\(\int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx\) [1216]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=-\frac {2 i}{7 a^2 (a-i a x)^{7/4} \sqrt [4]{a+i a x}}-\frac {8 i}{21 a^3 (a-i a x)^{3/4} \sqrt [4]{a+i a x}}+\frac {16 i \sqrt [4]{a-i a x}}{21 a^4 \sqrt [4]{a+i a x}} \]

[Out]

-2/7*I/a^2/(a-I*a*x)^(7/4)/(a+I*a*x)^(1/4)-8/21*I/a^3/(a-I*a*x)^(3/4)/(a+I*a*x)^(1/4)+16/21*I*(a-I*a*x)^(1/4)/
a^4/(a+I*a*x)^(1/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {47, 37} \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\frac {16 i \sqrt [4]{a-i a x}}{21 a^4 \sqrt [4]{a+i a x}}-\frac {8 i}{21 a^3 \sqrt [4]{a+i a x} (a-i a x)^{3/4}}-\frac {2 i}{7 a^2 \sqrt [4]{a+i a x} (a-i a x)^{7/4}} \]

[In]

Int[1/((a - I*a*x)^(11/4)*(a + I*a*x)^(5/4)),x]

[Out]

((-2*I)/7)/(a^2*(a - I*a*x)^(7/4)*(a + I*a*x)^(1/4)) - ((8*I)/21)/(a^3*(a - I*a*x)^(3/4)*(a + I*a*x)^(1/4)) +
(((16*I)/21)*(a - I*a*x)^(1/4))/(a^4*(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}{7 a^2 (a-i a x)^{7/4} \sqrt [4]{a+i a x}}+\frac {4 \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{5/4}} \, dx}{7 a} \\ & = -\frac {2 i}{7 a^2 (a-i a x)^{7/4} \sqrt [4]{a+i a x}}-\frac {8 i}{21 a^3 (a-i a x)^{3/4} \sqrt [4]{a+i a x}}+\frac {8 \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx}{21 a^2} \\ & = -\frac {2 i}{7 a^2 (a-i a x)^{7/4} \sqrt [4]{a+i a x}}-\frac {8 i}{21 a^3 (a-i a x)^{3/4} \sqrt [4]{a+i a x}}+\frac {16 i \sqrt [4]{a-i a x}}{21 a^4 \sqrt [4]{a+i a x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50 \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\frac {-2+24 i x+16 x^2}{21 a^3 (i+x) (a-i a x)^{3/4} \sqrt [4]{a+i a x}} \]

[In]

Integrate[1/((a - I*a*x)^(11/4)*(a + I*a*x)^(5/4)),x]

[Out]

(-2 + (24*I)*x + 16*x^2)/(21*a^3*(I + x)*(a - I*a*x)^(3/4)*(a + I*a*x)^(1/4))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.43

method	result	size
gosper	\(\frac {2 \left (x +i\right ) \left (-x +i\right ) \left (8 i x^{2}-12 x -i\right )}{21 \left (-i a x +a \right )^{\frac {11}{4}} \left (i a x +a \right )^{\frac {5}{4}}}\)	\(43\)
risch	\(\frac {\frac {16}{21} x^{2}+\frac {8}{7} i x -\frac {2}{21}}{a^{3} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x +i\right )}\)	\(44\)

[In]

int(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

2/21*(x+I)*(-x+I)*(8*I*x^2-I-12*x)/(a-I*a*x)^(11/4)/(a+I*a*x)^(5/4)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (8 \, x^{2} + 12 i \, x - 1\right )}}{21 \, {\left (a^{5} x^{3} + i \, a^{5} x^{2} + a^{5} x + i \, a^{5}\right )}} \]

[In]

integrate(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

2/21*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(8*x^2 + 12*I*x - 1)/(a^5*x^3 + I*a^5*x^2 + a^5*x + I*a^5)

Sympy [F]

\[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}} \left (- i a \left (x + i\right )\right )^{\frac {11}{4}}}\, dx \]

[In]

integrate(1/(a-I*a*x)**(11/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral(1/((I*a*(x - I))**(5/4)*(-I*a*(x + I))**(11/4)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(11/4)/(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 = 0.83 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.46 \[ \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{5/4}} \, dx=-\frac {{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,\left (8\,x^2+x\,12{}\mathrm {i}-1\right )\,2{}\mathrm {i}}{21\,a^4\,{\left (-1+x\,1{}\mathrm {i}\right )}^2\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

[In]

int(1/((a - a*x*1i)^(11/4)*(a + a*x*1i)^(5/4)),x)

[Out]

-((-a*(x*1i - 1))^(1/4)*(x*12i + 8*x^2 - 1)*2i)/(21*a^4*(x*1i - 1)^2*(a*(x*1i + 1))^(1/4))