3.6.54 \(\int \frac {\sqrt {d+i c d x} (a+b \sinh ^{-1}(c x))}{\sqrt {f-i c f x}} \, dx\) [554]

Optimal. Leaf size=158 \[ -\frac {i b d x \sqrt {1+c^2 x^2}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i d \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {d \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

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

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Rubi [A]
time = 0.19, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5796, 5838, 5783, 5798, 8} \begin {gather*} \frac {d \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i d \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i b d x \sqrt {c^2 x^2+1}}{\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]*(a + b*ArcSinh[c*x]))/Sqrt[f - I*c*f*x],x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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]

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5838

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

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Mathematica [A]
time = 0.30, size = 227, normalized size = 1.44 \begin {gather*} \frac {-2 i \sqrt {d+i c d x} \sqrt {f-i c f x} \left (b c x-a \sqrt {1+c^2 x^2}\right )+2 i b \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+b \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2+2 a \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 )}{2 c f \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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="maxima")

[Out]

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

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i d \left (c x - i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\sqrt {- i f \left (c x + i\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))*(a + b*asinh(c*x))/sqrt(-I*f*(c*x + I)), x)

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Giac [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="giac")

[Out]

integrate(sqrt(I*c*d*x + d)*(b*arcsinh(c*x) + a)/sqrt(-I*c*f*x + f), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\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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