\(\int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx\) [1177]

Optimal result

Integrand size = 25, antiderivative size = 148 \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=-\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{39 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

-4/39*I/a^3/(a-I*a*x)^(5/4)/(a+I*a*x)^(1/4)-2/13*I*(a+I*a*x)^(3/4)/a^2/(a-I*a*x)^(13/4)-10/117*I*(a+I*a*x)^(3/
4)/a^3/(a-I*a*x)^(9/4)+2/39*(x^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*EllipticE(sin(1/2*ar
ctan(x)),2^(1/2))/a^4/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {53, 48, 42, 203, 202} \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\frac {2 \sqrt [4]{x^2+1} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{39 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}-\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}} \]

[In]

Int[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(1/4)),x]

[Out]

((-4*I)/39)/(a^3*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) - (((2*I)/13)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(13/4)
) - (((10*I)/117)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(9/4)) + (2*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(
39*a^4*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4))

Rule 42

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[(a + b*x)^FracPart[m]*((c + d*x)^Frac
Part[m]/(a*c + b*d*x^2)^FracPart[m]), Int[(a*c + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c +
a*d, 0] &&  !IntegerQ[2*m]

Rule 48

Int[1/(((a_) + (b_.)*(x_))^(9/4)*((c_) + (d_.)*(x_))^(1/4)), x_Symbol] :> Simp[-4/(5*b*(a + b*x)^(5/4)*(c + d*
x)^(1/4)), x] - Dist[d/(5*b), Int[1/((a + b*x)^(5/4)*(c + d*x)^(5/4)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ
[b*c + a*d, 0] && NegQ[a^2*b^2]

Rule 53

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

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}+\frac {5 \int \frac {1}{(a-i a x)^{13/4} \sqrt [4]{a+i a x}} \, dx}{13 a} \\ & = -\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {5 \int \frac {1}{(a-i a x)^{9/4} \sqrt [4]{a+i a x}} \, dx}{39 a^2} \\ & = -\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {\int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx}{39 a^2} \\ & = -\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {\sqrt [4]{a^2+a^2 x^2} \int \frac {1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{39 a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {\sqrt [4]{1+x^2} \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx}{39 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {4 i}{39 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i (a+i a x)^{3/4}}{13 a^2 (a-i a x)^{13/4}}-\frac {10 i (a+i a x)^{3/4}}{117 a^3 (a-i a x)^{9/4}}+\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{39 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.47 \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=-\frac {2 i 2^{3/4} \sqrt [4]{1+i x} \operatorname {Hypergeometric2F1}\left (-\frac {13}{4},\frac {1}{4},-\frac {9}{4},\frac {1}{2}-\frac {i x}{2}\right )}{13 a (a-i a x)^{13/4} \sqrt [4]{a+i a x}} \]

[In]

Integrate[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(1/4)),x]

[Out]

(((-2*I)/13)*2^(3/4)*(1 + I*x)^(1/4)*Hypergeometric2F1[-13/4, 1/4, -9/4, 1/2 - (I/2)*x])/(a*(a - I*a*x)^(13/4)
*(a + I*a*x)^(1/4))

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {\frac {2}{39} x^{4}+\frac {2}{13} i x^{3}-\frac {16}{117} x^{2}-\frac {40}{117}}{\left (x +i\right )^{3} a^{4} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{39 \left (a^{2}\right )^{\frac {1}{4}} a^{4} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(114\)

[In]

int(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

2/117*(9*I*x^3+3*x^4-20-8*x^2)/(x+I)^3/a^4/(-a*(I*x-1))^(1/4)/(a*(I*x+1))^(1/4)-1/39/(a^2)^(1/4)*x*hypergeom([
1/4,1/2],[3/2],-x^2)/a^4*(-a^2*(I*x-1)*(I*x+1))^(1/4)/(-a*(I*x-1))^(1/4)/(a*(I*x+1))^(1/4)

Fricas [F]

\[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {17}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(1/4),x, algorithm="fricas")

[Out]

1/117*(2*(3*x^3 + 12*I*x^2 - 20*x - 20*I)*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4) + 117*(a^6*x^4 + 4*I*a^6*x^3 -
6*a^6*x^2 - 4*I*a^6*x + a^6)*integral(-1/39*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^6*x^2 + a^6), x))/(a^6*x^4
+ 4*I*a^6*x^3 - 6*a^6*x^2 - 4*I*a^6*x + a^6)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\text {Timed out} \]

[In]

integrate(1/(a-I*a*x)**(17/4)/(a+I*a*x)**(1/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {17}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(17/4)), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(17/4)/(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 [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{17/4} \sqrt [4]{a+i a x}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{17/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{1/4}} \,d x \]

[In]

int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(1/4)),x)

[Out]

int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(1/4)), x)