3.15.66 \(\int \frac {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=136 \[ \frac {243 d^3 (c+d x)^{2/3}}{440 (a+b x)^{2/3} (b c-a d)^4}-\frac {81 d^2 (c+d x)^{2/3}}{220 (a+b x)^{5/3} (b c-a d)^3}+\frac {27 d (c+d x)^{2/3}}{88 (a+b x)^{8/3} (b c-a d)^2}-\frac {3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)} \]

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Rubi [A]  time = 0.03, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {243 d^3 (c+d x)^{2/3}}{440 (a+b x)^{2/3} (b c-a d)^4}-\frac {81 d^2 (c+d x)^{2/3}}{220 (a+b x)^{5/3} (b c-a d)^3}+\frac {27 d (c+d x)^{2/3}}{88 (a+b x)^{8/3} (b c-a d)^2}-\frac {3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x]

[Out]

(-3*(c + d*x)^(2/3))/(11*(b*c - a*d)*(a + b*x)^(11/3)) + (27*d*(c + d*x)^(2/3))/(88*(b*c - a*d)^2*(a + b*x)^(8
/3)) - (81*d^2*(c + d*x)^(2/3))/(220*(b*c - a*d)^3*(a + b*x)^(5/3)) + (243*d^3*(c + d*x)^(2/3))/(440*(b*c - a*
d)^4*(a + b*x)^(2/3))

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 {1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx &=-\frac {3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}-\frac {(9 d) \int \frac {1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx}{11 (b c-a d)}\\ &=-\frac {3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac {27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}+\frac {\left (27 d^2\right ) \int \frac {1}{(a+b x)^{8/3} \sqrt [3]{c+d x}} \, dx}{44 (b c-a d)^2}\\ &=-\frac {3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac {27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}-\frac {81 d^2 (c+d x)^{2/3}}{220 (b c-a d)^3 (a+b x)^{5/3}}-\frac {\left (81 d^3\right ) \int \frac {1}{(a+b x)^{5/3} \sqrt [3]{c+d x}} \, dx}{220 (b c-a d)^3}\\ &=-\frac {3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac {27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}-\frac {81 d^2 (c+d x)^{2/3}}{220 (b c-a d)^3 (a+b x)^{5/3}}+\frac {243 d^3 (c+d x)^{2/3}}{440 (b c-a d)^4 (a+b x)^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 118, normalized size = 0.87 \begin {gather*} \frac {3 (c+d x)^{2/3} \left (220 a^3 d^3+132 a^2 b d^2 (3 d x-2 c)+33 a b^2 d \left (5 c^2-6 c d x+9 d^2 x^2\right )+b^3 \left (-40 c^3+45 c^2 d x-54 c d^2 x^2+81 d^3 x^3\right )\right )}{440 (a+b x)^{11/3} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x]

[Out]

(3*(c + d*x)^(2/3)*(220*a^3*d^3 + 132*a^2*b*d^2*(-2*c + 3*d*x) + 33*a*b^2*d*(5*c^2 - 6*c*d*x + 9*d^2*x^2) + b^
3*(-40*c^3 + 45*c^2*d*x - 54*c*d^2*x^2 + 81*d^3*x^3)))/(440*(b*c - a*d)^4*(a + b*x)^(11/3))

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IntegrateAlgebraic [A]  time = 0.18, size = 95, normalized size = 0.70 \begin {gather*} \frac {3 (c+d x)^{11/3} \left (\frac {165 b^2 d (a+b x)}{c+d x}+\frac {220 d^3 (a+b x)^3}{(c+d x)^3}-\frac {264 b d^2 (a+b x)^2}{(c+d x)^2}-40 b^3\right )}{440 (a+b x)^{11/3} (b c-a d)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x]

[Out]

(3*(c + d*x)^(11/3)*(-40*b^3 + (220*d^3*(a + b*x)^3)/(c + d*x)^3 - (264*b*d^2*(a + b*x)^2)/(c + d*x)^2 + (165*
b^2*d*(a + b*x))/(c + d*x)))/(440*(b*c - a*d)^4*(a + b*x)^(11/3))

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fricas [B]  time = 1.41, size = 420, normalized size = 3.09 \begin {gather*} \frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} - 40 \, b^{3} c^{3} + 165 \, a b^{2} c^{2} d - 264 \, a^{2} b c d^{2} + 220 \, a^{3} d^{3} - 27 \, {\left (2 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{2} + 9 \, {\left (5 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 44 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{440 \, {\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} + {\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \, {\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \, {\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \, {\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

3/440*(81*b^3*d^3*x^3 - 40*b^3*c^3 + 165*a*b^2*c^2*d - 264*a^2*b*c*d^2 + 220*a^3*d^3 - 27*(2*b^3*c*d^2 - 11*a*
b^2*d^3)*x^2 + 9*(5*b^3*c^2*d - 22*a*b^2*c*d^2 + 44*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^4*b^4*c^4
- 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^
2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3
+ a^5*b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2
+ 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {14}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(14/3)*(d*x + c)^(1/3)), x)

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maple [A]  time = 0.01, size = 171, normalized size = 1.26 \begin {gather*} \frac {3 \left (d x +c \right )^{\frac {2}{3}} \left (81 b^{3} d^{3} x^{3}+297 a \,b^{2} d^{3} x^{2}-54 b^{3} c \,d^{2} x^{2}+396 a^{2} b \,d^{3} x -198 a \,b^{2} c \,d^{2} x +45 b^{3} c^{2} d x +220 a^{3} d^{3}-264 a^{2} b c \,d^{2}+165 a \,b^{2} c^{2} d -40 b^{3} c^{3}\right )}{440 \left (b x +a \right )^{\frac {11}{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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x)

[Out]

3/440*(d*x+c)^(2/3)*(81*b^3*d^3*x^3+297*a*b^2*d^3*x^2-54*b^3*c*d^2*x^2+396*a^2*b*d^3*x-198*a*b^2*c*d^2*x+45*b^
3*c^2*d*x+220*a^3*d^3-264*a^2*b*c*d^2+165*a*b^2*c^2*d-40*b^3*c^3)/(b*x+a)^(11/3)/(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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x + a\right )}^{\frac {14}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(14/3)*(d*x + c)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+b\,x\right )}^{14/3}\,{\left (c+d\,x\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x)

[Out]

int(1/((a + b*x)^(14/3)*(c + d*x)^(1/3)), x)

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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(1/(b*x+a)**(14/3)/(d*x+c)**(1/3),x)

[Out]

Timed out

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