Integrand size = 25, antiderivative size = 339 \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\frac {e^{11/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}-\frac {e^{11/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a^2 d}+\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a^2 d}+\frac {2 e^6 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{3 a^2 d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 \sqrt {e \tan (c+d x)}}{a^2 d}-\frac {4 e^5 \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 a^2 d}+\frac {2 e^3 (e \tan (c+d x))^{5/2}}{5 a^2 d} \]
1/2*e^(11/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)- 1/2*e^(11/2)*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a^2/d*2^(1/2)+ 1/4*e^(11/2)*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^(11/2)*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)-2/3*e^6*(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) /a^2/d/(e*tan(d*x+c))^(1/2)+2*e^5*(e*tan(d*x+c))^(1/2)/a^2/d-4/3*e^5*sec(d *x+c)*(e*tan(d*x+c))^(1/2)/a^2/d+2/5*e^3*(e*tan(d*x+c))^(5/2)/a^2/d
\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx \]
Time = 0.77 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {(e \tan (c+d x))^{11/2}}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{11/2}}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 4376 |
\(\displaystyle \frac {e^4 \int (a-a \sec (c+d x))^2 (e \tan (c+d x))^{3/2}dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {e^4 \int \left (-e \cot \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \frac {e^4 \int \left (\sec ^2(c+d x) (e \tan (c+d x))^{3/2} a^2-2 \sec (c+d x) (e \tan (c+d x))^{3/2} a^2+(e \tan (c+d x))^{3/2} a^2\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^4 \left (\frac {a^2 e^{3/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}-\frac {a^2 e^{3/2} \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a^2 e^{3/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {2 a^2 e^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{3 d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 (e \tan (c+d x))^{5/2}}{5 d e}+\frac {2 a^2 e \sqrt {e \tan (c+d x)}}{d}-\frac {4 a^2 e \sec (c+d x) \sqrt {e \tan (c+d x)}}{3 d}\right )}{a^4}\) |
(e^4*((a^2*e^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sq rt[2]*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)*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*E llipticF[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]*Sqrt[e*Tan[c + d*x]])/(3*d) + (2*a^2*(e*Tan[c + d*x])^(5/2))/(5*d*e)))/ a^4
3.2.28.3.1 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 = 4.51 (sec) , antiderivative size = 1007, normalized size of antiderivative = 2.97
-1/30/a^2/d*2^(1/2)*(15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc( d*x+c))^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^( 1/2))*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*cos(d*x+c)^3-15*I*(cot(d*x+c)-csc(d* x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2 )*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d* x+c)^3+15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)* EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(csc(d*x +c)-cot(d*x+c)+1)^(1/2)*cos(d*x+c)^2-15*I*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)* (cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi(( csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^2-15*(cot (d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*EllipticPi((csc( d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^(1/2))*(csc(d*x+c)-cot(d*x+c)+1 )^(1/2)*cos(d*x+c)^3-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d* x+c))^(1/2)*(csc(d*x+c)-cot(d*x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x +c)+1)^(1/2),1/2-1/2*I,1/2*2^(1/2))*cos(d*x+c)^3+50*(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)*El lipticF((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2*2^(1/2))*cos(d*x+c)^3-15*(cot( d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(d*x+c))^(1/2)*(csc(d*x+c)-cot(d *x+c)+1)^(1/2)*EllipticPi((csc(d*x+c)-cot(d*x+c)+1)^(1/2),1/2+1/2*I,1/2*2^ (1/2))*cos(d*x+c)^2-15*(cot(d*x+c)-csc(d*x+c)+1)^(1/2)*(cot(d*x+c)-csc(...
Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {\left (e \tan \left (d x + c\right )\right )^{\frac {11}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {(e \tan (c+d x))^{11/2}}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{11/2}}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]