3.5.13 \(\int \frac {x^5 (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\) [413]

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

[Out]

14/27*(d*x^3+c)^(3/2)/d^2+8/27*(d*x^3+c)^(5/2)/d^2/(-d*x^3+8*c)-42*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2)
)/d^2+14*c*(d*x^3+c)^(1/2)/d^2

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Rubi [A]
time = 0.05, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 = {457, 79, 52, 65, 212} \begin {gather*} -\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} \end {gather*}

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 52

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 65

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 79

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] || L
tQ[p, n]))))

Rule 212

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

Rule 457

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]]

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} \text {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 \text {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) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.11, size = 81, normalized size = 0.84 \begin {gather*} \frac {2 \sqrt {c+d x^3} \left (-524 c^2+44 c d x^3+d^2 x^6\right )}{9 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 {gather*}

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))/(9*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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.40, size = 903, normalized size = 9.31

method	result	size
elliptic	\(\frac {24 c^{2} \sqrt {d \,x^{3}+c}}{d^{2} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d}+\frac {104 c \sqrt {d \,x^{3}+c}}{9 d^{2}}+\frac {7 i c \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{d^{4}}\)	\(473\)
risch	\(\text {Expression too large to display}\)	\(890\)
default	\(\text {Expression too large to display}\)	\(903\)

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,method=_RETURNVERBOSE)

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

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Maxima [A]
time = 0.50, size = 93, normalized size = 0.96 \begin {gather*} \frac {189 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 102 \, \sqrt {d x^{3} + c} c - \frac {216 \, \sqrt {d x^{3} + c} c^{2}}{d x^{3} - 8 \, c}}{9 \, d^{2}} \end {gather*}

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]

1/9*(189*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 2*(d*x^3 + c)^(3/2) + 102*
sqrt(d*x^3 + c)*c - 216*sqrt(d*x^3 + c)*c^2/(d*x^3 - 8*c))/d^2

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Fricas [A]
time = 4.15, size = 192, normalized size = 1.98 \begin {gather*} \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 {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \end {gather*}

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]

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

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Giac [A]
time = 0.55, size = 93, normalized size = 0.96 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 4.04, size = 107, normalized size = 1.10 \begin {gather*} \frac {104\,c\,\sqrt {d\,x^3+c}}{9\,d^2}+\frac {21\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^2}+\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d}+\frac {24\,c^2\,\sqrt {d\,x^3+c}}{d^2\,\left (8\,c-d\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(104*c*(c + d*x^3)^(1/2))/(9*d^2) + (21*c^(3/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)
))/d^2 + (2*x^3*(c + d*x^3)^(1/2))/(9*d) + (24*c^2*(c + d*x^3)^(1/2))/(d^2*(8*c - d*x^3))

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