Integrand size = 25, antiderivative size = 278 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=-\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a^2 \arctan \left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}-\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {2 a^2 \operatorname {EllipticF}\left (c-\frac {\pi }{4}+d x,2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 \sqrt {e \tan (c+d x)}}{d e} \]
-1/2*a^2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)/e^(1/2)+ 1/2*a^2*arctan(1+2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)/e^(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*2^(1/ 2)/e^(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*2^(1/2)/e^(1/2)-2*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 *tan(d*x+c))^(1/2)+2*a^2*(e*tan(d*x+c))^(1/2)/d/e
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 13.01 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=\frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \arctan (\tan (c+d x))\right ) \left (-2 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+8 \sqrt {\tan (c+d x)}+16 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}}{4 d \sqrt {e \tan (c+d x)}} \]
(a^2*Cos[(c + d*x)/2]^4*Sec[ArcTan[Tan[c + d*x]]/2]^4*(-2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] - Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + Sqr t[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 8*Sqrt[Tan[c + d *x]] + 16*Hypergeometric2F1[1/4, 1/2, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d *x]])*Sqrt[Tan[c + d*x]])/(4*d*Sqrt[e*Tan[c + d*x]])
Time = 0.59 (sec) , antiderivative size = 278, 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.
\(\displaystyle \int \frac {(a \sec (c+d x)+a)^2}{\sqrt {e \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}{\sqrt {-e \cot \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4374 |
\(\displaystyle \int \left (\frac {a^2}{\sqrt {e \tan (c+d x)}}+\frac {a^2 \sec ^2(c+d x)}{\sqrt {e \tan (c+d x)}}+\frac {2 a^2 \sec (c+d x)}{\sqrt {e \tan (c+d x)}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 \arctan \left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a^2 \arctan \left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d \sqrt {e}}+\frac {2 a^2 \sqrt {e \tan (c+d x)}}{d e}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) \operatorname {EllipticF}\left (c+d x-\frac {\pi }{4},2\right )}{d \sqrt {e \tan (c+d x)}}\) |
-((a^2*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*d*Sqrt [e])) + (a^2*ArcTan[1 + (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]* d*Sqrt[e]) - (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Tan[ c + d*x]]])/(2*Sqrt[2]*d*Sqrt[e]) + (a^2*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x ] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/(2*Sqrt[2]*d*Sqrt[e]) + (2*a^2*Elliptic F[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(d*Sqrt[e*Tan[c + d*x]]) + (2*a^2*Sqrt[e*Tan[c + d*x]])/(d*e)
3.2.14.3.1 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 = 5.46 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.97
method | result | size |
parts | \(\frac {a^{2} \left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \tan \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \tan \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \tan \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d e}+\frac {2 a^{2} \sqrt {e \tan \left (d x +c \right )}}{d e}+\frac {2 a^{2} \sqrt {2}\, \operatorname {EllipticF}\left (\sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )}\, \sqrt {\cot \left (d x +c \right )-\csc \left (d x +c \right )+1}\, \sqrt {\csc \left (d x +c \right )-\cot \left (d x +c \right )+1}\, \left (1+\sec \left (d x +c \right )\right )}{d \sqrt {e \tan \left (d x +c \right )}}\) | \(269\) |
default | \(\text {Expression too large to display}\) | \(931\) |
1/4*a^2/d/e*(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*a^2*(e*tan(d* x+c))^(1/2)/d/e+2*a^2/d*2^(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))^(1/2)*(cot(d*x+c)-csc(d*x+c)+1)^(1/2) *(csc(d*x+c)-cot(d*x+c)+1)^(1/2)/(e*tan(d*x+c))^(1/2)*(1+sec(d*x+c))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=a^{2} \left (\int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx\right ) \]
a**2*(Integral(1/sqrt(e*tan(c + d*x)), x) + Integral(2*sec(c + d*x)/sqrt(e *tan(c + d*x)), x) + Integral(sec(c + d*x)**2/sqrt(e*tan(c + d*x)), x))
Exception generated. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=\int { \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \tan \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]