3.6.68 \(\int \frac {a+b \sinh ^{-1}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{5/2}} \, dx\) [568]

Optimal. Leaf size=282 \[ \frac {i b d \left (1+c^2 x^2\right )^{5/2}}{6 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d \left (1+c^2 x^2\right )^{5/2} \text {ArcTan}(c x)}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

1/6*I*b*d*(c^2*x^2+1)^(5/2)/c/(c*x+I)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*d*(I-c*x)*(c^2*x^2+1)*(a+b*arcsi
nh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*d*x*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I
*c*f*x)^(5/2)+1/6*I*b*d*(c^2*x^2+1)^(5/2)*arctan(c*x)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*b*d*(c^2*x^2+1
)^(5/2)*ln(c^2*x^2+1)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)

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Rubi [A]
time = 0.20, antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {5796, 653, 197, 5837, 641, 46, 209, 266} \begin {gather*} \frac {2 d x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {d (-c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d \left (c^2 x^2+1\right )^{5/2} \text {ArcTan}(c x)}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d \left (c^2 x^2+1\right )^{5/2}}{6 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(5/2)),x]

[Out]

((I/6)*b*d*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (d*(I - c*x)*(1 + c^2*
x^2)*(a + b*ArcSinh[c*x]))/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (2*d*x*(1 + c^2*x^2)^2*(a + b*ArcSi
nh[c*x]))/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((I/6)*b*d*(1 + c^2*x^2)^(5/2)*ArcTan[c*x])/(c*(d + I*
c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*d*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(3*c*(d + I*c*d*x)^(5/2)*(f - I
*c*f*x)^(5/2))

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 209

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 653

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)/(2*a*c*(p + 1)))*(a + c*x
^2)^(p + 1), x] + Dist[d*((2*p + 3)/(2*a*(p + 1))), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x]
&& LtQ[p, -1] && NeQ[p, -3/2]

Rule 5796

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :>
Dist[(d + e*x)^q*((f + g*x)^q/(1 + c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n,
x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 + e^2, 0] && HalfIntegerQ[p,
q] && GeQ[p - q, 0]

Rule 5837

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[p + 1/2, 0]
&& GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x) \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {d (i-c x)}{3 c \left (1+c^2 x^2\right )^2}+\frac {2 d x}{3 \left (1+c^2 x^2\right )}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (b d \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {i-c x}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 b c d \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (b d \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{(-i-c x)^2 (i-c x)} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (b d \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {i}{2 (i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i b d \left (1+c^2 x^2\right )^{5/2}}{6 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i b d \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{6 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i b d \left (1+c^2 x^2\right )^{5/2}}{6 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {d (i-c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b d \left (1+c^2 x^2\right )^{5/2} \tan ^{-1}(c x)}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b d \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 202, normalized size = 0.72 \begin {gather*} \frac {\sqrt {f-i c f x} \left (4 i a-8 a c x+8 i a c^2 x^2-2 b \sqrt {1+c^2 x^2}+4 i b \left (1+2 i c x+2 c^2 x^2\right ) \sinh ^{-1}(c x)+5 b (1-i c x) \sqrt {1+c^2 x^2} \log (d (-1+i c x))+3 b \sqrt {1+c^2 x^2} \log (d+i c d x)-3 i b c x \sqrt {1+c^2 x^2} \log (d+i c d x)\right )}{12 c d f^3 (i+c x)^2 \sqrt {d+i c d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(5/2)),x]

[Out]

(Sqrt[f - I*c*f*x]*((4*I)*a - 8*a*c*x + (8*I)*a*c^2*x^2 - 2*b*Sqrt[1 + c^2*x^2] + (4*I)*b*(1 + (2*I)*c*x + 2*c
^2*x^2)*ArcSinh[c*x] + 5*b*(1 - I*c*x)*Sqrt[1 + c^2*x^2]*Log[d*(-1 + I*c*x)] + 3*b*Sqrt[1 + c^2*x^2]*Log[d + I
*c*d*x] - (3*I)*b*c*x*Sqrt[1 + c^2*x^2]*Log[d + I*c*d*x]))/(12*c*d*f^3*(I + c*x)^2*Sqrt[d + I*c*d*x])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {a +b \arcsinh \left (c x \right )}{\left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(5/2),x)

[Out]

int((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(5/2),x)

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Maxima [A]
time = 0.30, size = 237, normalized size = 0.84 \begin {gather*} \frac {1}{12} \, b c {\left (\frac {2 i \, \sqrt {d} \sqrt {f}}{c^{3} d^{2} f^{3} x + i \, c^{2} d^{2} f^{3}} - \frac {5 \, \log \left (c x + i\right )}{c^{2} d^{\frac {3}{2}} f^{\frac {5}{2}}} - \frac {3 \, \log \left (c x - i\right )}{c^{2} d^{\frac {3}{2}} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (\frac {3 i}{-3 i \, \sqrt {c^{2} d f x^{2} + d f} c^{2} d f^{2} x + 3 \, \sqrt {c^{2} d f x^{2} + d f} c d f^{2}} - \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d f^{2}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (\frac {3 i}{-3 i \, \sqrt {c^{2} d f x^{2} + d f} c^{2} d f^{2} x + 3 \, \sqrt {c^{2} d f x^{2} + d f} c d f^{2}} - \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d f^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(5/2),x, algorithm="maxima")

[Out]

1/12*b*c*(2*I*sqrt(d)*sqrt(f)/(c^3*d^2*f^3*x + I*c^2*d^2*f^3) - 5*log(c*x + I)/(c^2*d^(3/2)*f^(5/2)) - 3*log(c
*x - I)/(c^2*d^(3/2)*f^(5/2))) - 1/3*b*(3*I/(-3*I*sqrt(c^2*d*f*x^2 + d*f)*c^2*d*f^2*x + 3*sqrt(c^2*d*f*x^2 + d
*f)*c*d*f^2) - 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d*f^2))*arcsinh(c*x) - 1/3*a*(3*I/(-3*I*sqrt(c^2*d*f*x^2 + d*f)*c^
2*d*f^2*x + 3*sqrt(c^2*d*f*x^2 + d*f)*c*d*f^2) - 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d*f^2))

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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((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(5/2),x, algorithm="fricas")

[Out]

-1/24*(4*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x - 8*(2*b*c^2*x^2 + 2*I*b*c*x + b)*sqrt(I
*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) - 3*(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^2
*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*d^3*f^5))*log(-(I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c
*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) + I*b*c^2*x^3 + I*b*x)/(b*c^3*x^3 - I*b*c^2*x^2 + b*c*x - I*b)) + 5*(c^4*d^2*
f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^2*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*d^3*f^5))*log(-(I*sqrt(c^2*x^2 + 1)*s
qrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) - I*b*c^2*x^3 - I*b*x)/(b*c^3*x^3 + I*b*
c^2*x^2 + b*c*x + I*b)) + 3*(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^2*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*
d^3*f^5))*log(-(-I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) +
I*b*c^2*x^3 + I*b*x)/(b*c^3*x^3 - I*b*c^2*x^2 + b*c*x - I*b)) - 5*(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d
^2*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*d^3*f^5))*log(-(-I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f
)*c*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) - I*b*c^2*x^3 - I*b*x)/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + I*b)) + 8*(c^4*d
^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^2*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*d^3*f^5))*log((sqrt(c^2*x^2 + 1)*s
qrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) + b*c^2*x^3 + b*x)/(b*c^2*x^2 + b)) - 8*
(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^2*f^3*x + I*c*d^2*f^3)*sqrt(b^2/(c^2*d^3*f^5))*log(-(sqrt(c^2*x^2
+ 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f^2*x*sqrt(b^2/(c^2*d^3*f^5)) - b*c^2*x^3 - b*x)/(b*c^2*x^2 + b
)) - 8*(2*a*c^2*x^2 + 2*I*a*c*x + a)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - 24*(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^
3*x^2 + c^2*d^2*f^3*x + I*c*d^2*f^3)*integral(-1/6*sqrt(c^2*x^2 + 1)*(4*b*c*x - I*b)*sqrt(I*c*d*x + d)*sqrt(-I
*c*f*x + f)/(c^4*d^2*f^3*x^4 + 2*c^2*d^2*f^3*x^2 + d^2*f^3), x))/(c^4*d^2*f^3*x^3 + I*c^3*d^2*f^3*x^2 + c^2*d^
2*f^3*x + I*c*d^2*f^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(d+I*c*d*x)**(3/2)/(f-I*c*f*x)**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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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((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choi
ce was done

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))/((d + c*d*x*1i)^(3/2)*(f - c*f*x*1i)^(5/2)),x)

[Out]

int((a + b*asinh(c*x))/((d + c*d*x*1i)^(3/2)*(f - c*f*x*1i)^(5/2)), x)

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