3.15.58 \(\int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {27 d^2 (c+d x)^{4/3}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac {9 d (c+d x)^{4/3}}{35 (a+b x)^{7/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)} \]

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Rubi [A]  time = 0.02, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {45, 37} \begin {gather*} -\frac {27 d^2 (c+d x)^{4/3}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac {9 d (c+d x)^{4/3}}{35 (a+b x)^{7/3} (b c-a d)^2}-\frac {3 (c+d x)^{4/3}}{10 (a+b x)^{10/3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(13/3),x]

[Out]

(-3*(c + d*x)^(4/3))/(10*(b*c - a*d)*(a + b*x)^(10/3)) + (9*d*(c + d*x)^(4/3))/(35*(b*c - a*d)^2*(a + b*x)^(7/
3)) - (27*d^2*(c + d*x)^(4/3))/(140*(b*c - a*d)^3*(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 45

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*} \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{13/3}} \, dx &=-\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}-\frac {(3 d) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{10/3}} \, dx}{5 (b c-a d)}\\ &=-\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}+\frac {\left (9 d^2\right ) \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/3}} \, dx}{35 (b c-a d)^2}\\ &=-\frac {3 (c+d x)^{4/3}}{10 (b c-a d) (a+b x)^{10/3}}+\frac {9 d (c+d x)^{4/3}}{35 (b c-a d)^2 (a+b x)^{7/3}}-\frac {27 d^2 (c+d x)^{4/3}}{140 (b c-a d)^3 (a+b x)^{4/3}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.76 \begin {gather*} -\frac {3 (c+d x)^{4/3} \left (35 a^2 d^2+10 a b d (3 d x-4 c)+b^2 \left (14 c^2-12 c d x+9 d^2 x^2\right )\right )}{140 (a+b x)^{10/3} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(13/3),x]

[Out]

(-3*(c + d*x)^(4/3)*(35*a^2*d^2 + 10*a*b*d*(-4*c + 3*d*x) + b^2*(14*c^2 - 12*c*d*x + 9*d^2*x^2)))/(140*(b*c -
a*d)^3*(a + b*x)^(10/3))

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IntegrateAlgebraic [A]  time = 0.12, size = 73, normalized size = 0.72 \begin {gather*} -\frac {3 (c+d x)^{4/3} \left (\frac {14 b^2 (c+d x)^2}{(a+b x)^2}-\frac {40 b d (c+d x)}{a+b x}+35 d^2\right )}{140 (a+b x)^{4/3} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x)^(1/3)/(a + b*x)^(13/3),x]

[Out]

(-3*(c + d*x)^(4/3)*(35*d^2 - (40*b*d*(c + d*x))/(a + b*x) + (14*b^2*(c + d*x)^2)/(a + b*x)^2))/(140*(b*c - a*
d)^3*(a + b*x)^(4/3))

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fricas [B]  time = 0.87, size = 337, normalized size = 3.34 \begin {gather*} -\frac {3 \, {\left (9 \, b^{2} d^{3} x^{3} + 14 \, b^{2} c^{3} - 40 \, a b c^{2} d + 35 \, a^{2} c d^{2} - 3 \, {\left (b^{2} c d^{2} - 10 \, a b d^{3}\right )} x^{2} + {\left (2 \, b^{2} c^{2} d - 10 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{140 \, {\left (a^{4} b^{3} c^{3} - 3 \, a^{5} b^{2} c^{2} d + 3 \, a^{6} b c d^{2} - a^{7} d^{3} + {\left (b^{7} c^{3} - 3 \, a b^{6} c^{2} d + 3 \, a^{2} b^{5} c d^{2} - a^{3} b^{4} d^{3}\right )} x^{4} + 4 \, {\left (a b^{6} c^{3} - 3 \, a^{2} b^{5} c^{2} d + 3 \, a^{3} b^{4} c d^{2} - a^{4} b^{3} d^{3}\right )} x^{3} + 6 \, {\left (a^{2} b^{5} c^{3} - 3 \, a^{3} b^{4} c^{2} d + 3 \, a^{4} b^{3} c d^{2} - a^{5} b^{2} d^{3}\right )} x^{2} + 4 \, {\left (a^{3} b^{4} c^{3} - 3 \, a^{4} b^{3} c^{2} d + 3 \, a^{5} b^{2} c d^{2} - a^{6} b d^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="fricas")

[Out]

-3/140*(9*b^2*d^3*x^3 + 14*b^2*c^3 - 40*a*b*c^2*d + 35*a^2*c*d^2 - 3*(b^2*c*d^2 - 10*a*b*d^3)*x^2 + (2*b^2*c^2
*d - 10*a*b*c*d^2 + 35*a^2*d^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(a^4*b^3*c^3 - 3*a^5*b^2*c^2*d + 3*a^6*b*c*
d^2 - a^7*d^3 + (b^7*c^3 - 3*a*b^6*c^2*d + 3*a^2*b^5*c*d^2 - a^3*b^4*d^3)*x^4 + 4*(a*b^6*c^3 - 3*a^2*b^5*c^2*d
+ 3*a^3*b^4*c*d^2 - a^4*b^3*d^3)*x^3 + 6*(a^2*b^5*c^3 - 3*a^3*b^4*c^2*d + 3*a^4*b^3*c*d^2 - a^5*b^2*d^3)*x^2
+ 4*(a^3*b^4*c^3 - 3*a^4*b^3*c^2*d + 3*a^5*b^2*c*d^2 - a^6*b*d^3)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(13/3), x)

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} \frac {3 \left (d x +c \right )^{\frac {4}{3}} \left (9 b^{2} x^{2} d^{2}+30 a b \,d^{2} x -12 b^{2} c d x +35 a^{2} d^{2}-40 a b c d +14 b^{2} c^{2}\right )}{140 \left (b x +a \right )^{\frac {10}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/3)/(b*x+a)^(13/3),x)

[Out]

3/140*(d*x+c)^(4/3)*(9*b^2*d^2*x^2+30*a*b*d^2*x-12*b^2*c*d*x+35*a^2*d^2-40*a*b*c*d+14*b^2*c^2)/(b*x+a)^(10/3)/
(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {13}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(13/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(13/3), x)

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mupad [B]  time = 1.02, size = 203, normalized size = 2.01 \begin {gather*} \frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {105\,a^2\,c\,d^2-120\,a\,b\,c^2\,d+42\,b^2\,c^3}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,\left (105\,a^2\,d^3-30\,a\,b\,c\,d^2+6\,b^2\,c^2\,d\right )}{140\,b^3\,{\left (a\,d-b\,c\right )}^3}+\frac {27\,d^3\,x^3}{140\,b\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d^2\,x^2\,\left (10\,a\,d-b\,c\right )}{140\,b^2\,{\left (a\,d-b\,c\right )}^3}\right )}{x^3\,{\left (a+b\,x\right )}^{1/3}+\frac {a^3\,{\left (a+b\,x\right )}^{1/3}}{b^3}+\frac {3\,a\,x^2\,{\left (a+b\,x\right )}^{1/3}}{b}+\frac {3\,a^2\,x\,{\left (a+b\,x\right )}^{1/3}}{b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^(1/3)/(a + b*x)^(13/3),x)

[Out]

((c + d*x)^(1/3)*((42*b^2*c^3 + 105*a^2*c*d^2 - 120*a*b*c^2*d)/(140*b^3*(a*d - b*c)^3) + (x*(105*a^2*d^3 + 6*b
^2*c^2*d - 30*a*b*c*d^2))/(140*b^3*(a*d - b*c)^3) + (27*d^3*x^3)/(140*b*(a*d - b*c)^3) + (9*d^2*x^2*(10*a*d -
b*c))/(140*b^2*(a*d - b*c)^3)))/(x^3*(a + b*x)^(1/3) + (a^3*(a + b*x)^(1/3))/b^3 + (3*a*x^2*(a + b*x)^(1/3))/b
+ (3*a^2*x*(a + b*x)^(1/3))/b^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(13/3),x)

[Out]

Timed out

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