Integrand size = 28, antiderivative size = 83 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] Output:
(2+2*I)*a^(3/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^ (1/2))/d-2*a*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(217\) vs. \(2(83)=166\).
Time = 3.07 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.61 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 a \left (\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}-\frac {\sqrt {a+i a \tan (c+d x)} \left (\sqrt {a} \sqrt {1+i \tan (c+d x)}-\sqrt [4]{-1} \sqrt {a} \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}+\text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)}\right )}{\sqrt {a} \sqrt {1+i \tan (c+d x)}}\right )}{d \sqrt {\tan (c+d x)}} \] Input:
Integrate[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(3/2),x]
Output:
(2*a*(Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]] - (Sqrt[a + I*a*Tan[c + d*x]]*(Sqrt[a]*Sqr t[1 + I*Tan[c + d*x]] - (-1)^(1/4)*Sqrt[a]*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*Sqrt[Tan[c + d*x]] + ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[ I*a*Tan[c + d*x]]))/(Sqrt[a]*Sqrt[1 + I*Tan[c + d*x]])))/(d*Sqrt[Tan[c + d *x]])
Time = 0.39 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 4028, 3042, 4027, 218}
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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4028 |
\(\displaystyle 2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {4 a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(2+2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
Input:
Int[(a + I*a*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(3/2),x]
Output:
((2 + 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I *a*Tan[c + d*x]]])/d - (2*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d* x]])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - b*d)) Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 ] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 319 vs. \(2 (69 ) = 138\).
Time = 1.58 (sec) , antiderivative size = 320, normalized size of antiderivative = 3.86
method | result | size |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )+\ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )+4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\) | \(320\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \left (i \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )+\ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, \sqrt {i a}\, a \tan \left (d x +c \right )+4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )+4 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{2 d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \sqrt {i a}}\) | \(320\) |
Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*a*(I*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan (d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2 )*(I*a)^(1/2)*a*tan(d*x+c)+ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I *tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*(I*a)^(1/2 )*a*tan(d*x+c)+4*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+ I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)+4*(-I*a)^(1/ 2)*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2))/tan(d*x+c)^(1/2)/(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 353 vs. \(2 (63) = 126\).
Time = 0.09 (sec) , antiderivative size = 353, normalized size of antiderivative = 4.25 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} {\left (i \, a e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - i \, \sqrt {\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x, algorithm="fricas")
Output:
-1/2*(4*sqrt(2)*(I*a*e^(3*I*d*x + 3*I*c) + I*a*e^(I*d*x + I*c))*sqrt(a/(e^ (2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2* I*c) + 1)) + (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(8*I*a^3/d^2)*log(1/2*(2*sqrt (2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I *e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)) + I*sqrt(8*I*a^3/d^2) *d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt (8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2* I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c ) + 1)) - I*sqrt(8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a))/(d*e ^(2*I*d*x + 2*I*c) - d)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+I*a*tan(d*x+c))**(3/2)/tan(d*x+c)**(3/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(3/2)/tan(c + d*x)**(3/2), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (63) = 126\).
Time = 0.23 (sec) , antiderivative size = 555, normalized size of antiderivative = 6.69 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")
Output:
-((-(2*I - 2)*a*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2 *d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2* d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + (I + 1)*a*log(cos(d*x + c)^2 + sin(d*x + c)^2 + sq rt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos( 1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2( sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin (2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arcta n2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin( 2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) - 2*(((I - 1)*a*cos (d*x + c) - (I + 1)*a*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos (2*d*x + 2*c) + 1)) + (-(I + 1)*a*cos(d*x + c) - (I - 1)*a*sin(d*x + c))*s in(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2 *d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*d)
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),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 TypeDone
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(3/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(3/2)/tan(c + d*x)^(3/2), x)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a \left (\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}+\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right ) d \right )}{\tan \left (d x +c \right ) d} \] Input:
int((a+I*a*tan(d*x+c))^(3/2)/tan(d*x+c)^(3/2),x)
Output:
(2*sqrt(a)*a*(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) + int((sqrt(tan( c + d*x))*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)*d))/(t an(c + d*x)*d)