3.6.72 \(\int \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \sinh ^{-1}(c x))^2 \, dx\) [572]

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

[Out]

1/4*b^2*x*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)+1/2*x*(a+b*arcsinh(c*x))^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1
/4*b^2*arcsinh(c*x)*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/c/(c^2*x^2+1)^(1/2)-1/2*b*c*x^2*(a+b*arcsinh(c*x))*(d+
I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+1/6*(a+b*arcsinh(c*x))^3*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2
)/b/c/(c^2*x^2+1)^(1/2)

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Rubi [A]
time = 0.22, antiderivative size = 244, normalized size of antiderivative = 1.00, 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 = {5796, 5785, 5783, 5776, 327, 221} \begin {gather*} \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} \end {gather*}

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 221

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]

Rule 327

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 5776

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 5783

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

Rule 5785

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[(1/2)*Simp[Sqrt[d + e*x^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/2)*Simp[Sqrt[d + e*x^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 5796

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]

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*}

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Mathematica [A]
time = 0.65, size = 352, normalized size = 1.44 \begin {gather*} \frac {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} \sinh ^{-1}(c x)^3-6 a b \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )+12 a^2 \sqrt {d} \sqrt {f} \sqrt {1+c^2 x^2} \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^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (2 \sinh ^{-1}(c x)\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 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 )}{24 c \sqrt {1+c^2 x^2}} \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \left (a +b \arcsinh \left (c x \right )\right )^{2} \sqrt {i c d x +d}\, \sqrt {-i c f x +f}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {i d \left (c x - i\right )} \sqrt {- i f \left (c x + i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}\, dx \end {gather*}

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]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument TypeError: Bad Argument TypeDone

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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