Integrand size = 28, antiderivative size = 82 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] Output:
(1+I)*a^(1/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1 /2))/d-2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)
Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}-2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] Input:
Integrate[Sqrt[a + I*a*Tan[c + d*x]]/Tan[c + d*x]^(3/2),x]
Output:
(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]] - 2*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]])
Time = 0.38 (sec) , antiderivative size = 82, 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, 4031, 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.
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan (c+d x)^{3/2}}dx\) |
\(\Big \downarrow \) 4031 |
\(\displaystyle i \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {2 a^2 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(1+i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}\) |
Input:
Int[Sqrt[a + I*a*Tan[c + d*x]]/Tan[c + d*x]^(3/2),x]
Output:
((1 + I)*Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a *Tan[c + d*x]]])/d - (2*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[(-d)*(a + b*Tan[e + f*x])^m*((c + d*Ta n[e + f*x])^(n + 1)/(f*m*(c^2 + d^2))), x] + Simp[a/(a*c - b*d) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d , e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^ 2, 0] && EqQ[m + n + 1, 0] && !LtQ[m, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 155 vs. \(2 (68 ) = 136\).
Time = 1.69 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.90
method | result | size |
derivativedivides | \(\frac {\left (\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{2 d \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}}\) | \(156\) |
default | \(\frac {\left (\sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{2 d \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}}\) | \(156\) |
Input:
int((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/2/d*(2^(1/2)*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))*a*tan(d*x+c)-4*(a*tan(d*x+c)*(1+ I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))*(a*(1+I*tan(d*x+c)))^(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)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 329 vs. \(2 (62) = 124\).
Time = 0.08 (sec) , antiderivative size = 329, normalized size of antiderivative = 4.01 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (i \, e^{\left (3 i \, d x + 3 i \, c\right )} + i \, e^{\left (i \, d x + i \, c\right )}\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - i \, d \sqrt {\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \] Input:
integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="fricas")
Output:
-1/2*(4*sqrt(2)*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*e^(3*I*d*x + 3*I*c) + I*e^(I*d*x + I*c)) + (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(2*I*a/d^2)*log((sqrt(2)*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))*(e^(2*I*d*x + 2*I*c) + 1) + I*d*sqrt(2*I*a/d^2)*e^(I*d*x + I* c))*e^(-I*d*x - I*c)) - (d*e^(2*I*d*x + 2*I*c) - d)*sqrt(2*I*a/d^2)*log((s qrt(2)*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))*(e^(2*I*d*x + 2*I*c) + 1) - I*d*sqrt(2*I*a/d^2 )*e^(I*d*x + I*c))*e^(-I*d*x - I*c)))/(d*e^(2*I*d*x + 2*I*c) - d)
\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(3/2),x)
Output:
Integral(sqrt(I*a*(tan(c + d*x) - I))/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. 549 vs. \(2 (62) = 124\).
Time = 0.21 (sec) , antiderivative size = 549, normalized size of antiderivative = 6.70 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx =\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")
Output:
-1/2*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^( 1/4)*sqrt(a)*(-(2*I - 2)*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)*log(cos(d*x + c)^2 + sin(d*x + c) ^2 + sqrt(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*a rctan2(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(si n(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/ 2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))))) - 4*(((I - 1)*cos(d *x + c) - (I + 1)*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d *x + 2*c) + 1)) + (-(I + 1)*cos(d*x + c) - (I - 1)*sin(d*x + c))*sin(1/2*a rctan2(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. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(1/2)/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. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \] Input:
int((a + a*tan(c + d*x)*1i)^(1/2)/tan(c + d*x)^(3/2),x)
Output:
int((a + a*tan(c + d*x)*1i)^(1/2)/tan(c + d*x)^(3/2), x)
\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \] Input:
int((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(3/2),x)
Output:
sqrt(a)*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2, x)