3.1.73 \(\int x (\pi +c^2 \pi x^2)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [73]

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

[Out]

-1/7*b*Pi^(5/2)*x/c-1/7*b*c*Pi^(5/2)*x^3-3/35*b*c^3*Pi^(5/2)*x^5-1/49*b*c^5*Pi^(5/2)*x^7+1/7*(Pi*c^2*x^2+Pi)^(
7/2)*(a+b*arcsinh(c*x))/c^2/Pi

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Rubi [A]
time = 0.06, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {5798, 200} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*(b*Pi^(5/2)*x)/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)

Rule 200

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]

Rule 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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]

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*}

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Mathematica [A]
time = 0.10, size = 80, normalized size = 0.86 \begin {gather*} \frac {\pi ^{5/2} \left (35 a \left (1+c^2 x^2\right )^{7/2}-b c x \left (35+35 c^2 x^2+21 c^4 x^4+5 c^6 x^6\right )+35 b \left (1+c^2 x^2\right )^{7/2} \sinh ^{-1}(c x)\right )}{245 c^2} \end {gather*}

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 [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

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]

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

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Maxima [A]
time = 0.27, size = 96, normalized size = 1.03 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (73) = 146\).
time = 0.37, size = 225, normalized size = 2.42 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (85) = 170\).
time = 37.57, size = 299, normalized size = 3.22 \begin {gather*} \begin {cases} \frac {\pi ^{\frac {5}{2}} a c^{4} x^{6} \sqrt {c^{2} x^{2} + 1}}{7} + \frac {3 \pi ^{\frac {5}{2}} a c^{2} x^{4} \sqrt {c^{2} x^{2} + 1}}{7} + \frac {3 \pi ^{\frac {5}{2}} a x^{2} \sqrt {c^{2} x^{2} + 1}}{7} + \frac {\pi ^{\frac {5}{2}} a \sqrt {c^{2} x^{2} + 1}}{7 c^{2}} - \frac {\pi ^{\frac {5}{2}} b c^{5} x^{7}}{49} + \frac {\pi ^{\frac {5}{2}} b c^{4} x^{6} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {3 \pi ^{\frac {5}{2}} b c^{3} x^{5}}{35} + \frac {3 \pi ^{\frac {5}{2}} b c^{2} x^{4} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {\pi ^{\frac {5}{2}} b c x^{3}}{7} + \frac {3 \pi ^{\frac {5}{2}} b x^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{7} - \frac {\pi ^{\frac {5}{2}} b x}{7 c} + \frac {\pi ^{\frac {5}{2}} b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{7 c^{2}} & \text {for}\: c \neq 0 \\\frac {\pi ^{\frac {5}{2}} a x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

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]

Piecewise((pi**(5/2)*a*c**4*x**6*sqrt(c**2*x**2 + 1)/7 + 3*pi**(5/2)*a*c**2*x**4*sqrt(c**2*x**2 + 1)/7 + 3*pi*
*(5/2)*a*x**2*sqrt(c**2*x**2 + 1)/7 + pi**(5/2)*a*sqrt(c**2*x**2 + 1)/(7*c**2) - pi**(5/2)*b*c**5*x**7/49 + pi
**(5/2)*b*c**4*x**6*sqrt(c**2*x**2 + 1)*asinh(c*x)/7 - 3*pi**(5/2)*b*c**3*x**5/35 + 3*pi**(5/2)*b*c**2*x**4*sq
rt(c**2*x**2 + 1)*asinh(c*x)/7 - pi**(5/2)*b*c*x**3/7 + 3*pi**(5/2)*b*x**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/7 -
pi**(5/2)*b*x/(7*c) + pi**(5/2)*b*sqrt(c**2*x**2 + 1)*asinh(c*x)/(7*c**2), Ne(c, 0)), (pi**(5/2)*a*x**2/2, Tru
e))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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