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

Optimal. Leaf size=180 \[ -\frac{d^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 i d^2 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 b d^2 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.399947, antiderivative size = 180, 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, 5833, 637, 5819, 12, 627, 31, 5675} \[ -\frac{d^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 i d^2 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{2 b d^2 \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5833

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(a + b*ArcSinh[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Fr
eeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]

Rule 637

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

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 31

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

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -1
]

Rubi steps

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

Mathematica [A]  time = 1.20919, size = 285, normalized size = 1.58 \[ \frac{\frac{4 a \sqrt{d+i c d x} \sqrt{f-i c f x}}{c x+i}-2 a \sqrt{d} \sqrt{f} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+\frac{b \sqrt{d+i c d x} \sqrt{f-i c f x} \left (2 \left (\sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+i \cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right ) \left (4 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (c^2 x^2+1\right )\right )+\sinh ^{-1}(c x)^2 \left (-\left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )+4 \sinh ^{-1}(c x) \left (\sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-i \cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\sqrt{c^2 x^2+1} \left (\cosh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-i \sinh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )}}{2 c f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(I + c*x) - 2*a*Sqrt[d]*Sqrt[f]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[
d + I*c*d*x]*Sqrt[f - I*c*f*x]] + (b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-(ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/
2] - I*Sinh[ArcSinh[c*x]/2])) + 4*ArcSinh[c*x]*((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2]) + 2*(4*ArcTa
n[Tanh[ArcSinh[c*x]/2]] + I*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2])))/(Sqrt[1 + c^2*
x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(2*c*f^2)

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Maple [F]  time = 0.299, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) )\sqrt{d+icdx} \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)^(1/2)/(f-I*c*f*x)^(3/2),x)

[Out]

int((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(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)^(1/2)/(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*}{\rm integral}\left (-\frac{\sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a}{c^{2} f^{2} x^{2} + 2 i \, c f^{2} x - f^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(3/2),x, algorithm="fricas")

[Out]

integral(-(sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(I*c*d*x + d)*sqrt(-I*c*f
*x + f)*a)/(c^2*f^2*x^2 + 2*I*c*f^2*x - f^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \left (i c x + 1\right )} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )}{\left (- f \left (i c x - 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))*(d+I*c*d*x)**(1/2)/(f-I*c*f*x)**(3/2),x)

[Out]

Integral(sqrt(d*(I*c*x + 1))*(a + b*asinh(c*x))/(-f*(I*c*x - 1))**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError