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

Optimal. Leaf size=266 \[ \frac{3 f^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{f^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{2 i f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{b c f^2 x^2 \sqrt{c^2 x^2+1}}{4 \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{2 i b f^2 x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

[Out]

((2*I)*b*f^2*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (b*c*f^2*x^2*Sqrt[1 + c^2*x^2])/(4*S
qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - ((2*I)*f^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]) - (f^2*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (3*f^2*S
qrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

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Rubi [A]  time = 0.464772, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5712, 5821, 5675, 5717, 8, 5758, 30} \[ \frac{3 f^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{f^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{2 i f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{b c f^2 x^2 \sqrt{c^2 x^2+1}}{4 \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{2 i b f^2 x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*b*f^2*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (b*c*f^2*x^2*Sqrt[1 + c^2*x^2])/(4*S
qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - ((2*I)*f^2*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]) - (f^2*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (3*f^2*S
qrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(4*b*c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

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 5821

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

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
]

Rule 5717

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 5758

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{(f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{d+i c d x}} \, dx &=\frac{\sqrt{1+c^2 x^2} \int \frac{(f-i c f x)^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}\\ &=\frac{\sqrt{1+c^2 x^2} \int \left (\frac{f^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}-\frac{2 i c f^2 x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}-\frac{c^2 f^2 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}}\right ) \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}\\ &=\frac{\left (f^2 \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{\left (2 i c f^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{\left (c^2 f^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt{1+c^2 x^2}} \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}\\ &=-\frac{2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{f^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{\left (f^2 \sqrt{1+c^2 x^2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{\left (2 i b f^2 \sqrt{1+c^2 x^2}\right ) \int 1 \, dx}{\sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{\left (b c f^2 \sqrt{1+c^2 x^2}\right ) \int x \, dx}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}\\ &=\frac{2 i b f^2 x \sqrt{1+c^2 x^2}}{\sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{b c f^2 x^2 \sqrt{1+c^2 x^2}}{4 \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{3 f^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}\\ \end{align*}

Mathematica [A]  time = 1.05654, size = 344, normalized size = 1.29 \[ \frac{12 a \sqrt{d} f^{3/2} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )-16 i a f \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-4 a c f x \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}-4 b f (c x+4 i) \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)+16 i b c f x \sqrt{d+i c d x} \sqrt{f-i c f x}+6 b f \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2+b f \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )}{8 c d \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*I)*b*c*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (16*I)*a*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 +
c^2*x^2] - 4*a*c*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 4*b*f*(4*I + c*x)*Sqrt[d + I*c*d*
x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 6*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2
+ b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + 12*a*Sqrt[d]*f^(3/2)*Sqrt[1 + c^2*x^2]*Log[c
*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/(8*c*d*Sqrt[1 + c^2*x^2])

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Maple [F]  time = 0.284, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( f-icfx \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{d+icdx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),x)

[Out]

int((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),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((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),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{{\left (b c f x + i \, b f\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (a c f x + i \, a f\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}}{c d x - i \, d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Exception raised: RuntimeError