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

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

[Out]

((I/6)*b*f*(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)) + (f*(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*f*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*f*(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*f*(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))

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Rubi [A]  time = 0.306403, antiderivative size = 282, normalized size of antiderivative = 1., 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 = {5712, 639, 191, 5819, 627, 44, 203, 260} \[ \frac{2 f x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{f (c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b f \left (c^2 x^2+1\right )^{5/2}}{6 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b f \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{i b f \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/6)*b*f*(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)) + (f*(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*f*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*f*(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*f*(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 5712

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 639

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*e - c*d*x)*(a + c*x^2)^(p + 1))/(2*a
*c*(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 191

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 5819

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])

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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 44

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] && L
tQ[m + n + 2, 0])

Rule 203

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

Rule 260

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]

Rubi steps

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

Mathematica [A]  time = 0.607269, size = 201, normalized size = 0.71 \[ \frac{\sqrt{f-i c f x} \left (8 i a c^2 x^2+8 a c x+4 i a-5 i b c x \sqrt{c^2 x^2+1} \log (d+i c d x)+3 b (-1-i c x) \sqrt{c^2 x^2+1} \log (d (-1+i c x))-5 b \sqrt{c^2 x^2+1} \log (d+i c d x)+2 b \sqrt{c^2 x^2+1}+4 b \left (2 i c^2 x^2+2 c x+i\right ) \sinh ^{-1}(c x)\right )}{12 d^2 f^2 \left (c^3 x^2+c\right ) \sqrt{d+i c d x}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.25, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{-{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(48*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x - 96*(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)) - 4*(9*c^4*d^3*f^2*x^3 - 9*I*c^3*d^3*f^2*x^2 + 9*
c^2*d^3*f^2*x - 9*I*c*d^3*f^2)*sqrt(b^2/(c^2*d^5*f^3))*log((4*I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*
f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5*f^3)) + 4*I*b*c^2*x^3 + 4*I*b*x)/(4*b*c^3*x^3 + 4*I*b*c^2*x^2 + 4*b*c*x +
4*I*b)) + 4*(15*c^4*d^3*f^2*x^3 - 15*I*c^3*d^3*f^2*x^2 + 15*c^2*d^3*f^2*x - 15*I*c*d^3*f^2)*sqrt(b^2/(c^2*d^5
*f^3))*log((4*I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5*f^3)) - 4*I
*b*c^2*x^3 - 4*I*b*x)/(4*b*c^3*x^3 - 4*I*b*c^2*x^2 + 4*b*c*x - 4*I*b)) + 4*(9*c^4*d^3*f^2*x^3 - 9*I*c^3*d^3*f^
2*x^2 + 9*c^2*d^3*f^2*x - 9*I*c*d^3*f^2)*sqrt(b^2/(c^2*d^5*f^3))*log((-4*I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)
*sqrt(-I*c*f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5*f^3)) + 4*I*b*c^2*x^3 + 4*I*b*x)/(4*b*c^3*x^3 + 4*I*b*c^2*x^2
+ 4*b*c*x + 4*I*b)) - 4*(15*c^4*d^3*f^2*x^3 - 15*I*c^3*d^3*f^2*x^2 + 15*c^2*d^3*f^2*x - 15*I*c*d^3*f^2)*sqrt(b
^2/(c^2*d^5*f^3))*log((-4*I*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5
*f^3)) - 4*I*b*c^2*x^3 - 4*I*b*x)/(4*b*c^3*x^3 - 4*I*b*c^2*x^2 + 4*b*c*x - 4*I*b)) + 4*(24*c^4*d^3*f^2*x^3 - 2
4*I*c^3*d^3*f^2*x^2 + 24*c^2*d^3*f^2*x - 24*I*c*d^3*f^2)*sqrt(b^2/(c^2*d^5*f^3))*log((sqrt(c^2*x^2 + 1)*sqrt(I
*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5*f^3)) + b*c^2*x^3 + b*x)/(b*c^2*x^2 + b)) - 4*(24*c
^4*d^3*f^2*x^3 - 24*I*c^3*d^3*f^2*x^2 + 24*c^2*d^3*f^2*x - 24*I*c*d^3*f^2)*sqrt(b^2/(c^2*d^5*f^3))*log(-(sqrt(
c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f*x*sqrt(b^2/(c^2*d^5*f^3)) - b*c^2*x^3 - b*x)/(b*c^2*
x^2 + b)) - 3*(64*a*c^2*x^2 - 64*I*a*c*x + 32*a)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - 3*(96*c^4*d^3*f^2*x^3
- 96*I*c^3*d^3*f^2*x^2 + 96*c^2*d^3*f^2*x - 96*I*c*d^3*f^2)*integral(-1/6*sqrt(c^2*x^2 + 1)*(4*b*c*x + I*b)*sq
rt(I*c*d*x + d)*sqrt(-I*c*f*x + f)/(c^4*d^3*f^2*x^4 + 2*c^2*d^3*f^2*x^2 + d^3*f^2), x))/(96*c^4*d^3*f^2*x^3 -
96*I*c^3*d^3*f^2*x^2 + 96*c^2*d^3*f^2*x - 96*I*c*d^3*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError