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

Optimal. Leaf size=146 \[ \frac {\sqrt {\pi c^2 x^2+\pi } \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^3 c^6}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^2 c^6 \sqrt {\pi c^2 x^2+\pi }}-\frac {a+b \sinh ^{-1}(c x)}{3 \pi c^6 \left (\pi c^2 x^2+\pi \right )^{3/2}}-\frac {11 b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^6}-\frac {b x}{\pi ^{5/2} c^5}+\frac {b x}{6 \pi ^{5/2} c^5 \left (c^2 x^2+1\right )} \]

[Out]

-b*x/c^5/Pi^(5/2)+1/6*b*x/c^5/Pi^(5/2)/(c^2*x^2+1)+1/3*(-a-b*arcsinh(c*x))/c^6/Pi/(Pi*c^2*x^2+Pi)^(3/2)-11/6*b
*arctan(c*x)/c^6/Pi^(5/2)+2*(a+b*arcsinh(c*x))/c^6/Pi^2/(Pi*c^2*x^2+Pi)^(1/2)+(a+b*arcsinh(c*x))*(Pi*c^2*x^2+P
i)^(1/2)/c^6/Pi^3

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Rubi [A]  time = 0.18, antiderivative size = 149, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {266, 43, 5732, 12, 1157, 388, 203} \[ \frac {\sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} c^6}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{\pi ^{5/2} c^6 \sqrt {c^2 x^2+1}}-\frac {a+b \sinh ^{-1}(c x)}{3 \pi ^{5/2} c^6 \left (c^2 x^2+1\right )^{3/2}}+\frac {b x}{6 \pi ^{5/2} c^5 \left (c^2 x^2+1\right )}-\frac {b x}{\pi ^{5/2} c^5}-\frac {11 b \tan ^{-1}(c x)}{6 \pi ^{5/2} c^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*x)/(c^5*Pi^(5/2))) + (b*x)/(6*c^5*Pi^(5/2)*(1 + c^2*x^2)) - (a + b*ArcSinh[c*x])/(3*c^6*Pi^(5/2)*(1 + c^2
*x^2)^(3/2)) + (2*(a + b*ArcSinh[c*x]))/(c^6*Pi^(5/2)*Sqrt[1 + c^2*x^2]) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c
*x]))/(c^6*Pi^(5/2)) - (11*b*ArcTan[c*x])/(6*c^6*Pi^(5/2))

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 388

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

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\left (\pi +c^2 \pi x^2\right )^{5/2}} \, dx &=-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2}}-\frac {(b c) \int \frac {8+12 c^2 x^2+3 c^4 x^4}{3 c^6 \left (1+c^2 x^2\right )^2} \, dx}{\pi ^{5/2}}\\ &=-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2}}-\frac {b \int \frac {8+12 c^2 x^2+3 c^4 x^4}{\left (1+c^2 x^2\right )^2} \, dx}{3 c^5 \pi ^{5/2}}\\ &=\frac {b x}{6 c^5 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2}}+\frac {b \int \frac {-17-6 c^2 x^2}{1+c^2 x^2} \, dx}{6 c^5 \pi ^{5/2}}\\ &=-\frac {b x}{c^5 \pi ^{5/2}}+\frac {b x}{6 c^5 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2}}-\frac {(11 b) \int \frac {1}{1+c^2 x^2} \, dx}{6 c^5 \pi ^{5/2}}\\ &=-\frac {b x}{c^5 \pi ^{5/2}}+\frac {b x}{6 c^5 \pi ^{5/2} \left (1+c^2 x^2\right )}-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 \pi ^{5/2} \left (1+c^2 x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2} \sqrt {1+c^2 x^2}}+\frac {\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 \pi ^{5/2}}-\frac {11 b \tan ^{-1}(c x)}{6 c^6 \pi ^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 132, normalized size = 0.90 \[ \frac {6 a c^4 x^4+24 a c^2 x^2+16 a-5 b c x \sqrt {c^2 x^2+1}-11 b \left (c^2 x^2+1\right )^{3/2} \tan ^{-1}(c x)+2 b \left (3 c^4 x^4+12 c^2 x^2+8\right ) \sinh ^{-1}(c x)-6 b c^3 x^3 \sqrt {c^2 x^2+1}}{6 \pi ^{5/2} c^6 \left (c^2 x^2+1\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*a + 24*a*c^2*x^2 + 6*a*c^4*x^4 - 5*b*c*x*Sqrt[1 + c^2*x^2] - 6*b*c^3*x^3*Sqrt[1 + c^2*x^2] + 2*b*(8 + 12*c
^2*x^2 + 3*c^4*x^4)*ArcSinh[c*x] - 11*b*(1 + c^2*x^2)^(3/2)*ArcTan[c*x])/(6*c^6*Pi^(5/2)*(1 + c^2*x^2)^(3/2))

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fricas [A]  time = 0.89, size = 218, normalized size = 1.49 \[ \frac {11 \, \sqrt {\pi } {\left (b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b\right )} \arctan \left (-\frac {2 \, \sqrt {\pi } \sqrt {\pi + \pi c^{2} x^{2}} \sqrt {c^{2} x^{2} + 1} c x}{\pi - \pi c^{4} x^{4}}\right ) + 4 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (3 \, b c^{4} x^{4} + 12 \, b c^{2} x^{2} + 8 \, b\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, \sqrt {\pi + \pi c^{2} x^{2}} {\left (6 \, a c^{4} x^{4} + 24 \, a c^{2} x^{2} - {\left (6 \, b c^{3} x^{3} + 5 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} + 16 \, a\right )}}{12 \, {\left (\pi ^{3} c^{10} x^{4} + 2 \, \pi ^{3} c^{8} x^{2} + \pi ^{3} c^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(11*sqrt(pi)*(b*c^4*x^4 + 2*b*c^2*x^2 + b)*arctan(-2*sqrt(pi)*sqrt(pi + pi*c^2*x^2)*sqrt(c^2*x^2 + 1)*c*x
/(pi - pi*c^4*x^4)) + 4*sqrt(pi + pi*c^2*x^2)*(3*b*c^4*x^4 + 12*b*c^2*x^2 + 8*b)*log(c*x + sqrt(c^2*x^2 + 1))
+ 2*sqrt(pi + pi*c^2*x^2)*(6*a*c^4*x^4 + 24*a*c^2*x^2 - (6*b*c^3*x^3 + 5*b*c*x)*sqrt(c^2*x^2 + 1) + 16*a))/(pi
^3*c^10*x^4 + 2*pi^3*c^8*x^2 + pi^3*c^6)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [C]  time = 0.48, size = 231, normalized size = 1.58 \[ \frac {a \,x^{4}}{\pi \,c^{2} \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {4 a \,x^{2}}{c^{4} \pi \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {8 a}{3 c^{6} \pi \left (\pi \,c^{2} x^{2}+\pi \right )^{\frac {3}{2}}}+\frac {b \arcsinh \left (c x \right ) \sqrt {c^{2} x^{2}+1}}{\pi ^{\frac {5}{2}} c^{6}}-\frac {b x}{c^{5} \pi ^{\frac {5}{2}}}+\frac {2 b \arcsinh \left (c x \right ) x^{2}}{\pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{4}}+\frac {b x}{6 c^{5} \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )}+\frac {5 b \arcsinh \left (c x \right )}{3 \pi ^{\frac {5}{2}} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{6}}+\frac {11 i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 c^{6} \pi ^{\frac {5}{2}}}-\frac {11 i b \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 c^{6} \pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x^4/Pi/c^2/(Pi*c^2*x^2+Pi)^(3/2)+4*a/c^4*x^2/Pi/(Pi*c^2*x^2+Pi)^(3/2)+8/3*a/c^6/Pi/(Pi*c^2*x^2+Pi)^(3/2)+b/P
i^(5/2)/c^6*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-b*x/c^5/Pi^(5/2)+2*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^4*arcsinh(c*x)*x^
2+1/6*b*x/c^5/Pi^(5/2)/(c^2*x^2+1)+5/3*b/Pi^(5/2)/(c^2*x^2+1)^(3/2)/c^6*arcsinh(c*x)+11/6*I*b/c^6/Pi^(5/2)*ln(
c*x+(c^2*x^2+1)^(1/2)-I)-11/6*I*b/c^6/Pi^(5/2)*ln(c*x+(c^2*x^2+1)^(1/2)+I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, b {\left (\frac {{\left (3 \, \sqrt {\pi } c^{4} x^{4} + 12 \, \sqrt {\pi } c^{2} x^{2} + 8 \, \sqrt {\pi }\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (\pi ^{3} c^{8} x^{2} + \pi ^{3} c^{6}\right )} \sqrt {c^{2} x^{2} + 1}} + 3 \, \int \frac {3 \, \sqrt {\pi } c^{4} x^{4} + 12 \, \sqrt {\pi } c^{2} x^{2} + 8 \, \sqrt {\pi }}{3 \, {\left (\pi ^{3} c^{11} x^{6} + 2 \, \pi ^{3} c^{9} x^{4} + \pi ^{3} c^{7} x^{2} + {\left (\pi ^{3} c^{10} x^{5} + 2 \, \pi ^{3} c^{8} x^{3} + \pi ^{3} c^{6} x\right )} \sqrt {c^{2} x^{2} + 1}\right )}}\,{d x} - 3 \, \int \frac {3 \, \sqrt {\pi } c^{4} x^{4} + 12 \, \sqrt {\pi } c^{2} x^{2} + 8 \, \sqrt {\pi }}{3 \, {\left (\pi ^{3} c^{8} x^{3} + \pi ^{3} c^{6} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} + \frac {1}{3} \, a {\left (\frac {3 \, x^{4}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{2}} + \frac {12 \, x^{2}}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{4}} + \frac {8}{\pi {\left (\pi + \pi c^{2} x^{2}\right )}^{\frac {3}{2}} c^{6}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*((3*sqrt(pi)*c^4*x^4 + 12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))*log(c*x + sqrt(c^2*x^2 + 1))/((pi^3*c^8*x^2 + p
i^3*c^6)*sqrt(c^2*x^2 + 1)) + 3*integrate(1/3*(3*sqrt(pi)*c^4*x^4 + 12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))/(pi^3*c^
11*x^6 + 2*pi^3*c^9*x^4 + pi^3*c^7*x^2 + (pi^3*c^10*x^5 + 2*pi^3*c^8*x^3 + pi^3*c^6*x)*sqrt(c^2*x^2 + 1)), x)
- 3*integrate(1/3*(3*sqrt(pi)*c^4*x^4 + 12*sqrt(pi)*c^2*x^2 + 8*sqrt(pi))/((pi^3*c^8*x^3 + pi^3*c^6*x)*sqrt(c^
2*x^2 + 1)), x)) + 1/3*a*(3*x^4/(pi*(pi + pi*c^2*x^2)^(3/2)*c^2) + 12*x^2/(pi*(pi + pi*c^2*x^2)^(3/2)*c^4) + 8
/(pi*(pi + pi*c^2*x^2)^(3/2)*c^6))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x^{5}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx + \int \frac {b x^{5} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} \sqrt {c^{2} x^{2} + 1} + 2 c^{2} x^{2} \sqrt {c^{2} x^{2} + 1} + \sqrt {c^{2} x^{2} + 1}}\, dx}{\pi ^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x**5/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2 + 1)), x) +
Integral(b*x**5*asinh(c*x)/(c**4*x**4*sqrt(c**2*x**2 + 1) + 2*c**2*x**2*sqrt(c**2*x**2 + 1) + sqrt(c**2*x**2
+ 1)), x))/pi**(5/2)

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