3.1.70 \(\int \frac {(\pi +c^2 \pi x^2)^{3/2} (a+b \sinh ^{-1}(c x))}{x^4} \, dx\) [70]

Optimal. Leaf size=115 \[ -\frac {b c \pi ^{3/2}}{6 x^2}-\frac {c^2 \pi \sqrt {\pi +c^2 \pi x^2} \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {\left (\pi +c^2 \pi x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {c^3 \pi ^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}+\frac {4}{3} b c^3 \pi ^{3/2} \log (x) \]

[Out]

-1/6*b*c*Pi^(3/2)/x^2-1/3*(Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))/x^3+1/2*c^3*Pi^(3/2)*(a+b*arcsinh(c*x))^2/b
+4/3*b*c^3*Pi^(3/2)*ln(x)-c^2*Pi*(a+b*arcsinh(c*x))*(Pi*c^2*x^2+Pi)^(1/2)/x

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Rubi [A]
time = 0.15, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {5807, 5805, 29, 5783, 14} \begin {gather*} \frac {\pi ^{3/2} c^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b}-\frac {\pi c^2 \sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{x}-\frac {\left (\pi c^2 x^2+\pi \right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 x^3}+\frac {4}{3} \pi ^{3/2} b c^3 \log (x)-\frac {\pi ^{3/2} b c}{6 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

-1/6*(b*c*Pi^(3/2))/x^2 - (c^2*Pi*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/x - ((Pi + c^2*Pi*x^2)^(3/2)*(a
+ b*ArcSinh[c*x]))/(3*x^3) + (c^3*Pi^(3/2)*(a + b*ArcSinh[c*x])^2)/(2*b) + (4*b*c^3*Pi^(3/2)*Log[x])/3

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

Rule 5805

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

Rule 5807

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

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 125, normalized size = 1.09 \begin {gather*} \frac {\pi ^{3/2} \left (-b c x-2 a \sqrt {1+c^2 x^2}-8 a c^2 x^2 \sqrt {1+c^2 x^2}+\left (6 a c^3 x^3-2 b \sqrt {1+c^2 x^2} \left (1+4 c^2 x^2\right )\right ) \sinh ^{-1}(c x)+3 b c^3 x^3 \sinh ^{-1}(c x)^2+8 b c^3 x^3 \log (c x)\right )}{6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((Pi + c^2*Pi*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/x^4,x]

[Out]

(Pi^(3/2)*(-(b*c*x) - 2*a*Sqrt[1 + c^2*x^2] - 8*a*c^2*x^2*Sqrt[1 + c^2*x^2] + (6*a*c^3*x^3 - 2*b*Sqrt[1 + c^2*
x^2]*(1 + 4*c^2*x^2))*ArcSinh[c*x] + 3*b*c^3*x^3*ArcSinh[c*x]^2 + 8*b*c^3*x^3*Log[c*x]))/(6*x^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(621\) vs. \(2(97)=194\).
time = 6.62, size = 622, normalized size = 5.41

method	result	size
default	\(-\frac {a \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{3 \pi \,x^{3}}-\frac {2 a \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {5}{2}}}{3 \pi x}+\frac {2 a \,c^{4} x \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}{3}+a \,c^{4} \pi x \sqrt {\pi \,c^{2} x^{2}+\pi }+\frac {a \,c^{4} \pi ^{2} \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{\sqrt {\pi \,c^{2}}}+\frac {b \,c^{3} \pi ^{\frac {3}{2}} \arcsinh \left (c x \right )^{2}}{2}-\frac {8 b \,c^{3} \pi ^{\frac {3}{2}} \arcsinh \left (c x \right )}{3}+\frac {32 b \,\pi ^{\frac {3}{2}} x^{4} \arcsinh \left (c x \right ) c^{7}}{24 c^{4} x^{4}+9 c^{2} x^{2}+1}-\frac {32 b \,\pi ^{\frac {3}{2}} x^{3} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{6}}{24 c^{4} x^{4}+9 c^{2} x^{2}+1}+\frac {8 b \,\pi ^{\frac {3}{2}} x^{4} c^{7}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right )}-\frac {8 b \,\pi ^{\frac {3}{2}} x^{2} \left (c^{2} x^{2}+1\right ) c^{5}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right )}+\frac {12 b \,\pi ^{\frac {3}{2}} x^{2} \arcsinh \left (c x \right ) c^{5}}{24 c^{4} x^{4}+9 c^{2} x^{2}+1}-\frac {20 b \,\pi ^{\frac {3}{2}} x \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{4}}{24 c^{4} x^{4}+9 c^{2} x^{2}+1}-\frac {4 b \,\pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right ) c^{3}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right )}+\frac {4 b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right ) c^{3}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right )}-\frac {13 b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}\, c^{2}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right ) x}-\frac {b \,\pi ^{\frac {3}{2}} \left (c^{2} x^{2}+1\right ) c}{6 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right ) x^{2}}-\frac {b \,\pi ^{\frac {3}{2}} \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{3 \left (24 c^{4} x^{4}+9 c^{2} x^{2}+1\right ) x^{3}}+\frac {4 b \,c^{3} \pi ^{\frac {3}{2}} \ln \left (\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2}-1\right )}{3}\)	\(622\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((Pi*c^2*x^2+Pi)^(3/2)*(a+b*arcsinh(c*x))/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a/Pi/x^3*(Pi*c^2*x^2+Pi)^(5/2)-2/3*a*c^2/Pi/x*(Pi*c^2*x^2+Pi)^(5/2)+2/3*a*c^4*x*(Pi*c^2*x^2+Pi)^(3/2)+a*c
^4*Pi*x*(Pi*c^2*x^2+Pi)^(1/2)+a*c^4*Pi^2*ln(Pi*c^2*x/(Pi*c^2)^(1/2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/2*
b*c^3*Pi^(3/2)*arcsinh(c*x)^2-8/3*b*c^3*Pi^(3/2)*arcsinh(c*x)+32*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^4*arcsi
nh(c*x)*c^7-32*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^6+8/3*b*Pi^(3/2)/(24*c
^4*x^4+9*c^2*x^2+1)*x^4*c^7-8/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x^2*(c^2*x^2+1)*c^5+12*b*Pi^(3/2)/(24*c^4*
x^4+9*c^2*x^2+1)*x^2*arcsinh(c*x)*c^5-20*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*
c^4-4/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*(c^2*x^2+1)*c^3+4/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)*arcsinh(c*
x)*c^3-13/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)/x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^2-1/6*b*Pi^(3/2)/(24*c^4*x^
4+9*c^2*x^2+1)/x^2*(c^2*x^2+1)*c-1/3*b*Pi^(3/2)/(24*c^4*x^4+9*c^2*x^2+1)/x^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+4/
3*b*c^3*Pi^(3/2)*ln((c*x+(c^2*x^2+1)^(1/2))^2-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x))/x^4,x, algorithm="fricas")

[Out]

integral(sqrt(pi + pi*c^2*x^2)*(pi*a*c^2*x^2 + pi*a + (pi*b*c^2*x^2 + pi*b)*arcsinh(c*x))/x^4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \pi ^{\frac {3}{2}} \left (\int \frac {a \sqrt {c^{2} x^{2} + 1}}{x^{4}}\, dx + \int \frac {a c^{2} \sqrt {c^{2} x^{2} + 1}}{x^{2}}\, dx + \int \frac {b \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{x^{4}}\, dx + \int \frac {b c^{2} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{x^{2}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((pi*c**2*x**2+pi)**(3/2)*(a+b*asinh(c*x))/x**4,x)

[Out]

pi**(3/2)*(Integral(a*sqrt(c**2*x**2 + 1)/x**4, x) + Integral(a*c**2*sqrt(c**2*x**2 + 1)/x**2, x) + Integral(b
*sqrt(c**2*x**2 + 1)*asinh(c*x)/x**4, x) + Integral(b*c**2*sqrt(c**2*x**2 + 1)*asinh(c*x)/x**2, x))

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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((pi*c^2*x^2+pi)^(3/2)*(a+b*arcsinh(c*x))/x^4,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 \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (\Pi \,c^2\,x^2+\Pi \right )}^{3/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(3/2))/x^4,x)

[Out]

int(((a + b*asinh(c*x))*(Pi + Pi*c^2*x^2)^(3/2))/x^4, x)

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