\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{5/3}} \, dx\) [83]

Optimal result

Integrand size = 21, antiderivative size = 146 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=-\frac {d (b c-2 a d) x \sqrt [3]{a+b x^3}}{2 a b^2}+\frac {(b c-a d) x \left (c+d x^3\right )}{2 a b \left (a+b x^3\right )^{2/3}}+\frac {\left (b^2 c^2+2 a b c d-2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 a b^2 \left (a+b x^3\right )^{2/3}} \]

[Out]

-1/2*d*(-2*a*d+b*c)*x*(b*x^3+a)^(1/3)/a/b^2+1/2*(-a*d+b*c)*x*(d*x^3+c)/a/b/(b*x^3+a)^(2/3)+1/2*(-2*a^2*d^2+2*a
*b*c*d+b^2*c^2)*x*(1+b*x^3/a)^(2/3)*hypergeom([1/3, 2/3],[4/3],-b*x^3/a)/a/b^2/(b*x^3+a)^(2/3)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {424, 396, 252, 251} \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (-2 a^2 d^2+2 a b c d+b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 a b^2 \left (a+b x^3\right )^{2/3}}-\frac {d x \sqrt [3]{a+b x^3} (b c-2 a d)}{2 a b^2}+\frac {x \left (c+d x^3\right ) (b c-a d)}{2 a b \left (a+b x^3\right )^{2/3}} \]

[In]

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

[Out]

-1/2*(d*(b*c - 2*a*d)*x*(a + b*x^3)^(1/3))/(a*b^2) + ((b*c - a*d)*x*(c + d*x^3))/(2*a*b*(a + b*x^3)^(2/3)) + (
(b^2*c^2 + 2*a*b*c*d - 2*a^2*d^2)*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(2*a
*b^2*(a + b*x^3)^(2/3))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 252

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^Fra
cPart[p]), Int[(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

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 424

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d) x \left (c+d x^3\right )}{2 a b \left (a+b x^3\right )^{2/3}}+\frac {\int \frac {c (b c+a d)-2 d (b c-2 a d) x^3}{\left (a+b x^3\right )^{2/3}} \, dx}{2 a b} \\ & = -\frac {d (b c-2 a d) x \sqrt [3]{a+b x^3}}{2 a b^2}+\frac {(b c-a d) x \left (c+d x^3\right )}{2 a b \left (a+b x^3\right )^{2/3}}+-\frac {(-2 a d (b c-2 a d)-2 b c (b c+a d)) \int \frac {1}{\left (a+b x^3\right )^{2/3}} \, dx}{4 a b^2} \\ & = -\frac {d (b c-2 a d) x \sqrt [3]{a+b x^3}}{2 a b^2}+\frac {(b c-a d) x \left (c+d x^3\right )}{2 a b \left (a+b x^3\right )^{2/3}}+-\frac {\left ((-2 a d (b c-2 a d)-2 b c (b c+a d)) \left (1+\frac {b x^3}{a}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {b x^3}{a}\right )^{2/3}} \, dx}{4 a b^2 \left (a+b x^3\right )^{2/3}} \\ & = -\frac {d (b c-2 a d) x \sqrt [3]{a+b x^3}}{2 a b^2}+\frac {(b c-a d) x \left (c+d x^3\right )}{2 a b \left (a+b x^3\right )^{2/3}}+\frac {\left (b^2 c^2+2 a b c d-2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};-\frac {b x^3}{a}\right )}{2 a b^2 \left (a+b x^3\right )^{2/3}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 12.87 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.17 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \left (4 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {10}{3},-\frac {b x^3}{a}\right )-b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {8}{3},\frac {13}{3},-\frac {b x^3}{a}\right )-3 b x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {4}{3},2,\frac {8}{3};1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{84 a^2 \left (a+b x^3\right )^{2/3} \operatorname {Gamma}\left (\frac {5}{3}\right )} \]

[In]

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

[Out]

(x*(1 + (b*x^3)/a)^(2/3)*Gamma[2/3]*(4*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)*Hypergeometric2F1[1/3, 5/3, 10/3, -(
(b*x^3)/a)] - b*x^3*(11*c^2 + 16*c*d*x^3 + 5*d^2*x^6)*Hypergeometric2F1[4/3, 8/3, 13/3, -((b*x^3)/a)] - 3*b*x^
3*(c + d*x^3)^2*HypergeometricPFQ[{4/3, 2, 8/3}, {1, 13/3}, -((b*x^3)/a)]))/(84*a^2*(a + b*x^3)^(2/3)*Gamma[5/
3])

Maple [F]

\[\int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {5}{3}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

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

[Out]

integral((d^2*x^6 + 2*c*d*x^3 + c^2)*(b*x^3 + a)^(1/3)/(b^2*x^6 + 2*a*b*x^3 + a^2), x)

Sympy [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int \frac {\left (c + d x^{3}\right )^{2}}{\left (a + b x^{3}\right )^{\frac {5}{3}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {5}{3}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{5/3}} \,d x \]

[In]

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

[Out]

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