\(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx\) [1581]

Optimal result

Integrand size = 19, antiderivative size = 136 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{13 (b c-a d) (a+b x)^{13/3}}+\frac {27 d (c+d x)^{4/3}}{130 (b c-a d)^2 (a+b x)^{10/3}}-\frac {81 d^2 (c+d x)^{4/3}}{455 (b c-a d)^3 (a+b x)^{7/3}}+\frac {243 d^3 (c+d x)^{4/3}}{1820 (b c-a d)^4 (a+b x)^{4/3}} \]

[Out]

-3/13*(d*x+c)^(4/3)/(-a*d+b*c)/(b*x+a)^(13/3)+27/130*d*(d*x+c)^(4/3)/(-a*d+b*c)^2/(b*x+a)^(10/3)-81/455*d^2*(d
*x+c)^(4/3)/(-a*d+b*c)^3/(b*x+a)^(7/3)+243/1820*d^3*(d*x+c)^(4/3)/(-a*d+b*c)^4/(b*x+a)^(4/3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\frac {243 d^3 (c+d x)^{4/3}}{1820 (a+b x)^{4/3} (b c-a d)^4}-\frac {81 d^2 (c+d x)^{4/3}}{455 (a+b x)^{7/3} (b c-a d)^3}+\frac {27 d (c+d x)^{4/3}}{130 (a+b x)^{10/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{13 (a+b x)^{13/3} (b c-a d)} \]

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(16/3),x]

[Out]

(-3*(c + d*x)^(4/3))/(13*(b*c - a*d)*(a + b*x)^(13/3)) + (27*d*(c + d*x)^(4/3))/(130*(b*c - a*d)^2*(a + b*x)^(
10/3)) - (81*d^2*(c + d*x)^(4/3))/(455*(b*c - a*d)^3*(a + b*x)^(7/3)) + (243*d^3*(c + d*x)^(4/3))/(1820*(b*c -
a*d)^4*(a + b*x)^(4/3))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {3 (c+d x)^{4/3}}{13 (b c-a d) (a+b x)^{13/3}}-\frac {(9 d) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx}{13 (b c-a d)} \\ & = -\frac {3 (c+d x)^{4/3}}{13 (b c-a d) (a+b x)^{13/3}}+\frac {27 d (c+d x)^{4/3}}{130 (b c-a d)^2 (a+b x)^{10/3}}+\frac {\left (27 d^2\right ) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx}{65 (b c-a d)^2} \\ & = -\frac {3 (c+d x)^{4/3}}{13 (b c-a d) (a+b x)^{13/3}}+\frac {27 d (c+d x)^{4/3}}{130 (b c-a d)^2 (a+b x)^{10/3}}-\frac {81 d^2 (c+d x)^{4/3}}{455 (b c-a d)^3 (a+b x)^{7/3}}-\frac {\left (81 d^3\right ) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx}{455 (b c-a d)^3} \\ & = -\frac {3 (c+d x)^{4/3}}{13 (b c-a d) (a+b x)^{13/3}}+\frac {27 d (c+d x)^{4/3}}{130 (b c-a d)^2 (a+b x)^{10/3}}-\frac {81 d^2 (c+d x)^{4/3}}{455 (b c-a d)^3 (a+b x)^{7/3}}+\frac {243 d^3 (c+d x)^{4/3}}{1820 (b c-a d)^4 (a+b x)^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\frac {3 (c+d x)^{4/3} \left (455 a^3 d^3+195 a^2 b d^2 (-4 c+3 d x)+39 a b^2 d \left (14 c^2-12 c d x+9 d^2 x^2\right )+b^3 \left (-140 c^3+126 c^2 d x-108 c d^2 x^2+81 d^3 x^3\right )\right )}{1820 (b c-a d)^4 (a+b x)^{13/3}} \]

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(16/3),x]

[Out]

(3*(c + d*x)^(4/3)*(455*a^3*d^3 + 195*a^2*b*d^2*(-4*c + 3*d*x) + 39*a*b^2*d*(14*c^2 - 12*c*d*x + 9*d^2*x^2) +
b^3*(-140*c^3 + 126*c^2*d*x - 108*c*d^2*x^2 + 81*d^3*x^3)))/(1820*(b*c - a*d)^4*(a + b*x)^(13/3))

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.26

method	result	size
gosper	\(\frac {3 \left (d x +c \right )^{\frac {4}{3}} \left (81 d^{3} x^{3} b^{3}+351 x^{2} a \,b^{2} d^{3}-108 x^{2} b^{3} c \,d^{2}+585 x \,a^{2} b \,d^{3}-468 x a \,b^{2} c \,d^{2}+126 x \,b^{3} c^{2} d +455 a^{3} d^{3}-780 a^{2} b c \,d^{2}+546 a \,b^{2} c^{2} d -140 b^{3} c^{3}\right )}{1820 \left (b x +a \right )^{\frac {13}{3}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )}\)	\(171\)

[In]

int((d*x+c)^(1/3)/(b*x+a)^(16/3),x,method=_RETURNVERBOSE)

[Out]

3/1820*(d*x+c)^(4/3)*(81*b^3*d^3*x^3+351*a*b^2*d^3*x^2-108*b^3*c*d^2*x^2+585*a^2*b*d^3*x-468*a*b^2*c*d^2*x+126
*b^3*c^2*d*x+455*a^3*d^3-780*a^2*b*c*d^2+546*a*b^2*c^2*d-140*b^3*c^3)/(b*x+a)^(13/3)/(a^4*d^4-4*a^3*b*c*d^3+6*
a^2*b^2*c^2*d^2-4*a*b^3*c^3*d+b^4*c^4)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 533 vs. \(2 (112) = 224\).

Time = 0.24 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.92 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\frac {3 \, {\left (81 \, b^{3} d^{4} x^{4} - 140 \, b^{3} c^{4} + 546 \, a b^{2} c^{3} d - 780 \, a^{2} b c^{2} d^{2} + 455 \, a^{3} c d^{3} - 27 \, {\left (b^{3} c d^{3} - 13 \, a b^{2} d^{4}\right )} x^{3} + 9 \, {\left (2 \, b^{3} c^{2} d^{2} - 13 \, a b^{2} c d^{3} + 65 \, a^{2} b d^{4}\right )} x^{2} - {\left (14 \, b^{3} c^{3} d - 78 \, a b^{2} c^{2} d^{2} + 195 \, a^{2} b c d^{3} - 455 \, a^{3} d^{4}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{1820 \, {\left (a^{5} b^{4} c^{4} - 4 \, a^{6} b^{3} c^{3} d + 6 \, a^{7} b^{2} c^{2} d^{2} - 4 \, a^{8} b c d^{3} + a^{9} d^{4} + {\left (b^{9} c^{4} - 4 \, a b^{8} c^{3} d + 6 \, a^{2} b^{7} c^{2} d^{2} - 4 \, a^{3} b^{6} c d^{3} + a^{4} b^{5} d^{4}\right )} x^{5} + 5 \, {\left (a b^{8} c^{4} - 4 \, a^{2} b^{7} c^{3} d + 6 \, a^{3} b^{6} c^{2} d^{2} - 4 \, a^{4} b^{5} c d^{3} + a^{5} b^{4} d^{4}\right )} x^{4} + 10 \, {\left (a^{2} b^{7} c^{4} - 4 \, a^{3} b^{6} c^{3} d + 6 \, a^{4} b^{5} c^{2} d^{2} - 4 \, a^{5} b^{4} c d^{3} + a^{6} b^{3} d^{4}\right )} x^{3} + 10 \, {\left (a^{3} b^{6} c^{4} - 4 \, a^{4} b^{5} c^{3} d + 6 \, a^{5} b^{4} c^{2} d^{2} - 4 \, a^{6} b^{3} c d^{3} + a^{7} b^{2} d^{4}\right )} x^{2} + 5 \, {\left (a^{4} b^{5} c^{4} - 4 \, a^{5} b^{4} c^{3} d + 6 \, a^{6} b^{3} c^{2} d^{2} - 4 \, a^{7} b^{2} c d^{3} + a^{8} b d^{4}\right )} x\right )}} \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(16/3),x, algorithm="fricas")

[Out]

3/1820*(81*b^3*d^4*x^4 - 140*b^3*c^4 + 546*a*b^2*c^3*d - 780*a^2*b*c^2*d^2 + 455*a^3*c*d^3 - 27*(b^3*c*d^3 - 1
3*a*b^2*d^4)*x^3 + 9*(2*b^3*c^2*d^2 - 13*a*b^2*c*d^3 + 65*a^2*b*d^4)*x^2 - (14*b^3*c^3*d - 78*a*b^2*c^2*d^2 +
195*a^2*b*c*d^3 - 455*a^3*d^4)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(a^5*b^4*c^4 - 4*a^6*b^3*c^3*d + 6*a^7*b^2*c
^2*d^2 - 4*a^8*b*c*d^3 + a^9*d^4 + (b^9*c^4 - 4*a*b^8*c^3*d + 6*a^2*b^7*c^2*d^2 - 4*a^3*b^6*c*d^3 + a^4*b^5*d^
4)*x^5 + 5*(a*b^8*c^4 - 4*a^2*b^7*c^3*d + 6*a^3*b^6*c^2*d^2 - 4*a^4*b^5*c*d^3 + a^5*b^4*d^4)*x^4 + 10*(a^2*b^7
*c^4 - 4*a^3*b^6*c^3*d + 6*a^4*b^5*c^2*d^2 - 4*a^5*b^4*c*d^3 + a^6*b^3*d^4)*x^3 + 10*(a^3*b^6*c^4 - 4*a^4*b^5*
c^3*d + 6*a^5*b^4*c^2*d^2 - 4*a^6*b^3*c*d^3 + a^7*b^2*d^4)*x^2 + 5*(a^4*b^5*c^4 - 4*a^5*b^4*c^3*d + 6*a^6*b^3*
c^2*d^2 - 4*a^7*b^2*c*d^3 + a^8*b*d^4)*x)

Sympy [F(-1)]

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

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(16/3),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {16}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(16/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(16/3), x)

Giac [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {16}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(16/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(16/3), x)

Mupad [B] (verification not implemented)

Time = 1.30 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{16/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {243\,d^4\,x^4}{1820\,b\,{\left (a\,d-b\,c\right )}^4}-\frac {-1365\,a^3\,c\,d^3+2340\,a^2\,b\,c^2\,d^2-1638\,a\,b^2\,c^3\,d+420\,b^3\,c^4}{1820\,b^4\,{\left (a\,d-b\,c\right )}^4}+\frac {x\,\left (1365\,a^3\,d^4-585\,a^2\,b\,c\,d^3+234\,a\,b^2\,c^2\,d^2-42\,b^3\,c^3\,d\right )}{1820\,b^4\,{\left (a\,d-b\,c\right )}^4}+\frac {81\,d^3\,x^3\,\left (13\,a\,d-b\,c\right )}{1820\,b^2\,{\left (a\,d-b\,c\right )}^4}+\frac {27\,d^2\,x^2\,\left (65\,a^2\,d^2-13\,a\,b\,c\,d+2\,b^2\,c^2\right )}{1820\,b^3\,{\left (a\,d-b\,c\right )}^4}\right )}{x^4\,{\left (a+b\,x\right )}^{1/3}+\frac {a^4\,{\left (a+b\,x\right )}^{1/3}}{b^4}+\frac {6\,a^2\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {4\,a\,x^3\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {4\,a^3\,x\,{\left (a+b\,x\right )}^{1/3}}{b^3}} \]

[In]

int((c + d*x)^(1/3)/(a + b*x)^(16/3),x)

[Out]

((c + d*x)^(1/3)*((243*d^4*x^4)/(1820*b*(a*d - b*c)^4) - (420*b^3*c^4 - 1365*a^3*c*d^3 + 2340*a^2*b*c^2*d^2 -
1638*a*b^2*c^3*d)/(1820*b^4*(a*d - b*c)^4) + (x*(1365*a^3*d^4 - 42*b^3*c^3*d + 234*a*b^2*c^2*d^2 - 585*a^2*b*c
*d^3))/(1820*b^4*(a*d - b*c)^4) + (81*d^3*x^3*(13*a*d - b*c))/(1820*b^2*(a*d - b*c)^4) + (27*d^2*x^2*(65*a^2*d
^2 + 2*b^2*c^2 - 13*a*b*c*d))/(1820*b^3*(a*d - b*c)^4)))/(x^4*(a + b*x)^(1/3) + (a^4*(a + b*x)^(1/3))/b^4 + (6
*a^2*x^2*(a + b*x)^(1/3))/b^2 + (4*a*x^3*(a + b*x)^(1/3))/b + (4*a^3*x*(a + b*x)^(1/3))/b^3)