3.6.36 \(\int \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \sinh ^{-1}(c x)) \, dx\) [536]

Optimal. Leaf size=147 \[ -\frac {b c x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {1+c^2 x^2}}+\frac {1}{2} x \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {1+c^2 x^2}} \]

[Out]

1/2*x*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1/4*b*c*x^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(
c^2*x^2+1)^(1/2)+1/4*(a+b*arcsinh(c*x))^2*(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 = 0.12, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {5796, 5785, 5783, 30} \begin {gather*} \frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \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 )-\frac {b c x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {c^2 x^2+1}} \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]),x]

[Out]

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && 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 ) \, 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 ) \, 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 )+\frac {\left (\sqrt {d+i c d x} \sqrt {f-i c f x}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\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 \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=-\frac {b c x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \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 )+\frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {1+c^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.31, size = 233, normalized size = 1.59 \begin {gather*} \frac {1}{2} a x \sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)}+\frac {a \sqrt {d} \sqrt {f} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)}\right )}{2 c}-\frac {b \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left (\cosh \left (2 \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)+\sinh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{8 c \sqrt {-((-i d+c d x) (i f+c f x))} \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]),x]

[Out]

(a*x*Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)])/2 + (a*Sqrt[d]*Sqrt[f]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[I*
d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]])/(2*c) - (b*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f
*(1 + c^2*x^2))]*(Cosh[2*ArcSinh[c*x]] - 2*ArcSinh[c*x]*(ArcSinh[c*x] + Sinh[2*ArcSinh[c*x]])))/(8*c*Sqrt[-(((
-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*x^2])

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))*(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 = 0.00, 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))*(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))*(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*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(I*c*d*x + d)*sqrt(-I*c*f*x
+ f)*a, 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 )\, dx \end {gather*}

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(I*d*(c*x - I))*sqrt(-I*f*(c*x + I))*(a + b*asinh(c*x)), 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))*(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 = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\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))*(d + c*d*x*1i)^(1/2)*(f - c*f*x*1i)^(1/2),x)

[Out]

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

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