\(\int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx\) [1225]

Optimal result

Integrand size = 25, antiderivative size = 33 \[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\frac {2 i (a-i a x)^{5/4}}{5 a^2 (a+i a x)^{5/4}} \]

[Out]

2/5*I*(a-I*a*x)^(5/4)/a^2/(a+I*a*x)^(5/4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, 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} \[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\frac {2 i (a-i a x)^{5/4}}{5 a^2 (a+i a x)^{5/4}} \]

[In]

Int[(a - I*a*x)^(1/4)/(a + I*a*x)^(9/4),x]

[Out]

(((2*I)/5)*(a - I*a*x)^(5/4))/(a^2*(a + I*a*x)^(5/4))

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 (a-i a x)^{5/4}}{5 a^2 (a+i a x)^{5/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\frac {2 i (a-i a x)^{5/4}}{5 a^2 (a+i a x)^{5/4}} \]

[In]

Integrate[(a - I*a*x)^(1/4)/(a + I*a*x)^(9/4),x]

[Out]

(((2*I)/5)*(a - I*a*x)^(5/4))/(a^2*(a + I*a*x)^(5/4))

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97

method	result	size
gosper	\(-\frac {2 i \left (x +i\right ) \left (-x +i\right ) \left (-i a x +a \right )^{\frac {1}{4}}}{5 \left (i a x +a \right )^{\frac {9}{4}}}\)	\(32\)
risch	\(\frac {2 \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (x^{2}+2 i x -1\right )}{5 a^{2} \left (i x -1\right ) \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x -i\right )}\)	\(50\)

[In]

int((a-I*a*x)^(1/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)

[Out]

-2/5*I*(x+I)*(-x+I)*(a-I*a*x)^(1/4)/(a+I*a*x)^(9/4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27 \[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=-\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (x + i\right )}}{5 \, {\left (a^{3} x^{2} - 2 i \, a^{3} x - a^{3}\right )}} \]

[In]

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(9/4),x, algorithm="fricas")

[Out]

-2/5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(x + I)/(a^3*x^2 - 2*I*a^3*x - a^3)

Sympy [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\int \frac {\sqrt [4]{- i a \left (x + i\right )}}{\left (i a \left (x - i\right )\right )^{\frac {9}{4}}}\, dx \]

[In]

integrate((a-I*a*x)**(1/4)/(a+I*a*x)**(9/4),x)

[Out]

Integral((-I*a*(x + I))**(1/4)/(I*a*(x - I))**(9/4), x)

Maxima [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {1}{4}}}{{\left (i \, a x + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(9/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(1/4)/(I*a*x + a)^(9/4), x)

Giac [F]

\[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {1}{4}}}{{\left (i \, a x + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate((a-I*a*x)^(1/4)/(a+I*a*x)^(9/4),x, algorithm="giac")

[Out]

integrate((-I*a*x + a)^(1/4)/(I*a*x + a)^(9/4), x)

Mupad [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{9/4}} \, dx=-\frac {2\,\left (-1+x\,1{}\mathrm {i}\right )\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}}{5\,a^2\,\left (x-\mathrm {i}\right )\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

[In]

int((a - a*x*1i)^(1/4)/(a + a*x*1i)^(9/4),x)

[Out]

-(2*(x*1i - 1)*(-a*(x*1i - 1))^(1/4))/(5*a^2*(x - 1i)*(a*(x*1i + 1))^(1/4))