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

Optimal. Leaf size=344 \[ \frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (c^2 x^2+1\right )}+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (c^2 x^2+1\right )^2}+\frac{5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \left (c^2 x^2+1\right )^{5/2}}+\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{5/2}}-\frac{25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{5/2}}-\frac{b \sqrt{c^2 x^2+1} (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{36 c} \]

[Out]

(-25*b*c*x^2*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (5*b*c^3*x^4*(d + I*c*d*x)^(5
/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (b*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*Sqrt[1 + c^2*x^
2])/(36*c) + (x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/6 + (5*x*(d + I*c*d*x)^(5/2)*(f
- I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/(16*(1 + c^2*x^2)^2) + (5*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a
+ b*ArcSinh[c*x]))/(24*(1 + c^2*x^2)) + (5*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(32
*b*c*(1 + c^2*x^2)^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.303869, antiderivative size = 344, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5712, 5684, 5682, 5675, 30, 14, 261} \[ \frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (c^2 x^2+1\right )}+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (c^2 x^2+1\right )^2}+\frac{5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \left (c^2 x^2+1\right )^{5/2}}+\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac{5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{5/2}}-\frac{25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{5/2}}-\frac{b \sqrt{c^2 x^2+1} (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{36 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-25*b*c*x^2*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (5*b*c^3*x^4*(d + I*c*d*x)^(5
/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (b*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*Sqrt[1 + c^2*x^
2])/(36*c) + (x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/6 + (5*x*(d + I*c*d*x)^(5/2)*(f
- I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/(16*(1 + c^2*x^2)^2) + (5*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a
+ b*ArcSinh[c*x]))/(24*(1 + c^2*x^2)) + (5*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(32
*b*c*(1 + c^2*x^2)^(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 5684

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(x*(d + e*x^2)^p*
(a + b*ArcSinh[c*x])^n)/(2*p + 1), x] + (Dist[(2*d*p)/(2*p + 1), Int[(d + e*x^2)^(p - 1)*(a + b*ArcSinh[c*x])^
n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/((2*p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[x*(1
+ c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && Gt
Q[n, 0] && GtQ[p, 0]

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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 261

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

Rubi steps

\begin{align*} \int (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left ((d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{\left (1+c^2 x^2\right )^{5/2}}\\ &=\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{6 \left (1+c^2 x^2\right )^{5/2}}-\frac{\left (b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \left (1+c^2 x^2\right )^2 \, dx}{6 \left (1+c^2 x^2\right )^{5/2}}\\ &=-\frac{b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt{1+c^2 x^2}}{36 c}+\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (1+c^2 x^2\right )}+\frac{\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{8 \left (1+c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \left (1+c^2 x^2\right ) \, dx}{24 \left (1+c^2 x^2\right )^{5/2}}\\ &=-\frac{b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt{1+c^2 x^2}}{36 c}+\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^2}+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (1+c^2 x^2\right )}+\frac{\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac{a+b \sinh ^{-1}(c x)}{\sqrt{1+c^2 x^2}} \, dx}{16 \left (1+c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (x+c^2 x^3\right ) \, dx}{24 \left (1+c^2 x^2\right )^{5/2}}-\frac{\left (5 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \, dx}{16 \left (1+c^2 x^2\right )^{5/2}}\\ &=-\frac{25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (1+c^2 x^2\right )^{5/2}}-\frac{5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (1+c^2 x^2\right )^{5/2}}-\frac{b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt{1+c^2 x^2}}{36 c}+\frac{1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^2}+\frac{5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (1+c^2 x^2\right )}+\frac{5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \left (1+c^2 x^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 1.15647, size = 481, normalized size = 1.4 \[ \frac{384 a c^5 d^2 f^2 x^5 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+1248 a c^3 d^2 f^2 x^3 \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+1584 a c d^2 f^2 x \sqrt{c^2 x^2+1} \sqrt{d+i c d x} \sqrt{f-i c f x}+720 a d^{5/2} f^{5/2} \sqrt{c^2 x^2+1} \log \left (c d f x+\sqrt{d} \sqrt{f} \sqrt{d+i c d x} \sqrt{f-i c f x}\right )+360 b d^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x)^2+12 b d^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \sinh ^{-1}(c x) \left (45 \sinh \left (2 \sinh ^{-1}(c x)\right )+9 \sinh \left (4 \sinh ^{-1}(c x)\right )+\sinh \left (6 \sinh ^{-1}(c x)\right )\right )-270 b d^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )-27 b d^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (4 \sinh ^{-1}(c x)\right )-2 b d^2 f^2 \sqrt{d+i c d x} \sqrt{f-i c f x} \cosh \left (6 \sinh ^{-1}(c x)\right )}{2304 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1584*a*c*d^2*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 1248*a*c^3*d^2*f^2*x^3*Sqrt[d + I*
c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 384*a*c^5*d^2*f^2*x^5*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1
+ c^2*x^2] + 360*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 - 270*b*d^2*f^2*Sqrt[d + I*c*d*x
]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] - 27*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[4*ArcSinh[c*x
]] - 2*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[6*ArcSinh[c*x]] + 720*a*d^(5/2)*f^(5/2)*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]] + 12*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]*ArcSinh[c*x]*(45*Sinh[2*ArcSinh[c*x]] + 9*Sinh[4*ArcSinh[c*x]] + Sinh[6*ArcSinh[c*x]]))/(2304*c*
Sqrt[1 + c^2*x^2])

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Maple [F]  time = 0.246, size = 0, normalized size = 0. \begin{align*} \int \left ( d+icdx \right ) ^{{\frac{5}{2}}} \left ( f-icfx \right ) ^{{\frac{5}{2}}} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b c^{4} d^{2} f^{2} x^{4} + 2 \, b c^{2} d^{2} f^{2} x^{2} + b d^{2} f^{2}\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 (a c^{4} d^{2} f^{2} x^{4} + 2 \, a c^{2} d^{2} f^{2} x^{2} + a d^{2} f^{2}\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*c^4*d^2*f^2*x^4 + 2*b*c^2*d^2*f^2*x^2 + b*d^2*f^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x +
sqrt(c^2*x^2 + 1)) + (a*c^4*d^2*f^2*x^4 + 2*a*c^2*d^2*f^2*x^2 + a*d^2*f^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f
), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError