\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{10/3}} \, dx\) [75]

Optimal result

Integrand size = 21, antiderivative size = 78 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {9 c^2 x}{14 a^3 \sqrt [3]{a+b x^3}}+\frac {3 c x \left (c+d x^3\right )}{14 a^2 \left (a+b x^3\right )^{4/3}}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}} \]

[Out]

9/14*c^2*x/a^3/(b*x^3+a)^(1/3)+3/14*c*x*(d*x^3+c)/a^2/(b*x^3+a)^(4/3)+1/7*x*(d*x^3+c)^2/a/(b*x^3+a)^(7/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {386, 197} \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {9 c^2 x}{14 a^3 \sqrt [3]{a+b x^3}}+\frac {3 c x \left (c+d x^3\right )}{14 a^2 \left (a+b x^3\right )^{4/3}}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}} \]

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(10/3),x]

[Out]

(9*c^2*x)/(14*a^3*(a + b*x^3)^(1/3)) + (3*c*x*(c + d*x^3))/(14*a^2*(a + b*x^3)^(4/3)) + (x*(c + d*x^3)^2)/(7*a
*(a + b*x^3)^(7/3))

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 386

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

Rubi steps \begin{align*} \text {integral}& = \frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}}+\frac {(6 c) \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx}{7 a} \\ & = \frac {3 c x \left (c+d x^3\right )}{14 a^2 \left (a+b x^3\right )^{4/3}}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}}+\frac {\left (9 c^2\right ) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{14 a^2} \\ & = \frac {9 c^2 x}{14 a^3 \sqrt [3]{a+b x^3}}+\frac {3 c x \left (c+d x^3\right )}{14 a^2 \left (a+b x^3\right )^{4/3}}+\frac {x \left (c+d x^3\right )^2}{7 a \left (a+b x^3\right )^{7/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {9 b^2 c^2 x^7+3 a b c x^4 \left (7 c+d x^3\right )+a^2 \left (14 c^2 x+7 c d x^4+2 d^2 x^7\right )}{14 a^3 \left (a+b x^3\right )^{7/3}} \]

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(10/3),x]

[Out]

(9*b^2*c^2*x^7 + 3*a*b*c*x^4*(7*c + d*x^3) + a^2*(14*c^2*x + 7*c*d*x^4 + 2*d^2*x^7))/(14*a^3*(a + b*x^3)^(7/3)
)

Maple [A] (verified)

Time = 4.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.91

method	result	size
pseudoelliptic	\(\frac {\left (2 a^{2} d^{2}+3 a b c d +9 b^{2} c^{2}\right ) x^{7}+7 \left (a^{2} c d +3 b \,c^{2} a \right ) x^{4}+14 a^{2} c^{2} x}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(71\)
gosper	\(\frac {x \left (2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+9 b^{2} c^{2} x^{6}+7 a^{2} c d \,x^{3}+21 a b \,c^{2} x^{3}+14 a^{2} c^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(76\)
trager	\(\frac {x \left (2 a^{2} d^{2} x^{6}+3 a b c d \,x^{6}+9 b^{2} c^{2} x^{6}+7 a^{2} c d \,x^{3}+21 a b \,c^{2} x^{3}+14 a^{2} c^{2}\right )}{14 \left (b \,x^{3}+a \right )^{\frac {7}{3}} a^{3}}\)	\(76\)

[In]

int((d*x^3+c)^2/(b*x^3+a)^(10/3),x,method=_RETURNVERBOSE)

[Out]

1/14*((2*a^2*d^2+3*a*b*c*d+9*b^2*c^2)*x^7+7*(a^2*c*d+3*a*b*c^2)*x^4+14*a^2*c^2*x)/(b*x^3+a)^(7/3)/a^3

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.32 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {{\left ({\left (9 \, b^{2} c^{2} + 3 \, a b c d + 2 \, a^{2} d^{2}\right )} x^{7} + 14 \, a^{2} c^{2} x + 7 \, {\left (3 \, a b c^{2} + a^{2} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{14 \, {\left (a^{3} b^{3} x^{9} + 3 \, a^{4} b^{2} x^{6} + 3 \, a^{5} b x^{3} + a^{6}\right )}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="fricas")

[Out]

1/14*((9*b^2*c^2 + 3*a*b*c*d + 2*a^2*d^2)*x^7 + 14*a^2*c^2*x + 7*(3*a*b*c^2 + a^2*c*d)*x^4)*(b*x^3 + a)^(2/3)/
(a^3*b^3*x^9 + 3*a^4*b^2*x^6 + 3*a^5*b*x^3 + a^6)

Sympy [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\int \frac {\left (c + d x^{3}\right )^{2}}{\left (a + b x^{3}\right )^{\frac {10}{3}}}\, dx \]

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(10/3),x)

[Out]

Integral((c + d*x**3)**2/(a + b*x**3)**(10/3), x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=-\frac {{\left (4 \, b - \frac {7 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} c d x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{2}} + \frac {d^{2} x^{7}}{7 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a} + \frac {{\left (2 \, b^{2} - \frac {7 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {14 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c^{2} x^{7}}{14 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}} a^{3}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="maxima")

[Out]

-1/14*(4*b - 7*(b*x^3 + a)/x^3)*c*d*x^7/((b*x^3 + a)^(7/3)*a^2) + 1/7*d^2*x^7/((b*x^3 + a)^(7/3)*a) + 1/14*(2*
b^2 - 7*(b*x^3 + a)*b/x^3 + 14*(b*x^3 + a)^2/x^6)*c^2*x^7/((b*x^3 + a)^(7/3)*a^3)

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {10}{3}}} \,d x } \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(10/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(10/3), x)

Mupad [B] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.90 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx=\frac {2\,a^4\,d^2\,x+2\,a^2\,d^2\,x\,{\left (b\,x^3+a\right )}^2+9\,b^2\,c^2\,x\,{\left (b\,x^3+a\right )}^2+2\,a^2\,b^2\,c^2\,x-4\,a^3\,d^2\,x\,\left (b\,x^3+a\right )+3\,a\,b^2\,c^2\,x\,\left (b\,x^3+a\right )-4\,a^3\,b\,c\,d\,x+3\,a\,b\,c\,d\,x\,{\left (b\,x^3+a\right )}^2+a^2\,b\,c\,d\,x\,\left (b\,x^3+a\right )}{14\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{7/3}} \]

[In]

int((c + d*x^3)^2/(a + b*x^3)^(10/3),x)

[Out]

(2*a^4*d^2*x + 2*a^2*d^2*x*(a + b*x^3)^2 + 9*b^2*c^2*x*(a + b*x^3)^2 + 2*a^2*b^2*c^2*x - 4*a^3*d^2*x*(a + b*x^
3) + 3*a*b^2*c^2*x*(a + b*x^3) - 4*a^3*b*c*d*x + 3*a*b*c*d*x*(a + b*x^3)^2 + a^2*b*c*d*x*(a + b*x^3))/(14*a^3*
b^2*(a + b*x^3)^(7/3))