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

Optimal. Leaf size=114 \[ -\frac{a+b \sinh ^{-1}(c x)}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b x}{6 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 c^2 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.0782282, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {5717, 199, 203} \[ -\frac{a+b \sinh ^{-1}(c x)}{3 c^2 d \left (c^2 d x^2+d\right )^{3/2}}+\frac{b x}{6 c d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}+\frac{b \sqrt{c^2 x^2+1} \tan ^{-1}(c x)}{6 c^2 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5717

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[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 199

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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])

Rubi steps

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

Mathematica [A]  time = 0.183021, size = 130, normalized size = 1.14 \[ \frac{\sqrt{c^2 d x^2+d} \left (-2 a \sqrt{c^2 x^2+1}+b c^3 x^3-2 b \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+b c x\right )}{6 c^2 d^3 \left (c^2 x^2+1\right )^{5/2}}+\frac{b \sqrt{d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{6 c^2 d^3 \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.114, size = 198, normalized size = 1.7 \begin{align*} -{\frac{a}{3\,{c}^{2}d} \left ({c}^{2}d{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}+{\frac{bx}{6\,{d}^{3}c}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) } \left ({c}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{b{\it Arcsinh} \left ( cx \right ) }{3\,{d}^{3} \left ({c}^{2}{x}^{2}+1 \right ) ^{2}{c}^{2}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }}+{\frac{{\frac{i}{6}}b}{{c}^{2}{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}+i \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}}-{\frac{{\frac{i}{6}}b}{{c}^{2}{d}^{3}}\sqrt{d \left ({c}^{2}{x}^{2}+1 \right ) }\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}+1}-i \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a/c^2/d/(c^2*d*x^2+d)^(3/2)+1/6*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c*x-1/3*b*(d*(c^2*x^2+1))^(
1/2)/d^3/(c^2*x^2+1)^2/c^2*arcsinh(c*x)+1/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*ln(c*x+(c^2*x^
2+1)^(1/2)+I)-1/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 3.12803, size = 370, normalized size = 3.25 \begin{align*} -\frac{{\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \sqrt{d} \arctan \left (\frac{2 \, \sqrt{c^{2} d x^{2} + d} \sqrt{c^{2} x^{2} + 1} c \sqrt{d} x}{c^{4} d x^{4} - d}\right ) + 4 \, \sqrt{c^{2} d x^{2} + d} b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) - 2 \, \sqrt{c^{2} d x^{2} + d}{\left (\sqrt{c^{2} x^{2} + 1} b c x - 2 \, a\right )}}{12 \,{\left (c^{6} d^{3} x^{4} + 2 \, c^{4} d^{3} x^{2} + c^{2} d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((b*c^4*x^4 + 2*b*c^2*x^2 + b)*sqrt(d)*arctan(2*sqrt(c^2*d*x^2 + d)*sqrt(c^2*x^2 + 1)*c*sqrt(d)*x/(c^4*d
*x^4 - d)) + 4*sqrt(c^2*d*x^2 + d)*b*log(c*x + sqrt(c^2*x^2 + 1)) - 2*sqrt(c^2*d*x^2 + d)*(sqrt(c^2*x^2 + 1)*b
*c*x - 2*a))/(c^6*d^3*x^4 + 2*c^4*d^3*x^2 + c^2*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Integral(x*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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