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

Optimal. Leaf size=158 \[ \frac{d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{i d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{i b d x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

[Out]

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

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Rubi [A]  time = 0.290386, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {5712, 5821, 5675, 5717, 8} \[ \frac{d \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{i d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{i b d x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.583145, size = 227, normalized size = 1.44 \[ \frac{-2 i \sqrt{d+i c d x} \sqrt{f-i c f x} \left (b c x-a \sqrt{c^2 x^2+1}\right )+2 a \sqrt{d} \sqrt{f} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+2 i b \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)+b \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2}{2 c f \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.298, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) )\sqrt{d+icdx}{\frac{1}{\sqrt{f-icfx}}}}\, 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)^(1/2),x)

[Out]

int((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(f-I*c*f*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((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(f-I*c*f*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{i \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + i \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a}{c f x + i \, f}, 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)^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

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

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

[Out]

Timed out