Integrand size = 25, antiderivative size = 335 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a^2 e^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}+\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e} \]
1/2*a^2*e^(3/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)-1 /2*a^2*e^(3/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)+1/ 4*a^2*e^(3/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/ d*2^(1/2)-1/4*a^2*e^(3/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)* tan(d*x+c))/d*2^(1/2)+2/3*a^2*e^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*tan(d*x+c))^(1/2)+2*a^2*e*(e*tan(d*x+c))^(1/2)/d+4/3*a^2*e*sec(d*x+c )*(e*tan(d*x+c))^(1/2)/d+2/5*a^2*(e*tan(d*x+c))^(5/2)/d/e
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 26.79 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.77 \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \arctan (\tan (c+d x))\right ) (e \tan (c+d x))^{3/2} \left (30 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-30 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+15 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-15 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+120 \sqrt {\tan (c+d x)}-80 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}+80 \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}+24 \tan ^{\frac {5}{2}}(c+d x)\right )}{60 d \tan ^{\frac {3}{2}}(c+d x)} \]
(a^2*Cos[(c + d*x)/2]^4*Sec[ArcTan[Tan[c + d*x]]/2]^4*(e*Tan[c + d*x])^(3/ 2)*(30*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 30*Sqrt[2]*ArcTan[ 1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 15*Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - 15*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Ta n[c + d*x]] + 120*Sqrt[Tan[c + d*x]] - 80*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d*x]] + 80*Sqrt[Sec[c + d*x]^2]*Sqrt[Tan[c + d*x]] + 24*Tan[c + d*x]^(5/2)))/(60*d*Tan[c + d*x]^(3/2))
Time = 0.68 (sec) , antiderivative size = 335, 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.
\(\displaystyle \int (a \sec (c+d x)+a)^2 (e \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \int \left (a^2 (e \tan (c+d x))^{3/2}+a^2 \sec ^2(c+d x) (e \tan (c+d x))^{3/2}+2 a^2 \sec (c+d x) (e \tan (c+d x))^{3/2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {2 a^2 e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}+\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d}\) |
(a^2*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]* d) - (a^2*e^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqr t[2]*d) + (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e *Tan[c + d*x]]])/(2*Sqrt[2]*d) - (a^2*e^(3/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d) - (2*a^2*e^2*Ellipti cF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*Sqrt[e*Tan [c + d*x]]) + (2*a^2*e*Sqrt[e*Tan[c + d*x]])/d + (4*a^2*e*Sec[c + d*x]*Sqr t[e*Tan[c + d*x]])/(3*d) + (2*a^2*(e*Tan[c + d*x])^(5/2))/(5*d*e)
3.2.12.3.1 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 = 6.59 (sec) , antiderivative size = 397, normalized size of antiderivative = 1.19
method | result | size |
parts | \(\frac {2 a^{2} e \left (\sqrt {e \tan \left (d x +c \right )}-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{d}+\frac {2 a^{2} \left (e \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}-\frac {2 a^{2} \sqrt {2}\, \left (-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )^{2}-\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (d x +c \right )+\sqrt {2}\, \sin \left (d x +c \right )\right ) e \tan \left (d x +c \right ) \sqrt {e \tan \left (d x +c \right )}}{3 d \left (\cos \left (d x +c \right )^{2}-1\right )}\) | \(397\) |
default | \(\text {Expression too large to display}\) | \(931\) |
2*a^2/d*e*((e*tan(d*x+c))^(1/2)-1/8*(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)))+2/5*a^2*(e*tan(d*x+c))^(5/2)/d/e-2/3*a^2/d*2^(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)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c) ^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)*EllipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/ 2))*cos(d*x+c)+2^(1/2)*sin(d*x+c))*e*tan(d*x+c)*(e*tan(d*x+c))^(1/2)/(cos( d*x+c)^2-1)
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=a^{2} \left (\int \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx + \int 2 \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx + \int \left (e \tan {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral((e*tan(c + d*x))**(3/2), x) + Integral(2*(e*tan(c + d*x))** (3/2)*sec(c + d*x), x) + Integral((e*tan(c + d*x))**(3/2)*sec(c + d*x)**2, x))
Exception generated. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{2} \left (e \tan \left (d x + c\right )\right )^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (a+a \sec (c+d x))^2 (e \tan (c+d x))^{3/2} \, dx=\int {\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2 \,d x \]