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

Optimal. Leaf size=93 \[ \frac{\left (\pi c^2 x^2+\pi \right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 \pi c^2}-\frac{1}{49} \pi ^{5/2} b c^5 x^7-\frac{3}{35} \pi ^{5/2} b c^3 x^5-\frac{1}{7} \pi ^{5/2} b c x^3-\frac{\pi ^{5/2} b x}{7 c} \]

[Out]

-(b*Pi^(5/2)*x)/(7*c) - (b*c*Pi^(5/2)*x^3)/7 - (3*b*c^3*Pi^(5/2)*x^5)/35 - (b*c^5*Pi^(5/2)*x^7)/49 + ((Pi + c^
2*Pi*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^2*Pi)

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Rubi [B]  time = 0.0868338, antiderivative size = 193, normalized size of antiderivative = 2.08, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {5717, 194} \[ \frac{\left (\pi c^2 x^2+\pi \right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 \pi c^2}-\frac{\pi ^2 b c^5 x^7 \sqrt{\pi c^2 x^2+\pi }}{49 \sqrt{c^2 x^2+1}}-\frac{3 \pi ^2 b c^3 x^5 \sqrt{\pi c^2 x^2+\pi }}{35 \sqrt{c^2 x^2+1}}-\frac{\pi ^2 b c x^3 \sqrt{\pi c^2 x^2+\pi }}{7 \sqrt{c^2 x^2+1}}-\frac{\pi ^2 b x \sqrt{\pi c^2 x^2+\pi }}{7 c \sqrt{c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Pi^2*x*Sqrt[Pi + c^2*Pi*x^2])/(7*c*Sqrt[1 + c^2*x^2]) - (b*c*Pi^2*x^3*Sqrt[Pi + c^2*Pi*x^2])/(7*Sqrt[1 + c
^2*x^2]) - (3*b*c^3*Pi^2*x^5*Sqrt[Pi + c^2*Pi*x^2])/(35*Sqrt[1 + c^2*x^2]) - (b*c^5*Pi^2*x^7*Sqrt[Pi + c^2*Pi*
x^2])/(49*Sqrt[1 + c^2*x^2]) + ((Pi + c^2*Pi*x^2)^(7/2)*(a + b*ArcSinh[c*x]))/(7*c^2*Pi)

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 194

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

Rubi steps

\begin{align*} \int x \left (\pi +c^2 \pi x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \, dx &=\frac{\left (\pi +c^2 \pi x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 \pi }-\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+c^2 x^2\right )^3 \, dx}{7 c \sqrt{1+c^2 x^2}}\\ &=\frac{\left (\pi +c^2 \pi x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 \pi }-\frac{\left (b \pi ^2 \sqrt{\pi +c^2 \pi x^2}\right ) \int \left (1+3 c^2 x^2+3 c^4 x^4+c^6 x^6\right ) \, dx}{7 c \sqrt{1+c^2 x^2}}\\ &=-\frac{b \pi ^2 x \sqrt{\pi +c^2 \pi x^2}}{7 c \sqrt{1+c^2 x^2}}-\frac{b c \pi ^2 x^3 \sqrt{\pi +c^2 \pi x^2}}{7 \sqrt{1+c^2 x^2}}-\frac{3 b c^3 \pi ^2 x^5 \sqrt{\pi +c^2 \pi x^2}}{35 \sqrt{1+c^2 x^2}}-\frac{b c^5 \pi ^2 x^7 \sqrt{\pi +c^2 \pi x^2}}{49 \sqrt{1+c^2 x^2}}+\frac{\left (\pi +c^2 \pi x^2\right )^{7/2} \left (a+b \sinh ^{-1}(c x)\right )}{7 c^2 \pi }\\ \end{align*}

Mathematica [A]  time = 0.144706, size = 80, normalized size = 0.86 \[ \frac{\pi ^{5/2} \left (35 a \left (c^2 x^2+1\right )^{7/2}-b c x \left (5 c^6 x^6+21 c^4 x^4+35 c^2 x^2+35\right )+35 b \left (c^2 x^2+1\right )^{7/2} \sinh ^{-1}(c x)\right )}{245 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Pi^(5/2)*(35*a*(1 + c^2*x^2)^(7/2) - b*c*x*(35 + 35*c^2*x^2 + 21*c^4*x^4 + 5*c^6*x^6) + 35*b*(1 + c^2*x^2)^(7
/2)*ArcSinh[c*x]))/(245*c^2)

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Maple [B]  time = 0.063, size = 170, normalized size = 1.8 \begin{align*}{\frac{a}{7\,\pi \,{c}^{2}} \left ( \pi \,{c}^{2}{x}^{2}+\pi \right ) ^{{\frac{7}{2}}}}+{\frac{b{\pi }^{{\frac{5}{2}}}}{245\,{c}^{2}} \left ( 35\,{\it Arcsinh} \left ( cx \right ){c}^{8}{x}^{8}+140\,{\it Arcsinh} \left ( cx \right ){c}^{6}{x}^{6}-5\,{c}^{7}{x}^{7}\sqrt{{c}^{2}{x}^{2}+1}+210\,{\it Arcsinh} \left ( cx \right ){c}^{4}{x}^{4}-21\,{c}^{5}{x}^{5}\sqrt{{c}^{2}{x}^{2}+1}+140\,{\it Arcsinh} \left ( cx \right ){c}^{2}{x}^{2}-35\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}+35\,{\it Arcsinh} \left ( cx \right ) -35\,cx\sqrt{{c}^{2}{x}^{2}+1} \right ){\frac{1}{\sqrt{{c}^{2}{x}^{2}+1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*a/Pi/c^2*(Pi*c^2*x^2+Pi)^(7/2)+1/245*b/c^2*Pi^(5/2)/(c^2*x^2+1)^(1/2)*(35*arcsinh(c*x)*c^8*x^8+140*arcsinh
(c*x)*c^6*x^6-5*c^7*x^7*(c^2*x^2+1)^(1/2)+210*arcsinh(c*x)*c^4*x^4-21*c^5*x^5*(c^2*x^2+1)^(1/2)+140*arcsinh(c*
x)*c^2*x^2-35*c^3*x^3*(c^2*x^2+1)^(1/2)+35*arcsinh(c*x)-35*c*x*(c^2*x^2+1)^(1/2))

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Maxima [A]  time = 1.20053, size = 130, normalized size = 1.4 \begin{align*} \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{7}{2}} b \operatorname{arsinh}\left (c x\right )}{7 \, \pi c^{2}} + \frac{{\left (\pi + \pi c^{2} x^{2}\right )}^{\frac{7}{2}} a}{7 \, \pi c^{2}} - \frac{{\left (5 \, \pi ^{\frac{7}{2}} c^{6} x^{7} + 21 \, \pi ^{\frac{7}{2}} c^{4} x^{5} + 35 \, \pi ^{\frac{7}{2}} c^{2} x^{3} + 35 \, \pi ^{\frac{7}{2}} x\right )} b}{245 \, \pi c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*(pi + pi*c^2*x^2)^(7/2)*b*arcsinh(c*x)/(pi*c^2) + 1/7*(pi + pi*c^2*x^2)^(7/2)*a/(pi*c^2) - 1/245*(5*pi^(7/
2)*c^6*x^7 + 21*pi^(7/2)*c^4*x^5 + 35*pi^(7/2)*c^2*x^3 + 35*pi^(7/2)*x)*b/(pi*c)

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Fricas [B]  time = 2.60435, size = 508, normalized size = 5.46 \begin{align*} \frac{35 \, \sqrt{\pi + \pi c^{2} x^{2}}{\left (\pi ^{2} b c^{8} x^{8} + 4 \, \pi ^{2} b c^{6} x^{6} + 6 \, \pi ^{2} b c^{4} x^{4} + 4 \, \pi ^{2} b c^{2} x^{2} + \pi ^{2} b\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + \sqrt{\pi + \pi c^{2} x^{2}}{\left (35 \, \pi ^{2} a c^{8} x^{8} + 140 \, \pi ^{2} a c^{6} x^{6} + 210 \, \pi ^{2} a c^{4} x^{4} + 140 \, \pi ^{2} a c^{2} x^{2} + 35 \, \pi ^{2} a -{\left (5 \, \pi ^{2} b c^{7} x^{7} + 21 \, \pi ^{2} b c^{5} x^{5} + 35 \, \pi ^{2} b c^{3} x^{3} + 35 \, \pi ^{2} b c x\right )} \sqrt{c^{2} x^{2} + 1}\right )}}{245 \,{\left (c^{4} x^{2} + c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/245*(35*sqrt(pi + pi*c^2*x^2)*(pi^2*b*c^8*x^8 + 4*pi^2*b*c^6*x^6 + 6*pi^2*b*c^4*x^4 + 4*pi^2*b*c^2*x^2 + pi^
2*b)*log(c*x + sqrt(c^2*x^2 + 1)) + sqrt(pi + pi*c^2*x^2)*(35*pi^2*a*c^8*x^8 + 140*pi^2*a*c^6*x^6 + 210*pi^2*a
*c^4*x^4 + 140*pi^2*a*c^2*x^2 + 35*pi^2*a - (5*pi^2*b*c^7*x^7 + 21*pi^2*b*c^5*x^5 + 35*pi^2*b*c^3*x^3 + 35*pi^
2*b*c*x)*sqrt(c^2*x^2 + 1)))/(c^4*x^2 + c^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(x*(pi*c**2*x**2+pi)**(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: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError