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

Optimal. Leaf size=259 \[ \frac{2 i a b f x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{f \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{2 i b^2 f \left (c^2 x^2+1\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{2 i b^2 f x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

[Out]

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

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Rubi [A]  time = 0.507903, antiderivative size = 259, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {5712, 5821, 5675, 5717, 5653, 261} \[ \frac{2 i a b f x \sqrt{c^2 x^2+1}}{\sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{f \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}-\frac{2 i b^2 f \left (c^2 x^2+1\right )}{c \sqrt{d+i c d x} \sqrt{f-i c f x}}+\frac{2 i b^2 f x \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{\sqrt{d+i c d x} \sqrt{f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a*b*f*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - ((2*I)*b^2*f*(1 + c^2*x^2))/(c*Sqrt[
d + I*c*d*x]*Sqrt[f - I*c*f*x]) + ((2*I)*b^2*f*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*Sqrt[f - I
*c*f*x]) - (I*f*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (f*Sqrt[1 + c^
2*x^2]*(a + b*ArcSinh[c*x])^3)/(3*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 5653

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

Rule 261

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

Rubi steps

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

Mathematica [A]  time = 1.14369, size = 315, normalized size = 1.22 \[ \frac{-3 i \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a^2 \sqrt{c^2 x^2+1}-2 a b c x+2 b^2 \sqrt{c^2 x^2+1}\right )+3 a^2 \sqrt{d} \sqrt{f} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+3 b \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2 \left (a-i b \sqrt{c^2 x^2+1}\right )+6 i b \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x) \left (b c x-a \sqrt{c^2 x^2+1}\right )+b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^3}{3 c d \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.331, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}\sqrt{f-icfx}{\frac{1}{\sqrt{d+icdx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Exception raised: RuntimeError