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

Optimal. Leaf size=111 \[ \frac{f (c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{b f \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.241307, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {5712, 637, 5819, 12, 627, 31} \[ \frac{f (c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac{b f \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[p + 1/2, 0]
&& GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.397476, size = 113, normalized size = 1.02 \[ \frac{\sqrt{d+i c d x} \sqrt{f-i c f x} \left (a \sqrt{c^2 x^2+1}+b \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+b (-c x+i) \log (d+i c d x)\right )}{c d^2 f (c x-i) \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.255, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( d+icdx \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{f-icfx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31198, size = 1011, normalized size = 9.11 \begin{align*} \frac{2 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (c^{2} d^{2} f x - i \, c d^{2} f\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f}} \log \left (\frac{{\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} + 2 \,{\left (i \, c^{9} d^{2} f x^{4} + 2 \, c^{8} d^{2} f x^{3} + i \, c^{7} d^{2} f x^{2} + 2 \, c^{6} d^{2} f x\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f}}}{16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b}\right ) -{\left (c^{2} d^{2} f x - i \, c d^{2} f\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f}} \log \left (\frac{{\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} + 2 \,{\left (-i \, c^{9} d^{2} f x^{4} - 2 \, c^{8} d^{2} f x^{3} - i \, c^{7} d^{2} f x^{2} - 2 \, c^{6} d^{2} f x\right )} \sqrt{\frac{b^{2}}{c^{2} d^{3} f}}}{16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b}\right ) + 2 \, \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} a}{2 \,{\left (c^{2} d^{2} f x - i \, c d^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*log(c*x + sqrt(c^2*x^2 + 1)) + (c^2*d^2*f*x - I*c*d^2*f)*sqrt(b^
2/(c^2*d^3*f))*log(((-2*I*b*c^6*x^2 - 4*b*c^5*x + 4*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x
+ f) + 2*(I*c^9*d^2*f*x^4 + 2*c^8*d^2*f*x^3 + I*c^7*d^2*f*x^2 + 2*c^6*d^2*f*x)*sqrt(b^2/(c^2*d^3*f)))/(16*b*c
^3*x^3 - 16*I*b*c^2*x^2 + 16*b*c*x - 16*I*b)) - (c^2*d^2*f*x - I*c*d^2*f)*sqrt(b^2/(c^2*d^3*f))*log(((-2*I*b*c
^6*x^2 - 4*b*c^5*x + 4*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) + 2*(-I*c^9*d^2*f*x^4 -
2*c^8*d^2*f*x^3 - I*c^7*d^2*f*x^2 - 2*c^6*d^2*f*x)*sqrt(b^2/(c^2*d^3*f)))/(16*b*c^3*x^3 - 16*I*b*c^2*x^2 + 16
*b*c*x - 16*I*b)) + 2*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a)/(c^2*d^2*f*x - I*c*d^2*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asinh}{\left (c x \right )}}{\left (d \left (i c x + 1\right )\right )^{\frac{3}{2}} \sqrt{- f \left (i c x - 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError