c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.36 (sec) , antiderivative size = 812, normalized size of antiderivative = 2.62 \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc ^2(c+d x) \left (\frac {32 \cos \left (\frac {c}{2}\right ) \cos (d x) \sec (2 c) \sin \left (\frac {c}{2}\right )}{d}+\frac {16 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {16 \cos (c) \sec (2 c) \sin (d x)}{d}+\frac {16 \tan \left (\frac {c}{2}\right )}{d}\right ) (e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2}+\frac {e^{-2 i c} \left (-e^{4 i c} \sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )+2 \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {e^{-2 i c} \left (\sqrt {-1+e^{4 i (c+d x)}} \arctan \left (\sqrt {-1+e^{4 i (c+d x)}}\right )-2 e^{4 i c} \sqrt {-1+e^{2 i (c+d x)}} \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right ) \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)}-\frac {8 e^{i (c-d x)} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (3-3 e^{4 i (c+d x)}+e^{4 i d x} \left (1+e^{4 i c}\right ) \sqrt {1-e^{4 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},e^{4 i (c+d x)}\right )\right ) \sec (2 c) \sec ^2(c+d x) (e \tan (c+d x))^{5/2}}{3 d \sqrt {-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (1+e^{2 i (c+d x)}\right ) (a+a \sec (c+d x))^2 \tan ^{\frac {5}{2}}(c+d x)} \]
(Cos[c/2 + (d*x)/2]^4*Csc[c + d*x]^2*((32*Cos[c/2]*Cos[d*x]*Sec[2*c]*Sin[c /2])/d + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2])/d - (16*Cos[c]*Sec[ 2*c]*Sin[d*x])/d + (16*Tan[c/2])/d)*(e*Tan[c + d*x])^(5/2))/(a + a*Sec[c + d*x])^2 + ((-(E^((4*I)*c)*Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d*x))]]) + 2*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2* I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5 /2))/(d*E^((2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*( c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^ (5/2)) - ((Sqrt[-1 + E^((4*I)*(c + d*x))]*ArcTan[Sqrt[-1 + E^((4*I)*(c + d *x))]] - 2*E^((4*I)*c)*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)) )]])*Cos[c/2 + (d*x)/2]^4*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5/2))/ (d*E^((2*I)*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d *x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^(5/2) ) - (8*E^(I*(c - d*x))*Cos[c/2 + (d*x)/2]^4*(3 - 3*E^((4*I)*(c + d*x)) + E ^((4*I)*d*x)*(1 + E^((4*I)*c))*Sqrt[1 - E^((4*I)*(c + d*x))]*Hypergeometri c2F1[1/2, 3/4, 7/4, E^((4*I)*(c + d*x))])*Sec[2*c]*Sec[c + d*x]^2*(e*Tan[c + d*x])^(5/2))/(3*d*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)* (c + d*x)))]*(1 + E^((2*I)*(c + d*x)))*(a + a*Sec[c + d*x])^2*Tan[c + d...
Time = 0.76 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4376, 3042, 4374, 2009}
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.
| \(\displaystyle \int \frac {(e \tan (c+d x))^{5/2}}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4376 |
| \(\displaystyle \frac {e^4 \int \frac {(a-a \sec (c+d x))^2}{(e \tan (c+d x))^{3/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {e^4 \int \frac {\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}{\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{a^4}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \frac {e^4 \int \left (\frac {\sec ^2(c+d x) a^2}{(e \tan (c+d x))^{3/2}}-\frac {2 \sec (c+d x) a^2}{(e \tan (c+d x))^{3/2}}+\frac {a^2}{(e \tan (c+d x))^{3/2}}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \left (\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{3/2}}-\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{3/2}}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{3/2}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{3/2}}+\frac {4 a^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d e^2 \sqrt {\sin (2 c+2 d x)}}-\frac {4 a^2}{d e \sqrt {e \tan (c+d x)}}+\frac {4 a^2 \cos (c+d x)}{d e \sqrt {e \tan (c+d x)}}\right )}{a^4}\) |
(e^4*((a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d* e^(3/2)) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[ 2]*d*e^(3/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*T an[c + d*x]]])/(2*Sqrt[2]*d*e^(3/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(3/2)) - (4*a^2)/(d*e *Sqrt[e*Tan[c + d*x]]) + (4*a^2*Cos[c + d*x])/(d*e*Sqrt[e*Tan[c + d*x]]) + (4*a^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2]*Sqrt[e*Tan[c + d*x]])/(d *e^2*Sqrt[Sin[2*c + 2*d*x]])))/a^4
3.2.31.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Simp[a^(2*n)/e^(2*n) Int[(e*Cot[c + d*x])^(m + 2* n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[a ^2 - b^2, 0] && ILtQ[n, 0]
Result contains complex when optimal does not.
Time = 4.49 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.84
| method | result | size |
| default | \(-\frac {\sqrt {2}\, e^{2} \left (i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\operatorname {EllipticPi}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-8 \operatorname {EllipticE}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )+4 \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )\right ) \sqrt {e \tan \left (d x +c \right )}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sin \left (d x +c \right )}{2 a^{2} d \left (\cos \left (d x +c \right )-1\right )}\) | \(261\) |
-1/2/a^2/d*2^(1/2)*e^2*(I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1 /2*I,1/2*2^(1/2))-I*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1 /2*2^(1/2))+EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/ 2))+EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))-8*El lipticE((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))+4*EllipticF((csc(d*x+ c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2)))*(e*tan(d*x+c))^(1/2)*(csc(d*x+c)-cot( d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1 /2)*sin(d*x+c)/(cos(d*x+c)-1)
Timed out. \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{5/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]