\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{13/3}} \, dx\) [76]

Optimal result

Integrand size = 21, antiderivative size = 174 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {9 c^2 (9 b c-10 a d) x}{140 a^4 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {3 c (9 b c-10 a d) x \left (c+d x^3\right )}{140 a^3 (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {(9 b c-10 a d) x \left (c+d x^3\right )^2}{70 a^2 (b c-a d) \left (a+b x^3\right )^{7/3}}+\frac {b x \left (c+d x^3\right )^3}{10 a (b c-a d) \left (a+b x^3\right )^{10/3}} \]

[Out]

9/140*c^2*(-10*a*d+9*b*c)*x/a^4/(-a*d+b*c)/(b*x^3+a)^(1/3)+3/140*c*(-10*a*d+9*b*c)*x*(d*x^3+c)/a^3/(-a*d+b*c)/
(b*x^3+a)^(4/3)+1/70*(-10*a*d+9*b*c)*x*(d*x^3+c)^2/a^2/(-a*d+b*c)/(b*x^3+a)^(7/3)+1/10*b*x*(d*x^3+c)^3/a/(-a*d
+b*c)/(b*x^3+a)^(10/3)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {390, 386, 197} \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {9 c^2 x (9 b c-10 a d)}{140 a^4 \sqrt [3]{a+b x^3} (b c-a d)}+\frac {3 c x \left (c+d x^3\right ) (9 b c-10 a d)}{140 a^3 \left (a+b x^3\right )^{4/3} (b c-a d)}+\frac {x \left (c+d x^3\right )^2 (9 b c-10 a d)}{70 a^2 \left (a+b x^3\right )^{7/3} (b c-a d)}+\frac {b x \left (c+d x^3\right )^3}{10 a \left (a+b x^3\right )^{10/3} (b c-a d)} \]

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(13/3),x]

[Out]

(9*c^2*(9*b*c - 10*a*d)*x)/(140*a^4*(b*c - a*d)*(a + b*x^3)^(1/3)) + (3*c*(9*b*c - 10*a*d)*x*(c + d*x^3))/(140
*a^3*(b*c - a*d)*(a + b*x^3)^(4/3)) + ((9*b*c - 10*a*d)*x*(c + d*x^3)^2)/(70*a^2*(b*c - a*d)*(a + b*x^3)^(7/3)
) + (b*x*(c + d*x^3)^3)/(10*a*(b*c - a*d)*(a + b*x^3)^(10/3))

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]

Rule 390

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

Rubi steps \begin{align*} \text {integral}& = \frac {b x \left (c+d x^3\right )^3}{10 a (b c-a d) \left (a+b x^3\right )^{10/3}}+\frac {(9 b c-10 a d) \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{10/3}} \, dx}{10 a (b c-a d)} \\ & = \frac {(9 b c-10 a d) x \left (c+d x^3\right )^2}{70 a^2 (b c-a d) \left (a+b x^3\right )^{7/3}}+\frac {b x \left (c+d x^3\right )^3}{10 a (b c-a d) \left (a+b x^3\right )^{10/3}}+\frac {(3 c (9 b c-10 a d)) \int \frac {c+d x^3}{\left (a+b x^3\right )^{7/3}} \, dx}{35 a^2 (b c-a d)} \\ & = \frac {3 c (9 b c-10 a d) x \left (c+d x^3\right )}{140 a^3 (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {(9 b c-10 a d) x \left (c+d x^3\right )^2}{70 a^2 (b c-a d) \left (a+b x^3\right )^{7/3}}+\frac {b x \left (c+d x^3\right )^3}{10 a (b c-a d) \left (a+b x^3\right )^{10/3}}+\frac {\left (9 c^2 (9 b c-10 a d)\right ) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{140 a^3 (b c-a d)} \\ & = \frac {9 c^2 (9 b c-10 a d) x}{140 a^4 (b c-a d) \sqrt [3]{a+b x^3}}+\frac {3 c (9 b c-10 a d) x \left (c+d x^3\right )}{140 a^3 (b c-a d) \left (a+b x^3\right )^{4/3}}+\frac {(9 b c-10 a d) x \left (c+d x^3\right )^2}{70 a^2 (b c-a d) \left (a+b x^3\right )^{7/3}}+\frac {b x \left (c+d x^3\right )^3}{10 a (b c-a d) \left (a+b x^3\right )^{10/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.61 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x \left (81 b^3 c^2 x^9+18 a b^2 c x^6 \left (15 c+d x^3\right )+10 a^3 \left (14 c^2+7 c d x^3+2 d^2 x^6\right )+3 a^2 b x^3 \left (105 c^2+20 c d x^3+2 d^2 x^6\right )\right )}{140 a^4 \left (a+b x^3\right )^{10/3}} \]

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(13/3),x]

[Out]

(x*(81*b^3*c^2*x^9 + 18*a*b^2*c*x^6*(15*c + d*x^3) + 10*a^3*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6) + 3*a^2*b*x^3*(10
5*c^2 + 20*c*d*x^3 + 2*d^2*x^6)))/(140*a^4*(a + b*x^3)^(10/3))

Maple [A] (verified)

Time = 4.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.55

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {1}{7} d^{2} x^{6}+\frac {1}{2} c d \,x^{3}+c^{2}\right ) a^{3}+\frac {9 x^{3} b \left (\frac {2}{105} d^{2} x^{6}+\frac {4}{21} c d \,x^{3}+c^{2}\right ) a^{2}}{4}+\frac {27 x^{6} \left (\frac {d \,x^{3}}{15}+c \right ) b^{2} c a}{14}+\frac {81 b^{3} c^{2} x^{9}}{140}\right )}{\left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(96\)
gosper	\(\frac {x \left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 a^{3} c^{2}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(115\)
trager	\(\frac {x \left (6 a^{2} b \,d^{2} x^{9}+18 a \,b^{2} c d \,x^{9}+81 b^{3} c^{2} x^{9}+20 a^{3} d^{2} x^{6}+60 a^{2} b c d \,x^{6}+270 a \,b^{2} c^{2} x^{6}+70 a^{3} c d \,x^{3}+315 a^{2} b \,c^{2} x^{3}+140 a^{3} c^{2}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(115\)

[In]

int((d*x^3+c)^2/(b*x^3+a)^(13/3),x,method=_RETURNVERBOSE)

[Out]

x*((1/7*d^2*x^6+1/2*c*d*x^3+c^2)*a^3+9/4*x^3*b*(2/105*d^2*x^6+4/21*c*d*x^3+c^2)*a^2+27/14*x^6*(1/15*d*x^3+c)*b
^2*c*a+81/140*b^3*c^2*x^9)/(b*x^3+a)^(10/3)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.87 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {{\left (3 \, {\left (27 \, b^{3} c^{2} + 6 \, a b^{2} c d + 2 \, a^{2} b d^{2}\right )} x^{10} + 10 \, {\left (27 \, a b^{2} c^{2} + 6 \, a^{2} b c d + 2 \, a^{3} d^{2}\right )} x^{7} + 140 \, a^{3} c^{2} x + 35 \, {\left (9 \, a^{2} b c^{2} + 2 \, a^{3} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{4} b^{4} x^{12} + 4 \, a^{5} b^{3} x^{9} + 6 \, a^{6} b^{2} x^{6} + 4 \, a^{7} b x^{3} + a^{8}\right )}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(13/3),x, algorithm="fricas")

[Out]

1/140*(3*(27*b^3*c^2 + 6*a*b^2*c*d + 2*a^2*b*d^2)*x^10 + 10*(27*a*b^2*c^2 + 6*a^2*b*c*d + 2*a^3*d^2)*x^7 + 140
*a^3*c^2*x + 35*(9*a^2*b*c^2 + 2*a^3*c*d)*x^4)*(b*x^3 + a)^(2/3)/(a^4*b^4*x^12 + 4*a^5*b^3*x^9 + 6*a^6*b^2*x^6
+ 4*a^7*b*x^3 + a^8)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\text {Timed out} \]

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(13/3),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=-\frac {{\left (7 \, b - \frac {10 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} d^{2} x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{2}} + \frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} c d x^{10}}{70 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{3}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} c^{2} x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{4}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(13/3),x, algorithm="maxima")

[Out]

-1/70*(7*b - 10*(b*x^3 + a)/x^3)*d^2*x^10/((b*x^3 + a)^(10/3)*a^2) + 1/70*(14*b^2 - 40*(b*x^3 + a)*b/x^3 + 35*
(b*x^3 + a)^2/x^6)*c*d*x^10/((b*x^3 + a)^(10/3)*a^3) - 1/140*(14*b^3 - 60*(b*x^3 + a)*b^2/x^3 + 105*(b*x^3 + a
)^2*b/x^6 - 140*(b*x^3 + a)^3/x^9)*c^2*x^10/((b*x^3 + a)^(10/3)*a^4)

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {13}{3}}} \,d x } \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(13/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(13/3), x)

Mupad [B] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x\,\left (\frac {c^2}{10\,a}+\frac {a\,\left (\frac {d^2}{10\,b}-\frac {c\,d}{5\,a}\right )}{b}\right )}{{\left (b\,x^3+a\right )}^{10/3}}-\frac {x\,\left (\frac {d^2}{7\,b^2}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+9\,b^2\,c^2}{70\,a^2\,b^2}\right )}{{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (2\,a^2\,d^2+6\,a\,b\,c\,d+27\,b^2\,c^2\right )}{140\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (6\,a^2\,d^2+18\,a\,b\,c\,d+81\,b^2\,c^2\right )}{140\,a^4\,b^2\,{\left (b\,x^3+a\right )}^{1/3}} \]

[In]

int((c + d*x^3)^2/(a + b*x^3)^(13/3),x)

[Out]

(x*(c^2/(10*a) + (a*(d^2/(10*b) - (c*d)/(5*a)))/b))/(a + b*x^3)^(10/3) - (x*(d^2/(7*b^2) - (9*b^2*c^2 - a^2*d^
2 + 2*a*b*c*d)/(70*a^2*b^2)))/(a + b*x^3)^(7/3) + (x*(2*a^2*d^2 + 27*b^2*c^2 + 6*a*b*c*d))/(140*a^3*b^2*(a + b
*x^3)^(4/3)) + (x*(6*a^2*d^2 + 81*b^2*c^2 + 18*a*b*c*d))/(140*a^4*b^2*(a + b*x^3)^(1/3))