3.155 \(\int \frac {x^5 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=212 \[ \frac {\left (c^2 d x^2+d\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^3}-\frac {2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^2}-\frac {a+b \sinh ^{-1}(c x)}{c^6 d \sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 d x^2+d} \tan ^{-1}(c x)}{c^6 d^2 \sqrt {c^2 x^2+1}}+\frac {5 b x \sqrt {c^2 d x^2+d}}{3 c^5 d^2 \sqrt {c^2 x^2+1}}-\frac {b x^3 \sqrt {c^2 d x^2+d}}{9 c^3 d^2 \sqrt {c^2 x^2+1}} \]

[Out]

1/3*(c^2*d*x^2+d)^(3/2)*(a+b*arcsinh(c*x))/c^6/d^3+(-a-b*arcsinh(c*x))/c^6/d/(c^2*d*x^2+d)^(1/2)-2*(a+b*arcsin
h(c*x))*(c^2*d*x^2+d)^(1/2)/c^6/d^2+5/3*b*x*(c^2*d*x^2+d)^(1/2)/c^5/d^2/(c^2*x^2+1)^(1/2)-1/9*b*x^3*(c^2*d*x^2
+d)^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)+b*arctan(c*x)*(c^2*d*x^2+d)^(1/2)/c^6/d^2/(c^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.29, antiderivative size = 220, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {5751, 5758, 5717, 8, 30, 302, 203} \[ \frac {4 x^2 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}-\frac {8 \sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {c^2 d x^2+d}}-\frac {b x^3 \sqrt {c^2 x^2+1}}{9 c^3 d \sqrt {c^2 d x^2+d}}+\frac {5 b x \sqrt {c^2 x^2+1}}{3 c^5 d \sqrt {c^2 d x^2+d}}+\frac {b \sqrt {c^2 x^2+1} \tan ^{-1}(c x)}{c^6 d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

(5*b*x*Sqrt[1 + c^2*x^2])/(3*c^5*d*Sqrt[d + c^2*d*x^2]) - (b*x^3*Sqrt[1 + c^2*x^2])/(9*c^3*d*Sqrt[d + c^2*d*x^
2]) - (x^4*(a + b*ArcSinh[c*x]))/(c^2*d*Sqrt[d + c^2*d*x^2]) - (8*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3
*c^6*d^2) + (4*x^2*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(3*c^4*d^2) + (b*Sqrt[1 + c^2*x^2]*ArcTan[c*x])/(
c^6*d*Sqrt[d + c^2*d*x^2])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 5751

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

Rule 5758

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

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {4 \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+c^2 d x^2}} \, dx}{c^2 d}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {x^4}{1+c^2 x^2} \, dx}{c d \sqrt {d+c^2 d x^2}}\\ &=-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {d+c^2 d x^2}}+\frac {4 x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}-\frac {8 \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+c^2 d x^2}} \, dx}{3 c^4 d}-\frac {\left (4 b \sqrt {1+c^2 x^2}\right ) \int x^2 \, dx}{3 c^3 d \sqrt {d+c^2 d x^2}}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \left (-\frac {1}{c^4}+\frac {x^2}{c^2}+\frac {1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{c d \sqrt {d+c^2 d x^2}}\\ &=-\frac {b x \sqrt {1+c^2 x^2}}{c^5 d \sqrt {d+c^2 d x^2}}-\frac {b x^3 \sqrt {1+c^2 x^2}}{9 c^3 d \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {d+c^2 d x^2}}-\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}+\frac {4 x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}+\frac {\left (b \sqrt {1+c^2 x^2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{c^5 d \sqrt {d+c^2 d x^2}}+\frac {\left (8 b \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{3 c^5 d \sqrt {d+c^2 d x^2}}\\ &=\frac {5 b x \sqrt {1+c^2 x^2}}{3 c^5 d \sqrt {d+c^2 d x^2}}-\frac {b x^3 \sqrt {1+c^2 x^2}}{9 c^3 d \sqrt {d+c^2 d x^2}}-\frac {x^4 \left (a+b \sinh ^{-1}(c x)\right )}{c^2 d \sqrt {d+c^2 d x^2}}-\frac {8 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^6 d^2}+\frac {4 x^2 \sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{3 c^4 d^2}+\frac {b \sqrt {1+c^2 x^2} \tan ^{-1}(c x)}{c^6 d \sqrt {d+c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 148, normalized size = 0.70 \[ \frac {\sqrt {c^2 d x^2+d} \left (3 a \left (c^4 x^4-4 c^2 x^2-8\right )+b c x \sqrt {c^2 x^2+1} \left (15-c^2 x^2\right )+3 b \left (c^4 x^4-4 c^2 x^2-8\right ) \sinh ^{-1}(c x)\right )}{9 c^6 d^2 \left (c^2 x^2+1\right )}+\frac {b \sqrt {d \left (c^2 x^2+1\right )} \tan ^{-1}(c x)}{c^6 d^2 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(3/2),x]

[Out]

(Sqrt[d + c^2*d*x^2]*(b*c*x*(15 - c^2*x^2)*Sqrt[1 + c^2*x^2] + 3*a*(-8 - 4*c^2*x^2 + c^4*x^4) + 3*b*(-8 - 4*c^
2*x^2 + c^4*x^4)*ArcSinh[c*x]))/(9*c^6*d^2*(1 + c^2*x^2)) + (b*Sqrt[d*(1 + c^2*x^2)]*ArcTan[c*x])/(c^6*d^2*Sqr
t[1 + c^2*x^2])

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fricas [A]  time = 0.61, size = 197, normalized size = 0.93 \[ -\frac {9 \, {\left (b c^{2} x^{2} + b\right )} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {c^{2} d x^{2} + d} \sqrt {c^{2} x^{2} + 1} c \sqrt {d} x}{c^{4} d x^{4} - d}\right ) - 6 \, {\left (b c^{4} x^{4} - 4 \, b c^{2} x^{2} - 8 \, b\right )} \sqrt {c^{2} d x^{2} + d} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (3 \, a c^{4} x^{4} - 12 \, a c^{2} x^{2} - {\left (b c^{3} x^{3} - 15 \, b c x\right )} \sqrt {c^{2} x^{2} + 1} - 24 \, a\right )} \sqrt {c^{2} d x^{2} + d}}{18 \, {\left (c^{8} d^{2} x^{2} + c^{6} d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

-1/18*(9*(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)) -
6*(b*c^4*x^4 - 4*b*c^2*x^2 - 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(3*a*c^4*x^4 - 12*a*c^
2*x^2 - (b*c^3*x^3 - 15*b*c*x)*sqrt(c^2*x^2 + 1) - 24*a)*sqrt(c^2*d*x^2 + d))/(c^8*d^2*x^2 + c^6*d^2)

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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))/(c^2*d*x^2+d)^(3/2),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 = 0.29, size = 362, normalized size = 1.71 \[ \frac {a \,x^{4}}{3 c^{2} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {4 a \,x^{2}}{3 c^{4} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {8 a}{3 c^{6} d \sqrt {c^{2} d \,x^{2}+d}}-\frac {8 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{3 c^{6} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{6} d^{2}}-\frac {i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{\sqrt {c^{2} x^{2}+1}\, c^{6} d^{2}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{4}}{3 c^{2} d^{2} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x^{3}}{9 c^{3} d^{2} \sqrt {c^{2} x^{2}+1}}-\frac {4 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{3 c^{4} d^{2} \left (c^{2} x^{2}+1\right )}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{3 c^{5} d^{2} \sqrt {c^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x)

[Out]

1/3*a*x^4/c^2/d/(c^2*d*x^2+d)^(1/2)-4/3*a/c^4*x^2/d/(c^2*d*x^2+d)^(1/2)-8/3*a/c^6/d/(c^2*d*x^2+d)^(1/2)-8/3*b*
(d*(c^2*x^2+1))^(1/2)/c^6/d^2/(c^2*x^2+1)*arcsinh(c*x)+I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^2*ln(
c*x+(c^2*x^2+1)^(1/2)+I)-I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^2*ln(c*x+(c^2*x^2+1)^(1/2)-I)+1/3*b
*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcsinh(c*x)*x^4-1/9*b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1
/2)*x^3-4/3*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)*arcsinh(c*x)*x^2+5/3*b*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(
c^2*x^2+1)^(1/2)*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, a {\left (\frac {x^{4}}{\sqrt {c^{2} d x^{2} + d} c^{2} d} - \frac {4 \, x^{2}}{\sqrt {c^{2} d x^{2} + d} c^{4} d} - \frac {8}{\sqrt {c^{2} d x^{2} + d} c^{6} d}\right )} + \frac {1}{3} \, b {\left (\frac {{\left (c^{4} \sqrt {d} x^{4} - 4 \, c^{2} \sqrt {d} x^{2} - 8 \, \sqrt {d}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{\sqrt {c^{2} x^{2} + 1} c^{6} d^{2}} - \frac {\frac {1}{3} \, \sqrt {c^{2} x^{2} + 1} c^{2} \sqrt {d} x^{2} + 8 \, \sqrt {d} \operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right ) - \frac {14}{3} \, \sqrt {c^{2} x^{2} + 1} \sqrt {d}}{c^{6} d^{2}} + 3 \, \int \frac {c^{4} \sqrt {d} x^{4} - 4 \, c^{2} \sqrt {d} x^{2} - 8 \, \sqrt {d}}{3 \, {\left (c^{9} d^{2} x^{4} + c^{7} d^{2} x^{2} + {\left (c^{8} d^{2} x^{3} + c^{6} d^{2} x\right )} \sqrt {c^{2} x^{2} + 1}\right )}}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

1/3*a*(x^4/(sqrt(c^2*d*x^2 + d)*c^2*d) - 4*x^2/(sqrt(c^2*d*x^2 + d)*c^4*d) - 8/(sqrt(c^2*d*x^2 + d)*c^6*d)) +
1/3*b*((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))*log(c*x + sqrt(c^2*x^2 + 1))/(sqrt(c^2*x^2 + 1)*c^6*d
^2) - integrate((c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))/(sqrt(c^2*x^2 + 1)*x), x)/(c^6*d^2) + 3*inte
grate(1/3*(c^4*sqrt(d)*x^4 - 4*c^2*sqrt(d)*x^2 - 8*sqrt(d))/(c^9*d^2*x^4 + c^7*d^2*x^2 + (c^8*d^2*x^3 + c^6*d^
2*x)*sqrt(c^2*x^2 + 1)), x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2),x)

[Out]

int((x^5*(a + b*asinh(c*x)))/(d + c^2*d*x^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(3/2),x)

[Out]

Integral(x**5*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(3/2), x)

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