\(\int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx\) [1226]

Optimal result

Integrand size = 25, antiderivative size = 67 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}+\frac {4 i \sqrt [4]{a-i a x}}{5 a^3 \sqrt [4]{a+i a x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {47, 37} \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {4 i \sqrt [4]{a-i a x}}{5 a^3 \sqrt [4]{a+i a x}}+\frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}} \]

[In]

Int[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(9/4)),x]

[Out]

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

Rule 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}+\frac {2 \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx}{5 a} \\ & = \frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}+\frac {4 i \sqrt [4]{a-i a x}}{5 a^3 \sqrt [4]{a+i a x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 (3+2 i x) \sqrt [4]{a-i a x}}{5 a^3 (-i+x) \sqrt [4]{a+i a x}} \]

[In]

Integrate[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(9/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-\frac {2 \left (x +i\right ) \left (-x +i\right ) \left (-2 x +3 i\right )}{5 \left (-i a x +a \right )^{\frac {3}{4}} \left (i a x +a \right )^{\frac {9}{4}}}\)	\(37\)
risch	\(\frac {\frac {4}{5} x^{2}-\frac {2}{5} i x +\frac {6}{5}}{a^{2} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x -i\right )}\)	\(44\)

[In]

int(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)

[Out]

-2/5*(x+I)*(-x+I)*(-2*x+3*I)/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (2 \, x - 3 i\right )}}{5 \, {\left (a^{4} x^{2} - 2 i \, a^{4} x - a^{4}\right )}} \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {9}{4}} \left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}\, dx \]

[In]

integrate(1/(a-I*a*x)**(3/4)/(a+I*a*x)**(9/4),x)

[Out]

Integral(1/((I*a*(x - I))**(9/4)*(-I*a*(x + I))**(3/4)), x)

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {9}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(9/4)*(-I*a*x + a)^(3/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),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 = 0.68 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2\,\left (3+x\,2{}\mathrm {i}\right )\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}}{5\,a^3\,\left (x-\mathrm {i}\right )\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

[In]

int(1/((a - a*x*1i)^(3/4)*(a + a*x*1i)^(9/4)),x)

[Out]

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