Integrand size = 25, antiderivative size = 316 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}+\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac {4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}-\frac {2 a^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 d e^2 \sqrt {e \tan (c+d x)}} \]
1/2*a^2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)-1 /2*a^2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d/e^(5/2)*2^(1/2)+1/ 4*a^2*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d/e^(5/2 )*2^(1/2)-1/4*a^2*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+ c))/d/e^(5/2)*2^(1/2)+2/3*a^2*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x )*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/d/e ^2/(e*tan(d*x+c))^(1/2)-4/3*a^2/d/e/(e*tan(d*x+c))^(3/2)-4/3*a^2*sec(d*x+c )/d/e/(e*tan(d*x+c))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 15.82 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.71 \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=-\frac {a^2 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) \cot \left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \arctan (\tan (c+d x))\right ) \left (16 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {1}{4},-\tan ^2(c+d x)\right )+16 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\tan ^2(c+d x)\right )+3 \sqrt {2} \left (2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right ) \tan ^{\frac {3}{2}}(c+d x)\right )}{24 d e^2 \sqrt {e \tan (c+d x)}} \]
-1/24*(a^2*Cos[(c + d*x)/2]^2*Cos[c + d*x]*Cot[(c + d*x)/2]*Sec[ArcTan[Tan [c + d*x]]/2]^4*(16*Hypergeometric2F1[-3/4, 1/2, 1/4, -Tan[c + d*x]^2] + 1 6*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + 3*Sqrt[2]*(2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqr t[Tan[c + d*x]] + Tan[c + d*x]])*Tan[c + d*x]^(3/2)))/(d*e^2*Sqrt[e*Tan[c + d*x]])
Time = 0.69 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{(e \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}{\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \int \left (\frac {a^2}{(e \tan (c+d x))^{5/2}}+\frac {a^2 \sec ^2(c+d x)}{(e \tan (c+d x))^{5/2}}+\frac {2 a^2 \sec (c+d x)}{(e \tan (c+d x))^{5/2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d e^{5/2}}-\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d e^{5/2}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d e^{5/2}}-\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d e^2 \sqrt {e \tan (c+d x)}}-\frac {4 a^2}{3 d e (e \tan (c+d x))^{3/2}}-\frac {4 a^2 \sec (c+d x)}{3 d e (e \tan (c+d x))^{3/2}}\) |
(a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e^(5/2 )) - (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*e ^(5/2)) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*e^(5/2)) - (4*a^2)/(3*d*e*(e* Tan[c + d*x])^(3/2)) - (4*a^2*Sec[c + d*x])/(3*d*e*(e*Tan[c + d*x])^(3/2)) - (2*a^2*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]] )/(3*d*e^2*Sqrt[e*Tan[c + d*x]])
3.2.16.3.1 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]
Time = 5.55 (sec) , antiderivative size = 483, normalized size of antiderivative = 1.53
| method | result | size |
| parts | \(\frac {2 a^{2} e \left (-\frac {1}{3 e^{2} \left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e^{4}}\right )}{d}-\frac {2 a^{2}}{3 d e \left (e \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2} \sqrt {2}\, \left (1-\cos \left (d x +c \right )\right )^{2} \left (2 \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {2-2 \csc \left (d x +c \right )+2 \cot \left (d x +c \right )}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-\left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )^{4}+1\right ) \csc \left (d x +c \right )^{2}}{3 d \sqrt {\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\left (1-\cos \left (d x +c \right )\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \csc \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} \left (-\frac {e \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}}}\) | \(483\) |
| default | \(\frac {a^{2} \sqrt {2}\, \left (\cos \left (d x +c \right )+1\right ) \left (3 i \sin \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 )}\, \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 )-3 i \sin \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 )}\, \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 )-3 \sin \left (d x +c \right ) \sqrt {\cot \left (d x +c \right )-\csc \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}\, \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 )-3 \sin \left (d x +c \right ) \sqrt {\cot \left (d x +c \right )-\csc \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}\, \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 )+2 \sin \left (d x +c \right ) \sqrt {\cot \left (d x +c \right )-\csc \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}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )-4 \sqrt {2}\, \cos \left (d x +c \right )\right ) \csc \left (d x +c \right ) \sec \left (d x +c \right )}{6 e^{2} d \sqrt {e \tan \left (d x +c \right )}}\) | \(496\) |
2*a^2/d*e*(-1/3/e^2/(e*tan(d*x+c))^(3/2)-1/8/e^4*(e^2)^(1/4)*2^(1/2)*(ln(( e*tan(d*x+c)+(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*tan( d*x+c)-(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^( 1/2)/(e^2)^(1/4)*(e*tan(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e* tan(d*x+c))^(1/2)+1)))-2/3*a^2/d/e/(e*tan(d*x+c))^(3/2)+1/3*a^2/d*2^(1/2)* (1-cos(d*x+c))^2*(2*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*(2-2*csc(d*x+c)+2*cot( d*x+c))^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticF((csc(d*x+c)-cot(d*x+ c)+1)^(1/2),1/2*2^(1/2))*(-cot(d*x+c)+csc(d*x+c))-(1-cos(d*x+c))^4*csc(d*x +c)^4+1)/((1-cos(d*x+c))^3*csc(d*x+c)^3+cot(d*x+c)-csc(d*x+c))^(1/2)/((1-c os(d*x+c))*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c))^(1/2)/((1-cos(d*x +c))^2*csc(d*x+c)^2-1)^2/(-e/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-cot(d*x+c )+csc(d*x+c)))^(5/2)*csc(d*x+c)^2
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=a^{2} \left (\int \frac {1}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\left (e \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx\right ) \]
a**2*(Integral((e*tan(c + d*x))**(-5/2), x) + Integral(2*sec(c + d*x)/(e*t an(c + d*x))**(5/2), x) + Integral(sec(c + d*x)**2/(e*tan(c + d*x))**(5/2) , x))
Exception generated. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\left (e \tan \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{(e \tan (c+d x))^{5/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]