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

Optimal. Leaf size=244 \[ \frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{c^2 x^2+1}}-\frac{b c x^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{c^2 x^2+1}}+\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)}{4 c \sqrt{c^2 x^2+1}}+\frac{1}{4} b^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \]

[Out]

(b^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/4 - (b^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x])/(4*c*Sq
rt[1 + c^2*x^2]) - (b*c*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/(2*Sqrt[1 + c^2*x^2]) +
(x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/2 + (Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b
*ArcSinh[c*x])^3)/(6*b*c*Sqrt[1 + c^2*x^2])

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Rubi [A]  time = 0.352603, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {5712, 5682, 5675, 5661, 321, 215} \[ \frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{c^2 x^2+1}}-\frac{b c x^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{c^2 x^2+1}}+\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2-\frac{b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)}{4 c \sqrt{c^2 x^2+1}}+\frac{1}{4} b^2 x \sqrt{d+i c d x} \sqrt{f-i c f x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/4 - (b^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x])/(4*c*Sq
rt[1 + c^2*x^2]) - (b*c*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/(2*Sqrt[1 + c^2*x^2]) +
(x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/2 + (Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b
*ArcSinh[c*x])^3)/(6*b*c*Sqrt[1 + c^2*x^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 5682

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

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 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

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

Rubi steps

\begin{align*} \int \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac{\left (\sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx}{\sqrt{1+c^2 x^2}}\\ &=\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\left (\sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}-\frac{\left (b c \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt{1+c^2 x^2}}\\ &=-\frac{b c x^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}+\frac{\left (b^2 c^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{2 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}-\frac{b c x^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}-\frac{\left (b^2 \sqrt{d+i c d x} \sqrt{f-i c f x}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{4 \sqrt{1+c^2 x^2}}\\ &=\frac{1}{4} b^2 x \sqrt{d+i c d x} \sqrt{f-i c f x}-\frac{b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)}{4 c \sqrt{1+c^2 x^2}}-\frac{b c x^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt{1+c^2 x^2}}+\frac{1}{2} x \sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^3}{6 b c \sqrt{1+c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 1.10848, size = 352, normalized size = 1.44 \[ \frac{12 a^2 c x \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+12 a^2 \sqrt{d} \sqrt{f} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+6 b \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2 \left (2 a+b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )-6 b \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x) \left (b \cosh \left (2 \sinh ^{-1}(c x)\right )-2 a \sinh \left (2 \sinh ^{-1}(c x)\right )\right )-6 a b \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )+4 b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^3+3 b^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh \left (2 \sinh ^{-1}(c x)\right )}{24 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a^2*c*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 4*b^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*
ArcSinh[c*x]^3 - 6*a*b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + 12*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*b^2*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]*Sinh[2*ArcSinh[c*x]] - 6*b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]*(b*Cosh[2*ArcSinh[c*
x]] - 2*a*Sinh[2*ArcSinh[c*x]]) + 6*b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2*(2*a + b*Sinh[2*ArcSi
nh[c*x]]))/(24*c*Sqrt[1 + c^2*x^2])

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Maple [F]  time = 0.286, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}\sqrt{d+icdx}\sqrt{f-icfx}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(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*sqrt(I*c*d*x + d)*sqrt(-I
*c*f*x + f)*a*b*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2, 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((a+b*asinh(c*x))**2*(d+I*c*d*x)**(1/2)*(f-I*c*f*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: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError