Integrand size = 24, antiderivative size = 119 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {4 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d} \] Output:
-4*2^(1/2)*a^(5/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d +4*a^2*(a+I*a*tan(d*x+c))^(1/2)/d+2/3*a*(a+I*a*tan(d*x+c))^(3/2)/d+2/5*(a+ I*a*tan(d*x+c))^(5/2)/d
Time = 0.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {-60 \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (38+11 i \tan (c+d x)-3 \tan ^2(c+d x)\right )}{15 d} \] Input:
Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(-60*Sqrt[2]*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] + 2*a^2*Sqrt[a + I*a*Tan[c + d*x]]*(38 + (11*I)*Tan[c + d*x] - 3*Tan[c + d*x]^2))/(15*d)
Time = 0.50 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 4010, 3042, 3959, 3042, 3959, 3042, 3961, 219}
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.
\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2}dx\) |
\(\Big \downarrow \) 4010 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \int (i \tan (c+d x) a+a)^{5/2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \int (i \tan (c+d x) a+a)^{5/2}dx\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 3959 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 (a+i a \tan (c+d x))^{5/2}}{5 d}-i \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )\) |
Input:
Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2),x]
Output:
(2*(a + I*a*Tan[c + d*x])^(5/2))/(5*d) - I*((((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/d + 2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((2*I)*a*Sqrt[a + I*a*Tan[c + d*x]])/d))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a Int[(a + b*Tan[c + d* x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n , 1]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Simp [(b*c + a*d)/b Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e , f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && !LtQ[m, 0]
Time = 1.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d}\) | \(89\) |
default | \(\frac {\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{d}\) | \(89\) |
Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/d*(2/5*(a+I*a*tan(d*x+c))^(5/2)+2/3*a*(a+I*a*tan(d*x+c))^(3/2)+4*a^2*(a+ I*a*tan(d*x+c))^(1/2)-4*a^(5/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/ 2)*2^(1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (92) = 184\).
Time = 0.08 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.62 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 15 \, \sqrt {2} \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left (a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) - 2 \, \sqrt {2} {\left (26 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
Output:
-2/15*(15*sqrt(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*(a^3*e^(I*d*x + I*c) + sqrt(a^5/d^2)*(d*e^(2*I*d*x + 2*I *c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 15*sqr t(2)*sqrt(a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*l og(4*(a^3*e^(I*d*x + I*c) - sqrt(a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt (a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a^2) - 2*sqrt(2)*(26*a^2*e ^(5*I*d*x + 5*I*c) + 35*a^2*e^(3*I*d*x + 3*I*c) + 15*a^2*e^(I*d*x + I*c))* sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \tan {\left (c + d x \right )}\, dx \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))**(5/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(5/2)*tan(c + d*x), x)
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2 \, {\left (15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} + 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 30 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{4}\right )}}{15 \, a^{2} d} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
Output:
2/15*(15*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a ))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) + 3*(I*a*tan(d*x + c) + a)^(5/2)*a^2 + 5*(I*a*tan(d*x + c) + a)^(3/2)*a^3 + 30*sqrt(I*a*tan(d*x + c) + a)*a^4)/(a^2*d)
Exception generated. \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.82 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\frac {2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {4\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}+\frac {2\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {\sqrt {2}\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \] Input:
int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(5/2),x)
Output:
(2*(a + a*tan(c + d*x)*1i)^(5/2))/(5*d) + (4*a^2*(a + a*tan(c + d*x)*1i)^( 1/2))/d + (2*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) + (2^(1/2)*a^(5/2)*ata n((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(2*a^(1/2)))*4i)/d
\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{5/2} \, dx=\sqrt {a}\, a^{2} \left (-\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{3}d x \right )+2 \left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) \] Input:
int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(5/2),x)
Output:
sqrt(a)*a**2*( - int(sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**3,x) + 2*int(s qrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2,x)*i + int(sqrt(tan(c + d*x)*i + 1 )*tan(c + d*x),x))