\(\int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx\) [1210]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [C] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F(-2)]
   Giac [F(-2)]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 82 \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=-\frac {2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {6 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

-2/5*I/a^2/(a-I*a*x)^(5/4)/(a+I*a*x)^(1/4)+6/5*(x^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*E
llipticE(sin(1/2*arctan(x)),2^(1/2))/a^3/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {53, 42, 203, 202} \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\frac {6 \sqrt [4]{x^2+1} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

[In]

Int[1/((a - I*a*x)^(9/4)*(a + I*a*x)^(5/4)),x]

[Out]

((-2*I)/5)/(a^2*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) + (6*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(5*a^3*(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 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}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {3 \int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx}{5 a} \\ & = -\frac {2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {\left (3 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac {1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {\left (3 \sqrt [4]{1+x^2}\right ) \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {6 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{5 a^3 \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 = 0.85 \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=-\frac {i 2^{3/4} \sqrt [4]{1+i x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{4},-\frac {1}{4},\frac {1}{2}-\frac {i x}{2}\right )}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

[In]

Integrate[1/((a - I*a*x)^(9/4)*(a + I*a*x)^(5/4)),x]

[Out]

((-1/5*I)*2^(3/4)*(1 + I*x)^(1/4)*Hypergeometric2F1[-5/4, 5/4, -1/4, 1/2 - (I/2)*x])/(a^2*(a - I*a*x)^(5/4)*(a
 + I*a*x)^(1/4))

Maple [C] (verified)

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

Time = 0.40 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.30

method result size
risch \(\frac {\frac {6}{5} x^{2}+\frac {6}{5} i x +\frac {2}{5}}{\left (x +i\right ) a^{3} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {3 x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{5 \left (a^{2}\right )^{\frac {1}{4}} a^{3} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\) \(107\)

[In]

int(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {5}{4}} {\left (-i \, a x + a\right )}^{\frac {9}{4}}} \,d x } \]

[In]

integrate(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}} \left (- i a \left (x + i\right )\right )^{\frac {9}{4}}}\, dx \]

[In]

integrate(1/(a-I*a*x)**(9/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral(1/((I*a*(x - I))**(5/4)*(-I*a*(x + I))**(9/4)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/(a-I*a*x)^(9/4)/(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. \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/(a-I*a*x)^(9/4)/(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 [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{9/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \]

[In]

int(1/((a - a*x*1i)^(9/4)*(a + a*x*1i)^(5/4)),x)

[Out]

int(1/((a - a*x*1i)^(9/4)*(a + a*x*1i)^(5/4)), x)