3.650 \(\int \frac{a+b \sinh ^{-1}(c x)}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2+1}}{c \sqrt{d+e x^2}}\right )}{3 d^2 \sqrt{e}}-\frac{b c \sqrt{c^2 x^2+1}}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(3*d*(c^2*d - e)*Sqrt[d + e*x^2]) + (x*(a + b*ArcSinh[c*x]))/(3*d*(d + e*x^2)^(3/2))
+ (2*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[d + e*x^2]) - (2*b*ArcTanh[(Sqrt[e]*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + e*
x^2])])/(3*d^2*Sqrt[e])

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Rubi [A]  time = 0.168782, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {192, 191, 5704, 12, 571, 78, 63, 217, 206} \[ \frac{2 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}-\frac{2 b \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{c^2 x^2+1}}{c \sqrt{d+e x^2}}\right )}{3 d^2 \sqrt{e}}-\frac{b c \sqrt{c^2 x^2+1}}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*Sqrt[1 + c^2*x^2])/(3*d*(c^2*d - e)*Sqrt[d + e*x^2]) + (x*(a + b*ArcSinh[c*x]))/(3*d*(d + e*x^2)^(3/2))
+ (2*x*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[d + e*x^2]) - (2*b*ArcTanh[(Sqrt[e]*Sqrt[1 + c^2*x^2])/(c*Sqrt[d + e*
x^2])])/(3*d^2*Sqrt[e])

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[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 5704

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 + c^2*x^2], x], x], x]] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[e, c^2*d] && (IGtQ[p, 0] || ILtQ[p + 1/2, 0])

Rule 12

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

Rule 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 78

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

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

Rubi steps

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

Mathematica [C]  time = 0.335032, size = 139, normalized size = 0.95 \[ \frac{a x \left (3 d+2 e x^2\right )-b c x^2 \left (d+e x^2\right ) \sqrt{\frac{e x^2}{d}+1} F_1\left (1;\frac{1}{2},\frac{1}{2};2;-c^2 x^2,-\frac{e x^2}{d}\right )-\frac{b c d \sqrt{c^2 x^2+1} \left (d+e x^2\right )}{c^2 d-e}+b x \sinh ^{-1}(c x) \left (3 d+2 e x^2\right )}{3 d^2 \left (d+e x^2\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-((b*c*d*Sqrt[1 + c^2*x^2]*(d + e*x^2))/(c^2*d - e)) + a*x*(3*d + 2*e*x^2) - b*c*x^2*(d + e*x^2)*Sqrt[1 + (e*
x^2)/d]*AppellF1[1, 1/2, 1/2, 2, -(c^2*x^2), -((e*x^2)/d)] + b*x*(3*d + 2*e*x^2)*ArcSinh[c*x])/(3*d^2*(d + e*x
^2)^(3/2))

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Maple [F]  time = 0.325, size = 0, normalized size = 0. \begin{align*} \int{(a+b{\it Arcsinh} \left ( cx \right ) ) \left ( e{x}^{2}+d \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{e x^{2} + d} d^{2}} + \frac{x}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} d}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{{\left (e x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*(2*x/(sqrt(e*x^2 + d)*d^2) + x/((e*x^2 + d)^(3/2)*d)) + b*integrate(log(c*x + sqrt(c^2*x^2 + 1))/(e*x^2
+ d)^(5/2), x)

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Fricas [B]  time = 3.29745, size = 1503, normalized size = 10.29 \begin{align*} \left [\frac{{\left (b c^{2} d^{3} +{\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \,{\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt{e} \log \left (8 \, c^{4} e^{2} x^{4} + c^{4} d^{2} + 6 \, c^{2} d e + 8 \,{\left (c^{4} d e + c^{2} e^{2}\right )} x^{2} - 4 \,{\left (2 \, c^{3} e x^{2} + c^{3} d + c e\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{e x^{2} + d} \sqrt{e} + e^{2}\right ) + 2 \,{\left (2 \,{\left (b c^{2} d e^{2} - b e^{3}\right )} x^{3} + 3 \,{\left (b c^{2} d^{2} e - b d e^{2}\right )} x\right )} \sqrt{e x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (2 \,{\left (a c^{2} d e^{2} - a e^{3}\right )} x^{3} + 3 \,{\left (a c^{2} d^{2} e - a d e^{2}\right )} x -{\left (b c d e^{2} x^{2} + b c d^{2} e\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{e x^{2} + d}}{6 \,{\left (c^{2} d^{5} e - d^{4} e^{2} +{\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \,{\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}, \frac{{\left (b c^{2} d^{3} +{\left (b c^{2} d e^{2} - b e^{3}\right )} x^{4} - b d^{2} e + 2 \,{\left (b c^{2} d^{2} e - b d e^{2}\right )} x^{2}\right )} \sqrt{-e} \arctan \left (\frac{{\left (2 \, c^{2} e x^{2} + c^{2} d + e\right )} \sqrt{c^{2} x^{2} + 1} \sqrt{e x^{2} + d} \sqrt{-e}}{2 \,{\left (c^{3} e^{2} x^{4} + c d e +{\left (c^{3} d e + c e^{2}\right )} x^{2}\right )}}\right ) +{\left (2 \,{\left (b c^{2} d e^{2} - b e^{3}\right )} x^{3} + 3 \,{\left (b c^{2} d^{2} e - b d e^{2}\right )} x\right )} \sqrt{e x^{2} + d} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) +{\left (2 \,{\left (a c^{2} d e^{2} - a e^{3}\right )} x^{3} + 3 \,{\left (a c^{2} d^{2} e - a d e^{2}\right )} x -{\left (b c d e^{2} x^{2} + b c d^{2} e\right )} \sqrt{c^{2} x^{2} + 1}\right )} \sqrt{e x^{2} + d}}{3 \,{\left (c^{2} d^{5} e - d^{4} e^{2} +{\left (c^{2} d^{3} e^{3} - d^{2} e^{4}\right )} x^{4} + 2 \,{\left (c^{2} d^{4} e^{2} - d^{3} e^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*((b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(e)*log(8*c^4*e^2*
x^4 + c^4*d^2 + 6*c^2*d*e + 8*(c^4*d*e + c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d + c*e)*sqrt(c^2*x^2 + 1)*sqrt(e
*x^2 + d)*sqrt(e) + e^2) + 2*(2*(b*c^2*d*e^2 - b*e^3)*x^3 + 3*(b*c^2*d^2*e - b*d*e^2)*x)*sqrt(e*x^2 + d)*log(c
*x + sqrt(c^2*x^2 + 1)) + 2*(2*(a*c^2*d*e^2 - a*e^3)*x^3 + 3*(a*c^2*d^2*e - a*d*e^2)*x - (b*c*d*e^2*x^2 + b*c*
d^2*e)*sqrt(c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^2*d^5*e - d^4*e^2 + (c^2*d^3*e^3 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2
- d^3*e^3)*x^2), 1/3*((b*c^2*d^3 + (b*c^2*d*e^2 - b*e^3)*x^4 - b*d^2*e + 2*(b*c^2*d^2*e - b*d*e^2)*x^2)*sqrt(
-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d + e)*sqrt(c^2*x^2 + 1)*sqrt(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 + c*d*e + (c^
3*d*e + c*e^2)*x^2)) + (2*(b*c^2*d*e^2 - b*e^3)*x^3 + 3*(b*c^2*d^2*e - b*d*e^2)*x)*sqrt(e*x^2 + d)*log(c*x + s
qrt(c^2*x^2 + 1)) + (2*(a*c^2*d*e^2 - a*e^3)*x^3 + 3*(a*c^2*d^2*e - a*d*e^2)*x - (b*c*d*e^2*x^2 + b*c*d^2*e)*s
qrt(c^2*x^2 + 1))*sqrt(e*x^2 + d))/(c^2*d^5*e - d^4*e^2 + (c^2*d^3*e^3 - d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 - d^3*e
^3)*x^2)]

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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))/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \operatorname{arsinh}\left (c x\right ) + a}{{\left (e x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arcsinh(c*x) + a)/(e*x^2 + d)^(5/2), x)