\(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx\) [1578]

Optimal result

Integrand size = 19, antiderivative size = 32 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{4 (b c-a d) (a+b x)^{4/3}} \]

[Out]

-3/4*(d*x+c)^(4/3)/(-a*d+b*c)/(b*x+a)^(4/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{4 (a+b x)^{4/3} (b c-a d)} \]

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(7/3),x]

[Out]

(-3*(c + d*x)^(4/3))/(4*(b*c - a*d)*(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 (c+d x)^{4/3}}{4 (b c-a d) (a+b x)^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{4 (b c-a d) (a+b x)^{4/3}} \]

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(7/3),x]

[Out]

(-3*(c + d*x)^(4/3))/(4*(b*c - a*d)*(a + b*x)^(4/3))

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {3 \left (d x +c \right )^{\frac {4}{3}}}{4 \left (b x +a \right )^{\frac {4}{3}} \left (a d -b c \right )}\)	\(27\)

[In]

int((d*x+c)^(1/3)/(b*x+a)^(7/3),x,method=_RETURNVERBOSE)

[Out]

3/4/(b*x+a)^(4/3)*(d*x+c)^(4/3)/(a*d-b*c)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).

Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=-\frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {4}{3}}}{4 \, {\left (a^{2} b c - a^{3} d + {\left (b^{3} c - a b^{2} d\right )} x^{2} + 2 \, {\left (a b^{2} c - a^{2} b d\right )} x\right )}} \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/3),x, algorithm="fricas")

[Out]

-3/4*(b*x + a)^(2/3)*(d*x + c)^(4/3)/(a^2*b*c - a^3*d + (b^3*c - a*b^2*d)*x^2 + 2*(a*b^2*c - a^2*b*d)*x)

Sympy [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {7}{3}}}\, dx \]

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(7/3),x)

[Out]

Integral((c + d*x)**(1/3)/(a + b*x)**(7/3), x)

Maxima [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(7/3), x)

Giac [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(7/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(7/3), x)

Mupad [B] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.88 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx=-\frac {\left (\frac {3\,c}{4\,b^2\,c-4\,a\,b\,d}+\frac {3\,d\,x}{4\,b^2\,c-4\,a\,b\,d}\right )\,{\left (c+d\,x\right )}^{1/3}}{x\,{\left (a+b\,x\right )}^{1/3}-\frac {\left (4\,a^2\,d-4\,a\,b\,c\right )\,{\left (a+b\,x\right )}^{1/3}}{4\,b^2\,c-4\,a\,b\,d}} \]

[In]

int((c + d*x)^(1/3)/(a + b*x)^(7/3),x)

[Out]

-(((3*c)/(4*b^2*c - 4*a*b*d) + (3*d*x)/(4*b^2*c - 4*a*b*d))*(c + d*x)^(1/3))/(x*(a + b*x)^(1/3) - ((4*a^2*d -
4*a*b*c)*(a + b*x)^(1/3))/(4*b^2*c - 4*a*b*d))