3.567 \(\int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{d+i c d x} (f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=294 \[ \frac{d^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^2 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b d^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b d^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

((I/3)*b*d^2*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((2*I)/3)*d^2*(1 +
I*c*x)*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (d^2*x*(1 + c^2*x^2)^
2*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((I/3)*b*d^2*(1 + c^2*x^2)^(5/2)*ArcTan[
c*x])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*d^2*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(6*c*(d + I*c
*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

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Rubi [A]  time = 0.332912, antiderivative size = 294, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {5712, 653, 191, 5819, 627, 44, 203, 260} \[ \frac{d^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^2 (1+i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b d^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b d^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(5/2)),x]

[Out]

((I/3)*b*d^2*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (((2*I)/3)*d^2*(1 +
I*c*x)*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (d^2*x*(1 + c^2*x^2)^
2*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + ((I/3)*b*d^2*(1 + c^2*x^2)^(5/2)*ArcTan[
c*x])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*d^2*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(6*c*(d + I*c
*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

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 653

Int[((d_) + (e_.)*(x_))^2*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)*(a + c*x^2)^(p + 1))/(c*(
p + 1)), x] - Dist[(e^2*(p + 2))/(c*(p + 1)), Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, p}, x] &&
EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && LtQ[p, -1]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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 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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

Rubi steps

\begin{align*} \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{d+i c d x} (f-i c f x)^{5/2}} \, dx &=\frac{\left (1+c^2 x^2\right )^{5/2} \int \frac{(d+i c d x)^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac{2 i d^2 (1+i c x)}{3 c \left (1+c^2 x^2\right )^2}+\frac{d^2 x}{3 \left (1+c^2 x^2\right )}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1+i c x}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{\left (b c d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{x}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{(1-i c x)^2 (1+i c x)} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (2 i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac{1}{2 (i+c x)^2}+\frac{1}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{i b d^2 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{\left (i b d^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac{i b d^2 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{d^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i b d^2 \left (1+c^2 x^2\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{b d^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.438501, size = 139, normalized size = 0.47 \[ \frac{\sqrt{f-i c f x} \left ((c x+2 i) \left (i a c x+a+i b \sqrt{c^2 x^2+1}\right )+b (1-i c x) \sqrt{c^2 x^2+1} \log (d (-1+i c x))+i b \left (c^2 x^2+i c x+2\right ) \sinh ^{-1}(c x)\right )}{3 c f^3 (c x+i)^2 \sqrt{d+i c d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*(f - I*c*f*x)^(5/2)),x]

[Out]

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

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Maple [F]  time = 0.249, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( f-icfx \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{d+icdx}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2)/(d+I*c*d*x)^(1/2),x)

[Out]

int((a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2)/(d+I*c*d*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))/(f-I*c*f*x)^(5/2)/(d+I*c*d*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.92681, size = 1384, normalized size = 4.71 \begin{align*} -\frac{12 \, \sqrt{c^{2} x^{2} + 1} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} b c x - 12 \,{\left (b c^{2} x^{2} + i \, b c x + 2 \, b\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (3 \, c^{4} d f^{3} x^{3} + 3 i \, c^{3} d f^{3} x^{2} + 3 \, c^{2} d f^{3} x + 3 i \, c d f^{3}\right )} \sqrt{\frac{b^{2}}{c^{2} d f^{5}}} \log \left (\frac{3 \,{\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 (3 i \, c^{9} d f^{3} x^{4} - 6 \, c^{8} d f^{3} x^{3} + 3 i \, c^{7} d f^{3} x^{2} - 6 \, c^{6} d f^{3} x\right )} \sqrt{\frac{b^{2}}{c^{2} d f^{5}}}}{3 \,{\left (16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b\right )}}\right ) - 2 \,{\left (3 \, c^{4} d f^{3} x^{3} + 3 i \, c^{3} d f^{3} x^{2} + 3 \, c^{2} d f^{3} x + 3 i \, c d f^{3}\right )} \sqrt{\frac{b^{2}}{c^{2} d f^{5}}} \log \left (\frac{3 \,{\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 (-3 i \, c^{9} d f^{3} x^{4} + 6 \, c^{8} d f^{3} x^{3} - 3 i \, c^{7} d f^{3} x^{2} + 6 \, c^{6} d f^{3} x\right )} \sqrt{\frac{b^{2}}{c^{2} d f^{5}}}}{3 \,{\left (16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b\right )}}\right ) - 3 \,{\left (4 \, a c^{2} x^{2} + 4 i \, a c x + 8 \, a\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}}{3 \,{\left (12 \, c^{4} d f^{3} x^{3} + 12 i \, c^{3} d f^{3} x^{2} + 12 \, c^{2} d f^{3} x + 12 i \, c d f^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(f-I*c*f*x)^(5/2)/(d+I*c*d*x)^(1/2),x, algorithm="fricas")

[Out]

-1/3*(12*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x - 12*(b*c^2*x^2 + I*b*c*x + 2*b)*sqrt(I*
c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + 2*(3*c^4*d*f^3*x^3 + 3*I*c^3*d*f^3*x^2 + 3*c^2*d*
f^3*x + 3*I*c*d*f^3)*sqrt(b^2/(c^2*d*f^5))*log(1/3*(3*(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*(3*I*c^9*d*f^3*x^4 - 6*c^8*d*f^3*x^3 + 3*I*c^7*d*f^3*x^2 - 6*c^6*d*
f^3*x)*sqrt(b^2/(c^2*d*f^5)))/(16*b*c^3*x^3 + 16*I*b*c^2*x^2 + 16*b*c*x + 16*I*b)) - 2*(3*c^4*d*f^3*x^3 + 3*I*
c^3*d*f^3*x^2 + 3*c^2*d*f^3*x + 3*I*c*d*f^3)*sqrt(b^2/(c^2*d*f^5))*log(1/3*(3*(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*(-3*I*c^9*d*f^3*x^4 + 6*c^8*d*f^3*x^3 - 3*I
*c^7*d*f^3*x^2 + 6*c^6*d*f^3*x)*sqrt(b^2/(c^2*d*f^5)))/(16*b*c^3*x^3 + 16*I*b*c^2*x^2 + 16*b*c*x + 16*I*b)) -
3*(4*a*c^2*x^2 + 4*I*a*c*x + 8*a)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))/(12*c^4*d*f^3*x^3 + 12*I*c^3*d*f^3*x^2
+ 12*c^2*d*f^3*x + 12*I*c*d*f^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(f-I*c*f*x)**(5/2)/(d+I*c*d*x)**(1/2),x)

[Out]

Timed out

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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))/(f-I*c*f*x)^(5/2)/(d+I*c*d*x)^(1/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError