\(\int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx\) [1207]

Optimal result

Integrand size = 25, antiderivative size = 102 \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=-\frac {6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}+\frac {6 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {49, 42, 235, 233, 202} \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\frac {6 \sqrt [4]{x^2+1} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}} \]

[In]

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

[Out]

(-6*x)/((a - I*a*x)^(1/4)*(a + I*a*x)^(1/4)) + ((4*I)*(a - I*a*x)^(3/4))/(a*(a + I*a*x)^(1/4)) + (6*(1 + x^2)^
(1/4)*EllipticE[ArcTan[x]/2, 2])/((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 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && 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 233

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

Rule 235

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

Rubi steps \begin{align*} \text {integral}& = \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-3 \int \frac {1}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \, dx \\ & = \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {\left (3 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac {1}{\sqrt [4]{a^2+a^2 x^2}} \, dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {\left (3 \sqrt [4]{1+x^2}\right ) \int \frac {1}{\sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {4 i (a-i a x)^{3/4}}{a \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}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \\ & = -\frac {6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {4 i (a-i a x)^{3/4}}{a \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 )}{\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 = 0.69 \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\frac {i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{7/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2}-\frac {i x}{2}\right )}{7 a^2 \sqrt [4]{a+i a x}} \]

[In]

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

[Out]

((I/7)*2^(3/4)*(1 + I*x)^(1/4)*(a - I*a*x)^(7/4)*Hypergeometric2F1[5/4, 7/4, 11/4, 1/2 - (I/2)*x])/(a^2*(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 = 88, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {4 x +4 i}{\left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {3 x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (a^{2}\right )^{\frac {1}{4}} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(88\)

[In]

int((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

-(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)*(x - 3*I) - (a^2*x^2 - I*a^2*x)*integral(-6*(I*a*x + a)^(3/4)*(-I*a*x
+ a)^(3/4)/(a^2*x^4 + a^2*x^2), x))/(a^2*x^2 - I*a^2*x)

Sympy [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int \frac {\left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}}}\, dx \]

[In]

integrate((a-I*a*x)**(3/4)/(a+I*a*x)**(5/4),x)

[Out]

Integral((-I*a*(x + I))**(3/4)/(I*a*(x - I))**(5/4), x)

Maxima [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(5/4), x)

Giac [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \]

[In]

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(5/4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int \frac {{\left (a-a\,x\,1{}\mathrm {i}\right )}^{3/4}}{{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \]

[In]

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(5/4),x)

[Out]

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(5/4), x)