\(\int \frac {c+d x^3}{(a+b x^3)^{13/3}} \, dx\) [62]

Optimal result

Integrand size = 19, antiderivative size = 121 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {(b c-a d) x}{10 a b \left (a+b x^3\right )^{10/3}}+\frac {(9 b c+a d) x}{70 a^2 b \left (a+b x^3\right )^{7/3}}+\frac {3 (9 b c+a d) x}{140 a^3 b \left (a+b x^3\right )^{4/3}}+\frac {9 (9 b c+a d) x}{140 a^4 b \sqrt [3]{a+b x^3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {393, 198, 197} \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {9 x (a d+9 b c)}{140 a^4 b \sqrt [3]{a+b x^3}}+\frac {3 x (a d+9 b c)}{140 a^3 b \left (a+b x^3\right )^{4/3}}+\frac {x (a d+9 b c)}{70 a^2 b \left (a+b x^3\right )^{7/3}}+\frac {x (b c-a d)}{10 a b \left (a+b x^3\right )^{10/3}} \]

[In]

Int[(c + d*x^3)/(a + b*x^3)^(13/3),x]

[Out]

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

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])

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x \left (81 b^3 c x^9+35 a^3 \left (4 c+d x^3\right )+9 a b^2 x^6 \left (30 c+d x^3\right )+15 a^2 b x^3 \left (21 c+2 d x^3\right )\right )}{140 a^4 \left (a+b x^3\right )^{10/3}} \]

[In]

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

[Out]

(x*(81*b^3*c*x^9 + 35*a^3*(4*c + d*x^3) + 9*a*b^2*x^6*(30*c + d*x^3) + 15*a^2*b*x^3*(21*c + 2*d*x^3)))/(140*a^
4*(a + b*x^3)^(10/3))

Maple [A] (verified)

Time = 3.93 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.59

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {d \,x^{3}}{4}+c \right ) a^{3}+\frac {9 x^{3} b \left (\frac {2 d \,x^{3}}{21}+c \right ) a^{2}}{4}+\frac {27 x^{6} b^{2} \left (\frac {d \,x^{3}}{30}+c \right ) a}{14}+\frac {81 b^{3} c \,x^{9}}{140}\right )}{\left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(71\)
gosper	\(\frac {x \left (9 a \,b^{2} d \,x^{9}+81 b^{3} c \,x^{9}+30 a^{2} b d \,x^{6}+270 a \,b^{2} c \,x^{6}+35 a^{3} d \,x^{3}+315 a^{2} x^{3} b c +140 c \,a^{3}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(81\)
trager	\(\frac {x \left (9 a \,b^{2} d \,x^{9}+81 b^{3} c \,x^{9}+30 a^{2} b d \,x^{6}+270 a \,b^{2} c \,x^{6}+35 a^{3} d \,x^{3}+315 a^{2} x^{3} b c +140 c \,a^{3}\right )}{140 \left (b \,x^{3}+a \right )^{\frac {10}{3}} a^{4}}\)	\(81\)

[In]

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

[Out]

x/(b*x^3+a)^(10/3)*((1/4*d*x^3+c)*a^3+9/4*x^3*b*(2/21*d*x^3+c)*a^2+27/14*x^6*b^2*(1/30*d*x^3+c)*a+81/140*b^3*c
*x^9)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {{\left (9 \, {\left (9 \, b^{3} c + a b^{2} d\right )} x^{10} + 30 \, {\left (9 \, a b^{2} c + a^{2} b d\right )} x^{7} + 140 \, a^{3} c x + 35 \, {\left (9 \, a^{2} b c + a^{3} d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{140 \, {\left (a^{4} b^{4} x^{12} + 4 \, a^{5} b^{3} x^{9} + 6 \, a^{6} b^{2} x^{6} + 4 \, a^{7} b x^{3} + a^{8}\right )}} \]

[In]

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

[Out]

1/140*(9*(9*b^3*c + a*b^2*d)*x^10 + 30*(9*a*b^2*c + a^2*b*d)*x^7 + 140*a^3*c*x + 35*(9*a^2*b*c + a^3*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. \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.99 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {{\left (14 \, b^{2} - \frac {40 \, {\left (b x^{3} + a\right )} b}{x^{3}} + \frac {35 \, {\left (b x^{3} + a\right )}^{2}}{x^{6}}\right )} d x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{3}} - \frac {{\left (14 \, b^{3} - \frac {60 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {105 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {140 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} c x^{10}}{140 \, {\left (b x^{3} + a\right )}^{\frac {10}{3}} a^{4}} \]

[In]

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

[Out]

1/140*(14*b^2 - 40*(b*x^3 + a)*b/x^3 + 35*(b*x^3 + a)^2/x^6)*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*x^10/((b*x^3 + a)^(10/3)*a^4)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.51 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{13/3}} \, dx=\frac {x\,\left (\frac {c}{10\,a}-\frac {d}{10\,b}\right )}{{\left (b\,x^3+a\right )}^{10/3}}+\frac {x\,\left (a\,d+9\,b\,c\right )}{70\,a^2\,b\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (3\,a\,d+27\,b\,c\right )}{140\,a^3\,b\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (9\,a\,d+81\,b\,c\right )}{140\,a^4\,b\,{\left (b\,x^3+a\right )}^{1/3}} \]

[In]

int((c + d*x^3)/(a + b*x^3)^(13/3),x)

[Out]

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