\(\int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx\) [1182]

Optimal result

Integrand size = 25, antiderivative size = 100 \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}} \]

[Out]

-2/11*I*(a+I*a*x)^(3/4)/a^2/(a-I*a*x)^(11/4)-8/77*I*(a+I*a*x)^(3/4)/a^3/(a-I*a*x)^(7/4)-16/231*I*(a+I*a*x)^(3/
4)/a^4/(a-I*a*x)^(3/4)

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} \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=-\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}} \]

[In]

Int[1/((a - I*a*x)^(15/4)*(a + I*a*x)^(1/4)),x]

[Out]

(((-2*I)/11)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(11/4)) - (((8*I)/77)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(7/
4)) - (((16*I)/231)*(a + I*a*x)^(3/4))/(a^4*(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+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}+\frac {4 \int \frac {1}{(a-i a x)^{11/4} \sqrt [4]{a+i a x}} \, dx}{11 a} \\ & = -\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}+\frac {8 \int \frac {1}{(a-i a x)^{7/4} \sqrt [4]{a+i a x}} \, dx}{77 a^2} \\ & = -\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.85 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\frac {2 (a+i a x)^{3/4} \left (41 i+28 x-8 i x^2\right )}{231 a^4 (i+x)^2 (a-i a x)^{3/4}} \]

[In]

Integrate[1/((a - I*a*x)^(15/4)*(a + I*a*x)^(1/4)),x]

[Out]

(2*(a + I*a*x)^(3/4)*(41*I + 28*x - (8*I)*x^2))/(231*a^4*(I + x)^2*(a - I*a*x)^(3/4))

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.43

method	result	size
gosper	\(-\frac {2 \left (x +i\right ) \left (-x +i\right ) \left (8 i x^{2}-28 x -41 i\right )}{231 \left (-i a x +a \right )^{\frac {15}{4}} \left (i a x +a \right )^{\frac {1}{4}}}\)	\(43\)
risch	\(\frac {\frac {16}{231} x^{3}+\frac {40}{231} i x^{2}-\frac {26}{231} x +\frac {82}{231} i}{a^{3} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x +i\right )^{2}}\)	\(50\)

[In]

int(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-2/231*(x+I)*(-x+I)*(8*I*x^2-41*I-28*x)/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (8 \, x^{2} + 28 i \, x - 41\right )}}{231 \, {\left (a^{5} x^{3} + 3 i \, a^{5} x^{2} - 3 \, a^{5} x - i \, a^{5}\right )}} \]

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x, algorithm="fricas")

[Out]

2/231*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(8*x^2 + 28*I*x - 41)/(a^5*x^3 + 3*I*a^5*x^2 - 3*a^5*x - I*a^5)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a-I*a*x)**(15/4)/(a+I*a*x)**(1/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {15}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(15/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),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 = 0.81 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51 \[ \int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx=\frac {{\left (x-\mathrm {i}\right )}^4\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,\left (8\,x^2+x\,28{}\mathrm {i}-41\right )\,2{}\mathrm {i}}{231\,a^4\,{\left (x^2+1\right )}^3\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

[In]

int(1/((a - a*x*1i)^(15/4)*(a + a*x*1i)^(1/4)),x)

[Out]

((x - 1i)^4*(-a*(x*1i - 1))^(1/4)*(x*28i + 8*x^2 - 41)*2i)/(231*a^4*(x^2 + 1)^3*(a*(x*1i + 1))^(1/4))