\(\int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx\) [1214]

Optimal result

Integrand size = 25, antiderivative size = 31 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\frac {2 i \sqrt [4]{a-i a x}}{a^2 \sqrt [4]{a+i a x}} \]

[Out]

2*I*(a-I*a*x)^(1/4)/a^2/(a+I*a*x)^(1/4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {37} \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\frac {2 i \sqrt [4]{a-i a x}}{a^2 \sqrt [4]{a+i a x}} \]

[In]

Int[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(5/4)),x]

[Out]

((2*I)*(a - I*a*x)^(1/4))/(a^2*(a + I*a*x)^(1/4))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i \sqrt [4]{a-i a x}}{a^2 \sqrt [4]{a+i a x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\frac {2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}}{a^3 (-i+x)} \]

[In]

Integrate[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(5/4)),x]

[Out]

(2*(a - I*a*x)^(1/4)*(a + I*a*x)^(3/4))/(a^3*(-I + x))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {2 x +2 i}{a \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(31\)
gosper	\(-\frac {2 i \left (x +i\right ) \left (-x +i\right )}{\left (-i a x +a \right )^{\frac {3}{4}} \left (i a x +a \right )^{\frac {5}{4}}}\)	\(32\)

[In]

int(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

2/a/(-a*(I*x-1))^(3/4)/(a*(I*x+1))^(1/4)*(x+I)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{a^{3} x - i \, a^{3}} \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)/(a^3*x - I*a^3)

Sympy [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}} \left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate(1/(a-I*a*x)**(3/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral(1/((I*a*(x - I))**(5/4)*(-I*a*(x + I))**(3/4)), x)

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {5}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(5/4)*(-I*a*x + a)^(3/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:The choice was done assuming 0=[0,0]ext_reduce Error: Bad Argument TypeDone

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx=\frac {{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,2{}\mathrm {i}}{a^2\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

[In]

int(1/((a - a*x*1i)^(3/4)*(a + a*x*1i)^(5/4)),x)

[Out]

((-a*(x*1i - 1))^(1/4)*2i)/(a^2*(a*(x*1i + 1))^(1/4))