\(\int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx\) [3014]

Optimal result

Integrand size = 26, antiderivative size = 591 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{18 (b e-a f) (d e-c f)^2 (e+f x)^2}+\frac {\left (28 a^2 d^2 f^2-a b d f (51 d e+5 c f)+b^2 \left (18 d^2 e^2+15 c d e f-5 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{54 (b e-a f)^2 (d e-c f)^3 (e+f x)}+\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{27 \sqrt {3} (b e-a f)^{8/3} (d e-c f)^{10/3}}-\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \log (e+f x)}{162 (b e-a f)^{8/3} (d e-c f)^{10/3}}+\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{54 (b e-a f)^{8/3} (d e-c f)^{10/3}} \]

[Out]

1/3*(b*x+a)^(1/3)*(d*x+c)^(2/3)/(-c*f+d*e)/(f*x+e)^3+1/18*(-7*a*d*f+b*c*f+6*b*d*e)*(b*x+a)^(1/3)*(d*x+c)^(2/3)
/(-a*f+b*e)/(-c*f+d*e)^2/(f*x+e)^2+1/54*(28*a^2*d^2*f^2-a*b*d*f*(5*c*f+51*d*e)+b^2*(-5*c^2*f^2+15*c*d*e*f+18*d
^2*e^2))*(b*x+a)^(1/3)*(d*x+c)^(2/3)/(-a*f+b*e)^2/(-c*f+d*e)^3/(f*x+e)-1/162*(-a*d+b*c)*(14*a^2*d^2*f^2-4*a*b*
d*f*(-2*c*f+9*d*e)+b^2*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2))*ln(f*x+e)/(-a*f+b*e)^(8/3)/(-c*f+d*e)^(10/3)+1/54*(-
a*d+b*c)*(14*a^2*d^2*f^2-4*a*b*d*f*(-2*c*f+9*d*e)+b^2*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2))*ln(-(b*x+a)^(1/3)+(-a
*f+b*e)^(1/3)*(d*x+c)^(1/3)/(-c*f+d*e)^(1/3))/(-a*f+b*e)^(8/3)/(-c*f+d*e)^(10/3)+1/81*(-a*d+b*c)*(14*a^2*d^2*f
^2-4*a*b*d*f*(-2*c*f+9*d*e)+b^2*(5*c^2*f^2-18*c*d*e*f+27*d^2*e^2))*arctan(1/3*3^(1/2)+2/3*(-a*f+b*e)^(1/3)*(d*
x+c)^(1/3)/(-c*f+d*e)^(1/3)/(b*x+a)^(1/3)*3^(1/2))/(-a*f+b*e)^(8/3)/(-c*f+d*e)^(10/3)*3^(1/2)

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 591, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {101, 156, 12, 93} \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )\right ) \arctan \left (\frac {2 \sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt {3} \sqrt [3]{a+b x} \sqrt [3]{d e-c f}}+\frac {1}{\sqrt {3}}\right )}{27 \sqrt {3} (b e-a f)^{8/3} (d e-c f)^{10/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} \left (28 a^2 d^2 f^2-a b d f (5 c f+51 d e)+b^2 \left (-5 c^2 f^2+15 c d e f+18 d^2 e^2\right )\right )}{54 (e+f x) (b e-a f)^2 (d e-c f)^3}-\frac {(b c-a d) \log (e+f x) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )\right )}{162 (b e-a f)^{8/3} (d e-c f)^{10/3}}+\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (5 c^2 f^2-18 c d e f+27 d^2 e^2\right )\right ) \log \left (\frac {\sqrt [3]{c+d x} \sqrt [3]{b e-a f}}{\sqrt [3]{d e-c f}}-\sqrt [3]{a+b x}\right )}{54 (b e-a f)^{8/3} (d e-c f)^{10/3}}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3} (-7 a d f+b c f+6 b d e)}{18 (e+f x)^2 (b e-a f) (d e-c f)^2}+\frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (e+f x)^3 (d e-c f)} \]

[In]

Int[(a + b*x)^(1/3)/((c + d*x)^(1/3)*(e + f*x)^4),x]

[Out]

((a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*(d*e - c*f)*(e + f*x)^3) + ((6*b*d*e + b*c*f - 7*a*d*f)*(a + b*x)^(1/3)*(
c + d*x)^(2/3))/(18*(b*e - a*f)*(d*e - c*f)^2*(e + f*x)^2) + ((28*a^2*d^2*f^2 - a*b*d*f*(51*d*e + 5*c*f) + b^2
*(18*d^2*e^2 + 15*c*d*e*f - 5*c^2*f^2))*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(54*(b*e - a*f)^2*(d*e - c*f)^3*(e +
f*x)) + ((b*c - a*d)*(14*a^2*d^2*f^2 - 4*a*b*d*f*(9*d*e - 2*c*f) + b^2*(27*d^2*e^2 - 18*c*d*e*f + 5*c^2*f^2))*
ArcTan[1/Sqrt[3] + (2*(b*e - a*f)^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*(d*e - c*f)^(1/3)*(a + b*x)^(1/3))])/(27*Sqr
t[3]*(b*e - a*f)^(8/3)*(d*e - c*f)^(10/3)) - ((b*c - a*d)*(14*a^2*d^2*f^2 - 4*a*b*d*f*(9*d*e - 2*c*f) + b^2*(2
7*d^2*e^2 - 18*c*d*e*f + 5*c^2*f^2))*Log[e + f*x])/(162*(b*e - a*f)^(8/3)*(d*e - c*f)^(10/3)) + ((b*c - a*d)*(
14*a^2*d^2*f^2 - 4*a*b*d*f*(9*d*e - 2*c*f) + b^2*(27*d^2*e^2 - 18*c*d*e*f + 5*c^2*f^2))*Log[-(a + b*x)^(1/3) +
((b*e - a*f)^(1/3)*(c + d*x)^(1/3))/(d*e - c*f)^(1/3)])/(54*(b*e - a*f)^(8/3)*(d*e - c*f)^(10/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, Simp[(-Sqrt[3])*q*(ArcTan[1/Sqrt[3] + 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1
/3)))]/(d*e - c*f)), x] + (Simp[q*(Log[e + f*x]/(2*(d*e - c*f))), x] - Simp[3*q*(Log[q*(a + b*x)^(1/3) - (c +
d*x)^(1/3)]/(2*(d*e - c*f))), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*
x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}-\frac {\int \frac {\frac {1}{3} (b c-7 a d)-2 b d x}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^3} \, dx}{3 (d e-c f)} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{18 (b e-a f) (d e-c f)^2 (e+f x)^2}+\frac {\int \frac {\frac {1}{9} \left (-28 a^2 d^2 f-b^2 c (12 d e-5 c f)+5 a b d (6 d e+c f)\right )+\frac {1}{3} b d (6 b d e+b c f-7 a d f) x}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)^2} \, dx}{6 (b e-a f) (d e-c f)^2} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{18 (b e-a f) (d e-c f)^2 (e+f x)^2}+\frac {\left (28 a^2 d^2 f^2-a b d f (51 d e+5 c f)+b^2 \left (18 d^2 e^2+15 c d e f-5 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{54 (b e-a f)^2 (d e-c f)^3 (e+f x)}-\frac {\int \frac {2 (b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right )}{27 (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)} \, dx}{6 (b e-a f)^2 (d e-c f)^3} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{18 (b e-a f) (d e-c f)^2 (e+f x)^2}+\frac {\left (28 a^2 d^2 f^2-a b d f (51 d e+5 c f)+b^2 \left (18 d^2 e^2+15 c d e f-5 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{54 (b e-a f)^2 (d e-c f)^3 (e+f x)}-\frac {\left ((b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right )\right ) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)} \, dx}{81 (b e-a f)^2 (d e-c f)^3} \\ & = \frac {\sqrt [3]{a+b x} (c+d x)^{2/3}}{3 (d e-c f) (e+f x)^3}+\frac {(6 b d e+b c f-7 a d f) \sqrt [3]{a+b x} (c+d x)^{2/3}}{18 (b e-a f) (d e-c f)^2 (e+f x)^2}+\frac {\left (28 a^2 d^2 f^2-a b d f (51 d e+5 c f)+b^2 \left (18 d^2 e^2+15 c d e f-5 c^2 f^2\right )\right ) \sqrt [3]{a+b x} (c+d x)^{2/3}}{54 (b e-a f)^2 (d e-c f)^3 (e+f x)}+\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d e-c f} \sqrt [3]{a+b x}}\right )}{27 \sqrt {3} (b e-a f)^{8/3} (d e-c f)^{10/3}}-\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \log (e+f x)}{162 (b e-a f)^{8/3} (d e-c f)^{10/3}}+\frac {(b c-a d) \left (14 a^2 d^2 f^2-4 a b d f (9 d e-2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) \log \left (-\sqrt [3]{a+b x}+\frac {\sqrt [3]{b e-a f} \sqrt [3]{c+d x}}{\sqrt [3]{d e-c f}}\right )}{54 (b e-a f)^{8/3} (d e-c f)^{10/3}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.44 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\frac {\sqrt [3]{a+b x} \left (-18 f (-b e+a f) (-d e+c f) (a+b x) (c+d x)+3 f (-12 b d e+5 b c f+7 a d f) (a+b x) (c+d x) (e+f x)+\frac {2 \left (14 a^2 d^2 f^2+4 a b d f (-9 d e+2 c f)+b^2 \left (27 d^2 e^2-18 c d e f+5 c^2 f^2\right )\right ) (e+f x)^2 \left ((b e-a f) (c+d x)-(b c-a d) (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},1,\frac {4}{3},\frac {(d e-c f) (a+b x)}{(b e-a f) (c+d x)}\right )\right )}{(b e-a f) (d e-c f)}\right )}{54 (b e-a f)^2 (d e-c f)^2 \sqrt [3]{c+d x} (e+f x)^3} \]

[In]

Integrate[(a + b*x)^(1/3)/((c + d*x)^(1/3)*(e + f*x)^4),x]

[Out]

((a + b*x)^(1/3)*(-18*f*(-(b*e) + a*f)*(-(d*e) + c*f)*(a + b*x)*(c + d*x) + 3*f*(-12*b*d*e + 5*b*c*f + 7*a*d*f
)*(a + b*x)*(c + d*x)*(e + f*x) + (2*(14*a^2*d^2*f^2 + 4*a*b*d*f*(-9*d*e + 2*c*f) + b^2*(27*d^2*e^2 - 18*c*d*e
*f + 5*c^2*f^2))*(e + f*x)^2*((b*e - a*f)*(c + d*x) - (b*c - a*d)*(e + f*x)*Hypergeometric2F1[1/3, 1, 4/3, ((d
*e - c*f)*(a + b*x))/((b*e - a*f)*(c + d*x))]))/((b*e - a*f)*(d*e - c*f))))/(54*(b*e - a*f)^2*(d*e - c*f)^2*(c
+ d*x)^(1/3)*(e + f*x)^3)

Maple [F]

\[\int \frac {\left (b x +a \right )^{\frac {1}{3}}}{\left (d x +c \right )^{\frac {1}{3}} \left (f x +e \right )^{4}}d x\]

[In]

int((b*x+a)^(1/3)/(d*x+c)^(1/3)/(f*x+e)^4,x)

[Out]

int((b*x+a)^(1/3)/(d*x+c)^(1/3)/(f*x+e)^4,x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4464 vs. \(2 (535) = 1070\).

Time = 5.88 (sec) , antiderivative size = 9090, normalized size of antiderivative = 15.38 \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\text {Too large to display} \]

[In]

integrate((b*x+a)^(1/3)/(d*x+c)^(1/3)/(f*x+e)^4,x, algorithm="fricas")

[Out]

Too large to include

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\text {Timed out} \]

[In]

integrate((b*x+a)**(1/3)/(d*x+c)**(1/3)/(f*x+e)**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{4}} \,d x } \]

[In]

integrate((b*x+a)^(1/3)/(d*x+c)^(1/3)/(f*x+e)^4,x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/3)/((d*x + c)^(1/3)*(f*x + e)^4), x)

Giac [F]

\[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\int { \frac {{\left (b x + a\right )}^{\frac {1}{3}}}{{\left (d x + c\right )}^{\frac {1}{3}} {\left (f x + e\right )}^{4}} \,d x } \]

[In]

integrate((b*x+a)^(1/3)/(d*x+c)^(1/3)/(f*x+e)^4,x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/3)/((d*x + c)^(1/3)*(f*x + e)^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x} (e+f x)^4} \, dx=\int \frac {{\left (a+b\,x\right )}^{1/3}}{{\left (e+f\,x\right )}^4\,{\left (c+d\,x\right )}^{1/3}} \,d x \]

[In]

int((a + b*x)^(1/3)/((e + f*x)^4*(c + d*x)^(1/3)),x)

[Out]

int((a + b*x)^(1/3)/((e + f*x)^4*(c + d*x)^(1/3)), x)