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

Optimal. Leaf size=88 \[ -\frac{48 c^2 \sqrt{c+d x^3}}{d^2}+\frac{144 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2} \]

[Out]

(-48*c^2*Sqrt[c + d*x^3])/d^2 - (16*c*(c + d*x^3)^(3/2))/(9*d^2) - (2*(c + d*x^3)^(5/2))/(15*d^2) + (144*c^(5/
2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^2

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Rubi [A]  time = 0.0716315, antiderivative size = 88, 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, 80, 50, 63, 206} \[ -\frac{48 c^2 \sqrt{c+d x^3}}{d^2}+\frac{144 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-48*c^2*Sqrt[c + d*x^3])/d^2 - (16*c*(c + d*x^3)^(3/2))/(9*d^2) - (2*(c + d*x^3)^(5/2))/(15*d^2) + (144*c^(5/
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 80

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

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}}{8 c-d x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x (c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )\\ &=-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac{(8 c) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )}{3 d}\\ &=-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac{\left (24 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d}\\ &=-\frac{48 c^2 \sqrt{c+d x^3}}{d^2}-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac{\left (216 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{d}\\ &=-\frac{48 c^2 \sqrt{c+d x^3}}{d^2}-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac{\left (432 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^2}\\ &=-\frac{48 c^2 \sqrt{c+d x^3}}{d^2}-\frac{16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac{144 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^2}\\ \end{align*}

Mathematica [A]  time = 0.0359215, size = 70, normalized size = 0.8 \[ \frac{6480 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-2 \sqrt{c+d x^3} \left (1123 c^2+46 c d x^3+3 d^2 x^6\right )}{45 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(1123*c^2 + 46*c*d*x^3 + 3*d^2*x^6) + 6480*c^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(
45*d^2)

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Maple [C]  time = 0.01, size = 462, normalized size = 5.3 \begin{align*} -{\frac{2}{15\,{d}^{2}} \left ( d{x}^{3}+c \right ) ^{{\frac{5}{2}}}}-8\,{\frac{c}{d} \left ( 2/9\,{x}^{3}\sqrt{d{x}^{3}+c}+{\frac{56\,c\sqrt{d{x}^{3}+c}}{9\,d}}+{\frac{3\,ic\sqrt{2}}{{d}^{3}}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{3}d-8\,c \right ) }{\frac{\sqrt [3]{-{d}^{2}c} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{2/3} \right ) }{\sqrt{d{x}^{3}+c}}\sqrt{{\frac{i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{-i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{d}{-3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c}} \left ( x-{\frac{\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{-i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}{\it EllipticPi} \left ( 1/3\,\sqrt{3}\sqrt{{\frac{i\sqrt{3}d}{\sqrt [3]{-{d}^{2}c}} \left ( x+1/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}-{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) }},-1/18\,{\frac{2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd}{cd}},\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d} \left ( -3/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}+{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) ^{-1}}} \right ) }} \right ) } \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),x)

[Out]

-2/15*(d*x^3+c)^(5/2)/d^2-8/d*c*(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)))

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35718, size = 363, normalized size = 4.12 \begin{align*} \left [\frac{2 \,{\left (1620 \, c^{\frac{5}{2}} \log \left (\frac{d x^{3} + 6 \, \sqrt{d x^{3} + c} \sqrt{c} + 10 \, c}{d x^{3} - 8 \, c}\right ) -{\left (3 \, d^{2} x^{6} + 46 \, c d x^{3} + 1123 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{2}}, -\frac{2 \,{\left (3240 \, \sqrt{-c} c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (3 \, d^{2} x^{6} + 46 \, c d x^{3} + 1123 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{2}}\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),x, algorithm="fricas")

[Out]

[2/45*(1620*c^(5/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (3*d^2*x^6 + 46*c*d*x^3 +
1123*c^2)*sqrt(d*x^3 + c))/d^2, -2/45*(3240*sqrt(-c)*c^2*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (3*d^2*x^6 +
46*c*d*x^3 + 1123*c^2)*sqrt(d*x^3 + c))/d^2]

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Sympy [A]  time = 78.0689, size = 90, normalized size = 1.02 \begin{align*} - \frac{144 c^{3} \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{3 \sqrt{- c}} \right )}}{d^{2} \sqrt{- c}} - \frac{48 c^{2} \sqrt{c + d x^{3}}}{d^{2}} - \frac{16 c \left (c + d x^{3}\right )^{\frac{3}{2}}}{9 d^{2}} - \frac{2 \left (c + d x^{3}\right )^{\frac{5}{2}}}{15 d^{2}} \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),x)

[Out]

-144*c**3*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(d**2*sqrt(-c)) - 48*c**2*sqrt(c + d*x**3)/d**2 - 16*c*(c + d*x*
*3)**(3/2)/(9*d**2) - 2*(c + d*x**3)**(5/2)/(15*d**2)

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Giac [A]  time = 1.13788, size = 112, normalized size = 1.27 \begin{align*} -\frac{144 \, c^{3} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{\sqrt{-c} d^{2}} - \frac{2 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} d^{8} + 40 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c d^{8} + 1080 \, \sqrt{d x^{3} + c} c^{2} d^{8}\right )}}{45 \, d^{10}} \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),x, algorithm="giac")

[Out]

-144*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^2) - 2/45*(3*(d*x^3 + c)^(5/2)*d^8 + 40*(d*x^3 + c)^
(3/2)*c*d^8 + 1080*sqrt(d*x^3 + c)*c^2*d^8)/d^10