\(\int \frac {c+d x^3}{(a+b x^3)^{16/3}} \, dx\) [63]

Optimal result

Integrand size = 19, antiderivative size = 151 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac {81 (12 b c+a d) x}{1820 a^5 b \sqrt [3]{a+b x^3}} \]

[Out]

1/13*(-a*d+b*c)*x/a/b/(b*x^3+a)^(13/3)+1/130*(a*d+12*b*c)*x/a^2/b/(b*x^3+a)^(10/3)+9/910*(a*d+12*b*c)*x/a^3/b/
(b*x^3+a)^(7/3)+27/1820*(a*d+12*b*c)*x/a^4/b/(b*x^3+a)^(4/3)+81/1820*(a*d+12*b*c)*x/a^5/b/(b*x^3+a)^(1/3)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {393, 198, 197} \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {81 x (a d+12 b c)}{1820 a^5 b \sqrt [3]{a+b x^3}}+\frac {27 x (a d+12 b c)}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac {9 x (a d+12 b c)}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {x (a d+12 b c)}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {x (b c-a d)}{13 a b \left (a+b x^3\right )^{13/3}} \]

[In]

Int[(c + d*x^3)/(a + b*x^3)^(16/3),x]

[Out]

((b*c - a*d)*x)/(13*a*b*(a + b*x^3)^(13/3)) + ((12*b*c + a*d)*x)/(130*a^2*b*(a + b*x^3)^(10/3)) + (9*(12*b*c +
a*d)*x)/(910*a^3*b*(a + b*x^3)^(7/3)) + (27*(12*b*c + a*d)*x)/(1820*a^4*b*(a + b*x^3)^(4/3)) + (81*(12*b*c +
a*d)*x)/(1820*a^5*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}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) \int \frac {1}{\left (a+b x^3\right )^{13/3}} \, dx}{13 a b} \\ & = \frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {(9 (12 b c+a d)) \int \frac {1}{\left (a+b x^3\right )^{10/3}} \, dx}{130 a^2 b} \\ & = \frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {(27 (12 b c+a d)) \int \frac {1}{\left (a+b x^3\right )^{7/3}} \, dx}{455 a^3 b} \\ & = \frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac {(81 (12 b c+a d)) \int \frac {1}{\left (a+b x^3\right )^{4/3}} \, dx}{1820 a^4 b} \\ & = \frac {(b c-a d) x}{13 a b \left (a+b x^3\right )^{13/3}}+\frac {(12 b c+a d) x}{130 a^2 b \left (a+b x^3\right )^{10/3}}+\frac {9 (12 b c+a d) x}{910 a^3 b \left (a+b x^3\right )^{7/3}}+\frac {27 (12 b c+a d) x}{1820 a^4 b \left (a+b x^3\right )^{4/3}}+\frac {81 (12 b c+a d) x}{1820 a^5 b \sqrt [3]{a+b x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x \left (972 b^4 c x^{12}+455 a^4 \left (4 c+d x^3\right )+351 a^2 b^2 x^6 \left (20 c+d x^3\right )+81 a b^3 x^9 \left (52 c+d x^3\right )+195 a^3 b x^3 \left (28 c+3 d x^3\right )\right )}{1820 a^5 \left (a+b x^3\right )^{13/3}} \]

[In]

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

[Out]

(x*(972*b^4*c*x^12 + 455*a^4*(4*c + d*x^3) + 351*a^2*b^2*x^6*(20*c + d*x^3) + 81*a*b^3*x^9*(52*c + d*x^3) + 19
5*a^3*b*x^3*(28*c + 3*d*x^3)))/(1820*a^5*(a + b*x^3)^(13/3))

Maple [A] (verified)

Time = 4.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(\frac {x \left (\left (\frac {d \,x^{3}}{4}+c \right ) a^{4}+3 \left (\frac {3 d \,x^{3}}{28}+c \right ) x^{3} b \,a^{3}+\frac {27 x^{6} b^{2} \left (\frac {d \,x^{3}}{20}+c \right ) a^{2}}{7}+\frac {81 x^{9} \left (\frac {d \,x^{3}}{52}+c \right ) b^{3} a}{35}+\frac {243 b^{4} c \,x^{12}}{455}\right )}{\left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(90\)
gosper	\(\frac {x \left (81 a \,b^{3} d \,x^{12}+972 b^{4} c \,x^{12}+351 a^{2} b^{2} d \,x^{9}+4212 a \,b^{3} c \,x^{9}+585 a^{3} b d \,x^{6}+7020 a^{2} b^{2} c \,x^{6}+455 a^{4} d \,x^{3}+5460 a^{3} b c \,x^{3}+1820 a^{4} c \right )}{1820 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(105\)
trager	\(\frac {x \left (81 a \,b^{3} d \,x^{12}+972 b^{4} c \,x^{12}+351 a^{2} b^{2} d \,x^{9}+4212 a \,b^{3} c \,x^{9}+585 a^{3} b d \,x^{6}+7020 a^{2} b^{2} c \,x^{6}+455 a^{4} d \,x^{3}+5460 a^{3} b c \,x^{3}+1820 a^{4} c \right )}{1820 \left (b \,x^{3}+a \right )^{\frac {13}{3}} a^{5}}\)	\(105\)

[In]

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

[Out]

x*((1/4*d*x^3+c)*a^4+3*(3/28*d*x^3+c)*x^3*b*a^3+27/7*x^6*b^2*(1/20*d*x^3+c)*a^2+81/35*x^9*(1/52*d*x^3+c)*b^3*a
+243/455*b^4*c*x^12)/(b*x^3+a)^(13/3)/a^5

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {{\left (81 \, {\left (12 \, b^{4} c + a b^{3} d\right )} x^{13} + 351 \, {\left (12 \, a b^{3} c + a^{2} b^{2} d\right )} x^{10} + 585 \, {\left (12 \, a^{2} b^{2} c + a^{3} b d\right )} x^{7} + 1820 \, a^{4} c x + 455 \, {\left (12 \, a^{3} b c + a^{4} d\right )} x^{4}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{1820 \, {\left (a^{5} b^{5} x^{15} + 5 \, a^{6} b^{4} x^{12} + 10 \, a^{7} b^{3} x^{9} + 10 \, a^{8} b^{2} x^{6} + 5 \, a^{9} b x^{3} + a^{10}\right )}} \]

[In]

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

[Out]

1/1820*(81*(12*b^4*c + a*b^3*d)*x^13 + 351*(12*a*b^3*c + a^2*b^2*d)*x^10 + 585*(12*a^2*b^2*c + a^3*b*d)*x^7 +
1820*a^4*c*x + 455*(12*a^3*b*c + a^4*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. \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=-\frac {{\left (140 \, b^{3} - \frac {546 \, {\left (b x^{3} + a\right )} b^{2}}{x^{3}} + \frac {780 \, {\left (b x^{3} + a\right )}^{2} b}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3}}{x^{9}}\right )} d x^{13}}{1820 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{4}} + \frac {{\left (35 \, b^{4} - \frac {182 \, {\left (b x^{3} + a\right )} b^{3}}{x^{3}} + \frac {390 \, {\left (b x^{3} + a\right )}^{2} b^{2}}{x^{6}} - \frac {455 \, {\left (b x^{3} + a\right )}^{3} b}{x^{9}} + \frac {455 \, {\left (b x^{3} + a\right )}^{4}}{x^{12}}\right )} c x^{13}}{455 \, {\left (b x^{3} + a\right )}^{\frac {13}{3}} a^{5}} \]

[In]

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

[Out]

-1/1820*(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)*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*x^13/((b*x^3 + a)^(13/3)*a^5)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.87 \[ \int \frac {c+d x^3}{\left (a+b x^3\right )^{16/3}} \, dx=\frac {x\,\left (\frac {c}{13\,a}-\frac {d}{13\,b}\right )}{{\left (b\,x^3+a\right )}^{13/3}}+\frac {x\,\left (a\,d+12\,b\,c\right )}{130\,a^2\,b\,{\left (b\,x^3+a\right )}^{10/3}}+\frac {x\,\left (9\,a\,d+108\,b\,c\right )}{910\,a^3\,b\,{\left (b\,x^3+a\right )}^{7/3}}+\frac {x\,\left (27\,a\,d+324\,b\,c\right )}{1820\,a^4\,b\,{\left (b\,x^3+a\right )}^{4/3}}+\frac {x\,\left (81\,a\,d+972\,b\,c\right )}{1820\,a^5\,b\,{\left (b\,x^3+a\right )}^{1/3}} \]

[In]

int((c + d*x^3)/(a + b*x^3)^(16/3),x)

[Out]

(x*(c/(13*a) - d/(13*b)))/(a + b*x^3)^(13/3) + (x*(a*d + 12*b*c))/(130*a^2*b*(a + b*x^3)^(10/3)) + (x*(9*a*d +
108*b*c))/(910*a^3*b*(a + b*x^3)^(7/3)) + (x*(27*a*d + 324*b*c))/(1820*a^4*b*(a + b*x^3)^(4/3)) + (x*(81*a*d
+ 972*b*c))/(1820*a^5*b*(a + b*x^3)^(1/3))