3.2.66 \(\int \frac {x^5 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\) [166]

Optimal. Leaf size=210 \[ \frac {b x \sqrt {d+c^2 d x^2}}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b x \sqrt {d+c^2 d x^2}}{c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 d \left (d+c^2 d x^2\right )^{3/2}}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt {d+c^2 d x^2}}+\frac {\sqrt {d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^3}-\frac {11 b \sqrt {d+c^2 d x^2} \text {ArcTan}(c x)}{6 c^6 d^3 \sqrt {1+c^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 210, normalized size of antiderivative = 1.00, 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 = {272, 45, 5804, 12, 1171, 396, 209} \begin {gather*} \frac {\sqrt {c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^3}+\frac {2 \left (a+b \sinh ^{-1}(c x)\right )}{c^6 d^2 \sqrt {c^2 d x^2+d}}-\frac {a+b \sinh ^{-1}(c x)}{3 c^6 d \left (c^2 d x^2+d\right )^{3/2}}-\frac {11 b \text {ArcTan}(c x) \sqrt {c^2 d x^2+d}}{6 c^6 d^3 \sqrt {c^2 x^2+1}}-\frac {b x \sqrt {c^2 d x^2+d}}{c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {b x \sqrt {c^2 d x^2+d}}{6 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x*Sqrt[d + c^2*d*x^2])/(6*c^5*d^3*(1 + c^2*x^2)^(3/2)) - (b*x*Sqrt[d + c^2*d*x^2])/(c^5*d^3*Sqrt[1 + c^2*x^
2]) - (a + b*ArcSinh[c*x])/(3*c^6*d*(d + c^2*d*x^2)^(3/2)) + (2*(a + b*ArcSinh[c*x]))/(c^6*d^2*Sqrt[d + c^2*d*
x^2]) + (Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x]))/(c^6*d^3) - (11*b*Sqrt[d + c^2*d*x^2]*ArcTan[c*x])/(6*c^6*d
^3*Sqrt[1 + c^2*x^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 45

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 272

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 396

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 5804

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

Rubi steps

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

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Mathematica [A]
time = 0.16, size = 154, normalized size = 0.73 \begin {gather*} \frac {\sqrt {d+c^2 d x^2} \left (-b c x \sqrt {1+c^2 x^2} \left (5+6 c^2 x^2\right )+2 a \left (8+12 c^2 x^2+3 c^4 x^4\right )+2 b \left (8+12 c^2 x^2+3 c^4 x^4\right ) \sinh ^{-1}(c x)\right )}{6 c^6 d^3 \left (1+c^2 x^2\right )^2}-\frac {11 b \sqrt {d \left (1+c^2 x^2\right )} \text {ArcTan}(c x)}{6 c^6 d^3 \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains complex when optimal does not.
time = 3.14, size = 400, normalized size = 1.90

method	result	size
default	\(a \left (\frac {x^{4}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {4 \left (-\frac {x^{2}}{c^{2} d \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}-\frac {2}{3 d \,c^{4} \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\right )}{c^{2}}\right )+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{c^{4} d^{3} \left (c^{2} x^{2}+1\right )}-\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{c^{5} d^{3} \sqrt {c^{2} x^{2}+1}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{c^{6} d^{3} \left (c^{2} x^{2}+1\right )}+\frac {2 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right ) x^{2}}{d^{3} \left (c^{2} x^{2}+1\right )^{2} c^{4}}+\frac {b \sqrt {d \left (c^{2} x^{2}+1\right )}\, x}{6 d^{3} \left (c^{2} x^{2}+1\right )^{\frac {3}{2}} c^{5}}+\frac {5 b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \arcsinh \left (c x \right )}{3 d^{3} \left (c^{2} x^{2}+1\right )^{2} c^{6}}+\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}-i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{6} d^{3}}-\frac {11 i b \sqrt {d \left (c^{2} x^{2}+1\right )}\, \ln \left (c x +\sqrt {c^{2} x^{2}+1}+i\right )}{6 \sqrt {c^{2} x^{2}+1}\, c^{6} d^{3}}\)	\(400\)

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)^(5/2),x,method=_RETURNVERBOSE)

[Out]

a*(x^4/c^2/d/(c^2*d*x^2+d)^(3/2)-4/c^2*(-x^2/c^2/d/(c^2*d*x^2+d)^(3/2)-2/3/d/c^4/(c^2*d*x^2+d)^(3/2)))+b*(d*(c
^2*x^2+1))^(1/2)/c^4/d^3/(c^2*x^2+1)*arcsinh(c*x)*x^2-b*(d*(c^2*x^2+1))^(1/2)/c^5/d^3/(c^2*x^2+1)^(1/2)*x+b*(d
*(c^2*x^2+1))^(1/2)/c^6/d^3/(c^2*x^2+1)*arcsinh(c*x)+2*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/c^4*arcsinh(c
*x)*x^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^(3/2)/c^5*x+5/3*b*(d*(c^2*x^2+1))^(1/2)/d^3/(c^2*x^2+1)^2/
c^6*arcsinh(c*x)+11/6*I*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*ln(c*x+(c^2*x^2+1)^(1/2)-I)-11/6*I*b
*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^6/d^3*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.44, size = 219, normalized size = 1.04 \begin {gather*} \frac {11 \, {\left (b c^{4} x^{4} + 2 \, 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 ) + 4 \, {\left (3 \, b c^{4} x^{4} + 12 \, 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 (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 )} \sqrt {c^{2} d x^{2} + d}}{12 \, {\left (c^{10} d^{3} x^{4} + 2 \, c^{8} d^{3} x^{2} + c^{6} d^{3}\right )}} \end {gather*}

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)^(5/2),x, algorithm="fricas")

[Out]

1/12*(11*(b*c^4*x^4 + 2*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)) + 4*(3*b*c^4*x^4 + 12*b*c^2*x^2 + 8*b)*sqrt(c^2*d*x^2 + d)*log(c*x + sqrt(c^2*x^2 + 1)) + 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)*sqrt(c^2*d*x^2 + d))/(c^10*d^3*x^4
+ 2*c^8*d^3*x^2 + c^6*d^3)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {5}{2}}}\, dx \end {gather*}

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)**(5/2),x)

[Out]

Integral(x**5*(a + b*asinh(c*x))/(d*(c**2*x**2 + 1))**(5/2), 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(x^5*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),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 = -0.00 \begin {gather*} \int \frac {x^5\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{{\left (d\,c^2\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

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)^(5/2),x)

[Out]

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

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