\(\int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx\) [1197]

Optimal result

Integrand size = 25, antiderivative size = 65 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=-\frac {2 i}{a^2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}}+\frac {4 i (a-i a x)^{3/4}}{3 a^3 (a+i a x)^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, 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} \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\frac {4 i (a-i a x)^{3/4}}{3 a^3 (a+i a x)^{3/4}}-\frac {2 i}{a^2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}} \]

[In]

Int[1/((a - I*a*x)^(5/4)*(a + I*a*x)^(7/4)),x]

[Out]

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

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}{a^2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}}+\frac {2 \int \frac {1}{\sqrt [4]{a-i a x} (a+i a x)^{7/4}} \, dx}{a} \\ & = -\frac {2 i}{a^2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}}+\frac {4 i (a-i a x)^{3/4}}{3 a^3 (a+i a x)^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.58 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\frac {-2 i+4 x}{3 a^2 \sqrt [4]{a-i a x} (a+i a x)^{3/4}} \]

[In]

Integrate[1/((a - I*a*x)^(5/4)*(a + I*a*x)^(7/4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.51

method	result	size
risch	\(\frac {\frac {4 x}{3}-\frac {2 i}{3}}{a^{2} \left (a \left (i x +1\right )\right )^{\frac {3}{4}} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}}}\)	\(33\)
gosper	\(\frac {2 \left (x +i\right ) \left (-x +i\right ) \left (-2 x +i\right )}{3 \left (-i a x +a \right )^{\frac {5}{4}} \left (i a x +a \right )^{\frac {7}{4}}}\)	\(37\)

[In]

int(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(7/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}} {\left (2 \, x - i\right )}}{3 \, {\left (a^{4} x^{2} + a^{4}\right )}} \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(7/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {7}{4}} \left (- i a \left (x + i\right )\right )^{\frac {5}{4}}}\, dx \]

[In]

integrate(1/(a-I*a*x)**(5/4)/(a+I*a*x)**(7/4),x)

[Out]

Integral(1/((I*a*(x - I))**(7/4)*(-I*a*(x + I))**(5/4)), x)

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {7}{4}} {\left (-i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(7/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(7/4)*(-I*a*x + a)^(5/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(7/4),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 [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{7/4}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{5/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{7/4}} \,d x \]

[In]

int(1/((a - a*x*1i)^(5/4)*(a + a*x*1i)^(7/4)),x)

[Out]

int(1/((a - a*x*1i)^(5/4)*(a + a*x*1i)^(7/4)), x)