\(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx\) [1579]

Optimal result

Integrand size = 19, antiderivative size = 66 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=-\frac {3 (c+d x)^{4/3}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d (c+d x)^{4/3}}{28 (b c-a d)^2 (a+b x)^{4/3}} \]

[Out]

-3/7*(d*x+c)^(4/3)/(-a*d+b*c)/(b*x+a)^(7/3)+9/28*d*(d*x+c)^(4/3)/(-a*d+b*c)^2/(b*x+a)^(4/3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\frac {9 d (c+d x)^{4/3}}{28 (a+b x)^{4/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{7 (a+b x)^{7/3} (b c-a d)} \]

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(10/3),x]

[Out]

(-3*(c + d*x)^(4/3))/(7*(b*c - a*d)*(a + b*x)^(7/3)) + (9*d*(c + d*x)^(4/3))/(28*(b*c - a*d)^2*(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}}{7 (b c-a d) (a+b x)^{7/3}}-\frac {(3 d) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx}{7 (b c-a d)} \\ & = -\frac {3 (c+d x)^{4/3}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d (c+d x)^{4/3}}{28 (b c-a d)^2 (a+b x)^{4/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=-\frac {3 (c+d x)^{4/3} (4 b c-7 a d-3 b d x)}{28 (b c-a d)^2 (a+b x)^{7/3}} \]

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(10/3),x]

[Out]

(-3*(c + d*x)^(4/3)*(4*b*c - 7*a*d - 3*b*d*x))/(28*(b*c - a*d)^2*(a + b*x)^(7/3))

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {3 \left (d x +c \right )^{\frac {4}{3}} \left (3 b d x +7 a d -4 b c \right )}{28 \left (b x +a \right )^{\frac {7}{3}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(54\)

[In]

int((d*x+c)^(1/3)/(b*x+a)^(10/3),x,method=_RETURNVERBOSE)

[Out]

3/28*(d*x+c)^(4/3)*(3*b*d*x+7*a*d-4*b*c)/(b*x+a)^(7/3)/(a^2*d^2-2*a*b*c*d+b^2*c^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (54) = 108\).

Time = 0.23 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.65 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\frac {3 \, {\left (3 \, b d^{2} x^{2} - 4 \, b c^{2} + 7 \, a c d - {\left (b c d - 7 \, a d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{28 \, {\left (a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3} + 3 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{3} c^{2} - 2 \, a^{3} b^{2} c d + a^{4} b d^{2}\right )} x\right )}} \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(10/3),x, algorithm="fricas")

[Out]

3/28*(3*b*d^2*x^2 - 4*b*c^2 + 7*a*c*d - (b*c*d - 7*a*d^2)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(a^3*b^2*c^2 - 2*
a^4*b*c*d + a^5*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^3 + 3*(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)*
x^2 + 3*(a^2*b^3*c^2 - 2*a^3*b^2*c*d + a^4*b*d^2)*x)

Sympy [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {10}{3}}}\, dx \]

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(10/3),x)

[Out]

Integral((c + d*x)**(1/3)/(a + b*x)**(10/3), x)

Maxima [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {10}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(10/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(10/3), x)

Giac [F]

\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {10}{3}}} \,d x } \]

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(10/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(10/3), x)

Mupad [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.92 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx=\frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {x\,\left (21\,a\,d^2-3\,b\,c\,d\right )}{28\,b^2\,{\left (a\,d-b\,c\right )}^2}-\frac {12\,b\,c^2-21\,a\,c\,d}{28\,b^2\,{\left (a\,d-b\,c\right )}^2}+\frac {9\,d^2\,x^2}{28\,b\,{\left (a\,d-b\,c\right )}^2}\right )}{x^2\,{\left (a+b\,x\right )}^{1/3}+\frac {a^2\,{\left (a+b\,x\right )}^{1/3}}{b^2}+\frac {2\,a\,x\,{\left (a+b\,x\right )}^{1/3}}{b}} \]

[In]

int((c + d*x)^(1/3)/(a + b*x)^(10/3),x)

[Out]

((c + d*x)^(1/3)*((x*(21*a*d^2 - 3*b*c*d))/(28*b^2*(a*d - b*c)^2) - (12*b*c^2 - 21*a*c*d)/(28*b^2*(a*d - b*c)^
2) + (9*d^2*x^2)/(28*b*(a*d - b*c)^2)))/(x^2*(a + b*x)^(1/3) + (a^2*(a + b*x)^(1/3))/b^2 + (2*a*x*(a + b*x)^(1
/3))/b)