\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{16/3}} \, dx\) [77]

Optimal result

Integrand size = 21, antiderivative size = 211 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {2 (b c-a d) (3 b c+a d) x}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {\left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{455 a^3 b^2 \left (a+b x^3\right )^{7/3}}+\frac {3 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{910 a^4 b^2 \left (a+b x^3\right )^{4/3}}+\frac {9 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{910 a^5 b^2 \sqrt [3]{a+b x^3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{13 a b \left (a+b x^3\right )^{13/3}} \]

[Out]

2/65*(-a*d+b*c)*(a*d+3*b*c)*x/a^2/b^2/(b*x^3+a)^(10/3)+1/455*(2*a^2*d^2+9*a*b*c*d+54*b^2*c^2)*x/a^3/b^2/(b*x^3
+a)^(7/3)+3/910*(2*a^2*d^2+9*a*b*c*d+54*b^2*c^2)*x/a^4/b^2/(b*x^3+a)^(4/3)+9/910*(2*a^2*d^2+9*a*b*c*d+54*b^2*c
^2)*x/a^5/b^2/(b*x^3+a)^(1/3)+1/13*(-a*d+b*c)*x*(d*x^3+c)/a/b/(b*x^3+a)^(13/3)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {424, 393, 198, 197} \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {2 x (b c-a d) (a d+3 b c)}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {9 x \left (2 a^2 d^2+9 a b c d+54 b^2 c^2\right )}{910 a^5 b^2 \sqrt [3]{a+b x^3}}+\frac {3 x \left (2 a^2 d^2+9 a b c d+54 b^2 c^2\right )}{910 a^4 b^2 \left (a+b x^3\right )^{4/3}}+\frac {x \left (2 a^2 d^2+9 a b c d+54 b^2 c^2\right )}{455 a^3 b^2 \left (a+b x^3\right )^{7/3}}+\frac {x \left (c+d x^3\right ) (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}} \]

[In]

Int[(c + d*x^3)^2/(a + b*x^3)^(16/3),x]

[Out]

(2*(b*c - a*d)*(3*b*c + a*d)*x)/(65*a^2*b^2*(a + b*x^3)^(10/3)) + ((54*b^2*c^2 + 9*a*b*c*d + 2*a^2*d^2)*x)/(45
5*a^3*b^2*(a + b*x^3)^(7/3)) + (3*(54*b^2*c^2 + 9*a*b*c*d + 2*a^2*d^2)*x)/(910*a^4*b^2*(a + b*x^3)^(4/3)) + (9
*(54*b^2*c^2 + 9*a*b*c*d + 2*a^2*d^2)*x)/(910*a^5*b^2*(a + b*x^3)^(1/3)) + ((b*c - a*d)*x*(c + d*x^3))/(13*a*b
*(a + b*x^3)^(13/3))

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 198

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 )}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {\int \frac {c (12 b c+a d)+d (9 b c+4 a d) x^3}{\left (a+b x^3\right )^{13/3}} \, dx}{13 a b} \\ & = \frac {2 (b c-a d) (3 b c+a d) x}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {\left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx}{65 a^2 b^2} \\ & = \frac {2 (b c-a d) (3 b c+a d) x}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {\left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{455 a^3 b^2 \left (a+b x^3\right )^{7/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {\left (6 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right )\right ) \int \frac {1}{\left (a+b x^3\right )^{7/3}} \, dx}{455 a^3 b^2} \\ & = \frac {2 (b c-a d) (3 b c+a d) x}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {\left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{455 a^3 b^2 \left (a+b x^3\right )^{7/3}}+\frac {3 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{910 a^4 b^2 \left (a+b x^3\right )^{4/3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {\left (9 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right )\right ) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{910 a^4 b^2} \\ & = \frac {2 (b c-a d) (3 b c+a d) x}{65 a^2 b^2 \left (a+b x^3\right )^{10/3}}+\frac {\left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{455 a^3 b^2 \left (a+b x^3\right )^{7/3}}+\frac {3 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{910 a^4 b^2 \left (a+b x^3\right )^{4/3}}+\frac {9 \left (54 b^2 c^2+9 a b c d+2 a^2 d^2\right ) x}{910 a^5 b^2 \sqrt [3]{a+b x^3}}+\frac {(b c-a d) x \left (c+d x^3\right )}{13 a b \left (a+b x^3\right )^{13/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.56 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.65 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x \left (486 b^4 c^2 x^{12}+81 a b^3 c x^9 \left (26 c+d x^3\right )+65 a^4 \left (14 c^2+7 c d x^3+2 d^2 x^6\right )+39 a^3 b x^3 \left (70 c^2+15 c d x^3+2 d^2 x^6\right )+9 a^2 b^2 x^6 \left (390 c^2+39 c d x^3+2 d^2 x^6\right )\right )}{910 a^5 \left (a+b x^3\right )^{13/3}} \]

[In]

Integrate[(c + d*x^3)^2/(a + b*x^3)^(16/3),x]

[Out]

(x*(486*b^4*c^2*x^12 + 81*a*b^3*c*x^9*(26*c + d*x^3) + 65*a^4*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6) + 39*a^3*b*x^3*
(70*c^2 + 15*c*d*x^3 + 2*d^2*x^6) + 9*a^2*b^2*x^6*(390*c^2 + 39*c*d*x^3 + 2*d^2*x^6)))/(910*a^5*(a + b*x^3)^(1
3/3))

Maple [A] (verified)

Time = 4.06 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {1}{7} d^{2} x^{6}+\frac {1}{2} c d \,x^{3}+c^{2}\right ) a^{4}+3 x^{3} b \left (\frac {1}{35} d^{2} x^{6}+\frac {3}{14} c d \,x^{3}+c^{2}\right ) a^{3}+\frac {27 x^{6} b^{2} \left (\frac {1}{195} d^{2} x^{6}+\frac {1}{10} c d \,x^{3}+c^{2}\right ) a^{2}}{7}+\frac {81 x^{9} b^{3} c \left (\frac {d \,x^{3}}{26}+c \right ) a}{35}+\frac {243 b^{4} c^{2} x^{12}}{455}\right )}{\left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(126\)
gosper	\(\frac {x \left (18 a^{2} b^{2} d^{2} x^{12}+81 a \,b^{3} c d \,x^{12}+486 b^{4} c^{2} x^{12}+78 a^{3} b \,d^{2} x^{9}+351 a^{2} b^{2} c d \,x^{9}+2106 a \,b^{3} c^{2} x^{9}+130 a^{4} d^{2} x^{6}+585 a^{3} b c d \,x^{6}+3510 a^{2} b^{2} c^{2} x^{6}+455 a^{4} c d \,x^{3}+2730 a^{3} b \,c^{2} x^{3}+910 a^{4} c^{2}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(156\)
trager	\(\frac {x \left (18 a^{2} b^{2} d^{2} x^{12}+81 a \,b^{3} c d \,x^{12}+486 b^{4} c^{2} x^{12}+78 a^{3} b \,d^{2} x^{9}+351 a^{2} b^{2} c d \,x^{9}+2106 a \,b^{3} c^{2} x^{9}+130 a^{4} d^{2} x^{6}+585 a^{3} b c d \,x^{6}+3510 a^{2} b^{2} c^{2} x^{6}+455 a^{4} c d \,x^{3}+2730 a^{3} b \,c^{2} x^{3}+910 a^{4} c^{2}\right )}{910 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(156\)

[In]

int((d*x^3+c)^2/(b*x^3+a)^(16/3),x,method=_RETURNVERBOSE)

[Out]

x*((1/7*d^2*x^6+1/2*c*d*x^3+c^2)*a^4+3*x^3*b*(1/35*d^2*x^6+3/14*c*d*x^3+c^2)*a^3+27/7*x^6*b^2*(1/195*d^2*x^6+1
/10*c*d*x^3+c^2)*a^2+81/35*x^9*b^3*c*(1/26*d*x^3+c)*a+243/455*b^4*c^2*x^12)/(b*x^3+a)^(13/3)/a^5

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (9 \, {\left (54 \, b^{4} c^{2} + 9 \, a b^{3} c d + 2 \, a^{2} b^{2} d^{2}\right )} x^{13} + 39 \, {\left (54 \, a b^{3} c^{2} + 9 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{10} + 65 \, {\left (54 \, a^{2} b^{2} c^{2} + 9 \, a^{3} b c d + 2 \, a^{4} d^{2}\right )} x^{7} + 910 \, a^{4} c^{2} x + 455 \, {\left (6 \, a^{3} b c^{2} + a^{4} c d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{910 \, {\left (a^{5} b^{5} x^{15} + 5 \, a^{6} b^{4} x^{12} + 10 \, a^{7} b^{3} x^{9} + 10 \, a^{8} b^{2} x^{6} + 5 \, a^{9} b x^{3} + a^{10}\right )}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(16/3),x, algorithm="fricas")

[Out]

1/910*(9*(54*b^4*c^2 + 9*a*b^3*c*d + 2*a^2*b^2*d^2)*x^13 + 39*(54*a*b^3*c^2 + 9*a^2*b^2*c*d + 2*a^3*b*d^2)*x^1
0 + 65*(54*a^2*b^2*c^2 + 9*a^3*b*c*d + 2*a^4*d^2)*x^7 + 910*a^4*c^2*x + 455*(6*a^3*b*c^2 + a^4*c*d)*x^4)*(b*x^
3 + a)^(2/3)/(a^5*b^5*x^15 + 5*a^6*b^4*x^12 + 10*a^7*b^3*x^9 + 10*a^8*b^2*x^6 + 5*a^9*b*x^3 + a^10)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\text {Timed out} \]

[In]

integrate((d*x**3+c)**2/(b*x**3+a)**(16/3),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.00 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (35 \, b^{2} - \frac {91 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {65 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} d^{2} x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{3}} - \frac {{\left (140 \, b^{3} - \frac {546 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {780 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} c d x^{13}}{910 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{4}} + \frac {{\left (35 \, b^{4} - \frac {182 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {390 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac {455 \, {\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} c^{2} x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{5}} \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(16/3),x, algorithm="maxima")

[Out]

1/455*(35*b^2 - 91*(b*x^3 + a)*b/x^3 + 65*(b*x^3 + a)^2/x^6)*d^2*x^13/((b*x^3 + a)^(13/3)*a^3) - 1/910*(140*b^
3 - 546*(b*x^3 + a)*b^2/x^3 + 780*(b*x^3 + a)^2*b/x^6 - 455*(b*x^3 + a)^3/x^9)*c*d*x^13/((b*x^3 + a)^(13/3)*a^
4) + 1/455*(35*b^4 - 182*(b*x^3 + a)*b^3/x^3 + 390*(b*x^3 + a)^2*b^2/x^6 - 455*(b*x^3 + a)^3*b/x^9 + 455*(b*x^
3 + a)^4/x^12)*c^2*x^13/((b*x^3 + a)^(13/3)*a^5)

Giac [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {16}{3}}} \,d x } \]

[In]

integrate((d*x^3+c)^2/(b*x^3+a)^(16/3),x, algorithm="giac")

[Out]

integrate((d*x^3 + c)^2/(b*x^3 + a)^(16/3), x)

Mupad [B] (verification not implemented)

Time = 5.58 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.03 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x\,\left (\frac {c^2}{13\,a}+\frac {a\,\left (\frac {d^2}{13\,b}-\frac {2\,c\,d}{13\,a}\right )}{b}\right )}{{\left (b\,x^3+a\right )}^{13/3}}-\frac {x\,\left (\frac {d^2}{10\,b^2}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+12\,b^2\,c^2}{130\,a^2\,b^2}\right )}{{\left (b\,x^3+a\right )}^{10/3}}+\frac {x\,\left (2\,a^2\,d^2+9\,a\,b\,c\,d+54\,b^2\,c^2\right )}{455\,a^3\,b^2\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (6\,a^2\,d^2+27\,a\,b\,c\,d+162\,b^2\,c^2\right )}{910\,a^4\,b^2\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (18\,a^2\,d^2+81\,a\,b\,c\,d+486\,b^2\,c^2\right )}{910\,a^5\,b^2\,{\left (b\,x^3+a\right )}^{1/3}} \]

[In]

int((c + d*x^3)^2/(a + b*x^3)^(16/3),x)

[Out]

(x*(c^2/(13*a) + (a*(d^2/(13*b) - (2*c*d)/(13*a)))/b))/(a + b*x^3)^(13/3) - (x*(d^2/(10*b^2) - (12*b^2*c^2 - a
^2*d^2 + 2*a*b*c*d)/(130*a^2*b^2)))/(a + b*x^3)^(10/3) + (x*(2*a^2*d^2 + 54*b^2*c^2 + 9*a*b*c*d))/(455*a^3*b^2
*(a + b*x^3)^(7/3)) + (x*(6*a^2*d^2 + 162*b^2*c^2 + 27*a*b*c*d))/(910*a^4*b^2*(a + b*x^3)^(4/3)) + (x*(18*a^2*
d^2 + 486*b^2*c^2 + 81*a*b*c*d))/(910*a^5*b^2*(a + b*x^3)^(1/3))