3.13.23 \(\int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx\) [1223]

Optimal. Leaf size=154 \[ -\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {14 x}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x} \left (1+x^2\right )}+\frac {42 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

-2/13*I/a^2/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4)-2/13*I/a^3/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4)+14/65*x/a^6/(a-I*a*x)^
(1/4)/(a+I*a*x)^(1/4)/(x^2+1)+42/65*(x^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*EllipticE(si
n(1/2*arctan(x)),2^(1/2))/a^6/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {53, 42, 205, 203, 202} \begin {gather*} \frac {42 \sqrt [4]{x^2+1} E\left (\left .\frac {\text {ArcTan}(x)}{2}\right |2\right )}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {14 x}{65 a^6 \left (x^2+1\right ) \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(9/4)),x]

[Out]

((-2*I)/13)/(a^2*(a - I*a*x)^(13/4)*(a + I*a*x)^(5/4)) - ((2*I)/13)/(a^3*(a - I*a*x)^(9/4)*(a + I*a*x)^(5/4))
+ (14*x)/(65*a^6*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4)*(1 + x^2)) + (42*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2
])/(65*a^6*(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]

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rubi steps

\begin {align*} \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}+\frac {9 \int \frac {1}{(a-i a x)^{13/4} (a+i a x)^{9/4}} \, dx}{13 a}\\ &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {7 \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{9/4}} \, dx}{13 a^2}\\ &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {\left (7 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac {1}{\left (a^2+a^2 x^2\right )^{9/4}} \, dx}{13 a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {14 x}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x} \left (1+x^2\right )}+\frac {\left (21 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac {1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{65 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {14 x}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x} \left (1+x^2\right )}+\frac {\left (21 \sqrt [4]{1+x^2}\right ) \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {14 x}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x} \left (1+x^2\right )}+\frac {42 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 70, normalized size = 0.45 \begin {gather*} -\frac {i \sqrt [4]{1+i x} \, _2F_1\left (-\frac {13}{4},\frac {9}{4};-\frac {9}{4};\frac {1}{2}-\frac {i x}{2}\right )}{13 \sqrt [4]{2} a^3 (a-i a x)^{13/4} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(9/4)),x]

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.21, size = 130, normalized size = 0.84

method	result	size
risch	\(\frac {\frac {42}{65} x^{5}+\frac {84}{65} i x^{4}+\frac {14}{65} x^{3}+\frac {112}{65} i x^{2}-\frac {46}{65} x +\frac {4}{13} i}{\left (x -i\right ) \left (x +i\right )^{3} a^{6} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {21 x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{65 \left (a^{2}\right )^{\frac {1}{4}} a^{6} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)

[Out]

2/65*(42*I*x^4+21*x^5+56*I*x^2-23*x+7*x^3+10*I)/(x-I)/(x+I)^3/a^6/(-a*(-1+I*x))^(1/4)/(a*(1+I*x))^(1/4)-21/65/
(a^2)^(1/4)*x*hypergeom([1/4,1/2],[3/2],-x^2)/a^6*(-a^2*(-1+I*x)*(1+I*x))^(1/4)/(-a*(-1+I*x))^(1/4)/(a*(1+I*x)
)^(1/4)

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="fricas")

[Out]

1/65*(2*(21*x^5 + 42*I*x^4 + 7*x^3 + 56*I*x^2 - 23*x + 10*I)*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4) + 65*(a^8*x^
6 + 2*I*a^8*x^5 + a^8*x^4 + 4*I*a^8*x^3 - a^8*x^2 + 2*I*a^8*x - a^8)*integral(-21/65*(I*a*x + a)^(3/4)*(-I*a*x
+ a)^(3/4)/(a^8*x^2 + a^8), x))/(a^8*x^6 + 2*I*a^8*x^5 + a^8*x^4 + 4*I*a^8*x^3 - a^8*x^2 + 2*I*a^8*x - a^8)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(17/4)/(a+I*a*x)**(9/4),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:ext_reduce Error: Bad Argument TypeDone

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{17/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{9/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(9/4)),x)

[Out]

int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(9/4)), x)

________________________________________________________________________________________