3.413 \(\int \frac{x^5 (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\)

Optimal. Leaf size=97 \[ -\frac{42 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}+\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{14 c \sqrt{c+d x^3}}{d^2} \]

[Out]

(14*c*Sqrt[c + d*x^3])/d^2 + (14*(c + d*x^3)^(3/2))/(27*d^2) + (8*(c + d*x^3)^(5/2))/(27*d^2*(8*c - d*x^3)) -
(42*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^2

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Rubi [A]  time = 0.0748169, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {446, 78, 50, 63, 206} \[ -\frac{42 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}+\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{14 c \sqrt{c+d x^3}}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]

[Out]

(14*c*Sqrt[c + d*x^3])/d^2 + (14*(c + d*x^3)^(3/2))/(27*d^2) + (8*(c + d*x^3)^(5/2))/(27*d^2*(8*c - d*x^3)) -
(42*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^2

Rule 446

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

Rule 78

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{x^5 \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x (c+d x)^{3/2}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac{7 \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )}{9 d}\\ &=\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac{(7 c) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d}\\ &=\frac{14 c \sqrt{c+d x^3}}{d^2}+\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac{\left (63 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{d}\\ &=\frac{14 c \sqrt{c+d x^3}}{d^2}+\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac{\left (126 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^2}\\ &=\frac{14 c \sqrt{c+d x^3}}{d^2}+\frac{14 \left (c+d x^3\right )^{3/2}}{27 d^2}+\frac{8 \left (c+d x^3\right )^{5/2}}{27 d^2 \left (8 c-d x^3\right )}-\frac{42 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.0520667, size = 90, normalized size = 0.93 \[ \frac{2 \sqrt{c+d x^3} \left (-524 c^2+44 c d x^3+d^2 x^6\right )+378 c^{3/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^2 \left (d x^3-8 c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]

[Out]

(2*Sqrt[c + d*x^3]*(-524*c^2 + 44*c*d*x^3 + d^2*x^6) + 378*c^(3/2)*(8*c - d*x^3)*ArcTanh[Sqrt[c + d*x^3]/(3*Sq
rt[c])])/(9*d^2*(-8*c + d*x^3))

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Maple [C]  time = 0.012, size = 902, normalized size = 9.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)

[Out]

1/d*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*I/d^3*c*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(
-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+
I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))
^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/
3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*
3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha
+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/
2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+8/d*c*(-3/d*c*(d*x^3+c)^(1/2)/(d*x^3-8*c)+2/3*(d*x^3+
c)^(1/2)/d+1/2*I/d^3*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/
(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*
(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_a
lpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*
3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*
I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*
d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(
_Z^3*d-8*c)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.90032, size = 437, normalized size = 4.51 \begin{align*} \left [\frac{189 \,{\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt{c} \log \left (\frac{d x^{3} - 6 \, \sqrt{d x^{3} + c} \sqrt{c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 2 \,{\left (d^{2} x^{6} + 44 \, c d x^{3} - 524 \, c^{2}\right )} \sqrt{d x^{3} + c}}{9 \,{\left (d^{3} x^{3} - 8 \, c d^{2}\right )}}, \frac{2 \,{\left (189 \,{\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (d^{2} x^{6} + 44 \, c d x^{3} - 524 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{9 \,{\left (d^{3} x^{3} - 8 \, c d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")

[Out]

[1/9*(189*(c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 2*(d^2*x^6
+ 44*c*d*x^3 - 524*c^2)*sqrt(d*x^3 + c))/(d^3*x^3 - 8*c*d^2), 2/9*(189*(c*d*x^3 - 8*c^2)*sqrt(-c)*arctan(1/3*
sqrt(d*x^3 + c)*sqrt(-c)/c) + (d^2*x^6 + 44*c*d*x^3 - 524*c^2)*sqrt(d*x^3 + c))/(d^3*x^3 - 8*c*d^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.12488, size = 126, normalized size = 1.3 \begin{align*} \frac{42 \, c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} d^{2}} - \frac{24 \, \sqrt{d x^{3} + c} c^{2}}{{\left (d x^{3} - 8 \, c\right )} d^{2}} + \frac{2 \,{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} d^{4} + 51 \, \sqrt{d x^{3} + c} c d^{4}\right )}}{9 \, d^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")

[Out]

42*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^2) - 24*sqrt(d*x^3 + c)*c^2/((d*x^3 - 8*c)*d^2) + 2/9*
((d*x^3 + c)^(3/2)*d^4 + 51*sqrt(d*x^3 + c)*c*d^4)/d^6