3.542 \(\int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\)

Optimal. Leaf size=304 \[ \frac {f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {c^2 x^2+1}}-\frac {i f \left (c^2 x^2+1\right ) \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac {b c f x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {c^2 x^2+1}}+\frac {i b f x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {c^2 x^2+1}}+\frac {i b c^2 f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {c^2 x^2+1}} \]

[Out]

1/2*f*x*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1/3*I*f*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(d+I*c*d
*x)^(1/2)*(f-I*c*f*x)^(1/2)/c+1/3*I*b*f*x*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)-1/4*b*c*f*x^2*
(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+1/9*I*b*c^2*f*x^3*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c
^2*x^2+1)^(1/2)+1/4*f*(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 = 0.34, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5712, 5821, 5682, 5675, 30, 5717} \[ \frac {f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {c^2 x^2+1}}-\frac {i f \left (c^2 x^2+1\right ) \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )+\frac {i b c^2 f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {c^2 x^2+1}}-\frac {b c f x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {c^2 x^2+1}}+\frac {i b f x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*b*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[1 + c^2*x^2] - (b*c*f*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*
c*f*x])/(4*Sqrt[1 + c^2*x^2]) + ((I/9)*b*c^2*f*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/Sqrt[1 + c^2*x^2] + (f
*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/2 - ((I/3)*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*
(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/c + (f*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 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 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 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 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 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]))

Rubi steps

\begin {align*} \int \sqrt {d+i c d x} (f-i c f x)^{3/2} \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 (f-i c f x) \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (\sqrt {d+i c d x} \sqrt {f-i c f x}\right ) \int \left (f \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )-i c f x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {\left (f \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 {\left (i c f \sqrt {d+i c d x} \sqrt {f-i c f x}\right ) \int x \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\sqrt {1+c^2 x^2}}\\ &=\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac {i f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac {\left (f \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 (i b f \sqrt {d+i c d x} \sqrt {f-i c f x}\right ) \int \left (1+c^2 x^2\right ) \, dx}{3 \sqrt {1+c^2 x^2}}-\frac {\left (b c f \sqrt {d+i c d x} \sqrt {f-i c f x}\right ) \int x \, dx}{2 \sqrt {1+c^2 x^2}}\\ &=\frac {i b f x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {1+c^2 x^2}}-\frac {b c f x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {1+c^2 x^2}}+\frac {i b c^2 f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {1+c^2 x^2}}+\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+b \sinh ^{-1}(c x)\right )-\frac {i f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c}+\frac {f \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 = 1.59, size = 273, normalized size = 0.90 \[ \frac {12 a f \left (-2 i c^2 x^2+3 c x-2 i\right ) \sqrt {d+i c d x} \sqrt {f-i c f x}+36 a \sqrt {d} f^{3/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+\frac {9 b f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \sinh ^{-1}(c x)^2+2 \sinh \left (2 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)-\cosh \left (2 \sinh ^{-1}(c x)\right )\right )}{\sqrt {c^2 x^2+1}}+\frac {2 i b f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-3 \sinh ^{-1}(c x) \left (3 \sqrt {c^2 x^2+1}+\cosh \left (3 \sinh ^{-1}(c x)\right )\right )+9 c x+\sinh \left (3 \sinh ^{-1}(c x)\right )\right )}{\sqrt {c^2 x^2+1}}}{72 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-2*I + 3*c*x - (2*I)*c^2*x^2) + 36*a*Sqrt[d]*f^(3/2)*Log[c*d*f*x
+ Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] + (9*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(2*ArcSinh
[c*x]^2 - Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]]))/Sqrt[1 + c^2*x^2] + ((2*I)*b*f*Sqrt[d +
I*c*d*x]*Sqrt[f - I*c*f*x]*(9*c*x - 3*ArcSinh[c*x]*(3*Sqrt[1 + c^2*x^2] + Cosh[3*ArcSinh[c*x]]) + Sinh[3*ArcS
inh[c*x]]))/Sqrt[1 + c^2*x^2])/(72*c)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-i \, b c f x + b f\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (-i \, a c f x + a f\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \left (-i c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right ) \sqrt {i c d x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))*(d+I*c*d*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

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)^(3/2),x)

[Out]

int((a + b*asinh(c*x))*(d + c*d*x*1i)^(1/2)*(f - c*f*x*1i)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i d \left (c x - i\right )} \left (- i f \left (c x + i\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f-I*c*f*x)**(3/2)*(a+b*asinh(c*x))*(d+I*c*d*x)**(1/2),x)

[Out]

Integral(sqrt(I*d*(c*x - I))*(-I*f*(c*x + I))**(3/2)*(a + b*asinh(c*x)), x)

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