3.15.72 \(\int \frac {1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {27 d^2 \sqrt [3]{c+d x}}{14 \sqrt [3]{a+b x} (b c-a d)^3}+\frac {9 d \sqrt [3]{c+d x}}{14 (a+b x)^{4/3} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)} \]

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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 \sqrt [3]{c+d x}}{14 \sqrt [3]{a+b x} (b c-a d)^3}+\frac {9 d \sqrt [3]{c+d x}}{14 (a+b x)^{4/3} (b c-a d)^2}-\frac {3 \sqrt [3]{c+d x}}{7 (a+b x)^{7/3} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3))/(7*(b*c - a*d)*(a + b*x)^(7/3)) + (9*d*(c + d*x)^(1/3))/(14*(b*c - a*d)^2*(a + b*x)^(4/3)
) - (27*d^2*(c + d*x)^(1/3))/(14*(b*c - a*d)^3*(a + b*x)^(1/3))

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)^{10/3} (c+d x)^{2/3}} \, dx &=-\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}-\frac {(6 d) \int \frac {1}{(a+b x)^{7/3} (c+d x)^{2/3}} \, dx}{7 (b c-a d)}\\ &=-\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}+\frac {\left (9 d^2\right ) \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx}{14 (b c-a d)^2}\\ &=-\frac {3 \sqrt [3]{c+d x}}{7 (b c-a d) (a+b x)^{7/3}}+\frac {9 d \sqrt [3]{c+d x}}{14 (b c-a d)^2 (a+b x)^{4/3}}-\frac {27 d^2 \sqrt [3]{c+d x}}{14 (b c-a d)^3 \sqrt [3]{a+b x}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 0.74 \begin {gather*} -\frac {3 \sqrt [3]{c+d x} \left (14 a^2 d^2-7 a b d (c-3 d x)+b^2 \left (2 c^2-3 c d x+9 d^2 x^2\right )\right )}{14 (a+b x)^{7/3} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3)*(14*a^2*d^2 - 7*a*b*d*(c - 3*d*x) + b^2*(2*c^2 - 3*c*d*x + 9*d^2*x^2)))/(14*(b*c - a*d)^3*
(a + b*x)^(7/3))

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IntegrateAlgebraic [A]  time = 0.12, size = 83, normalized size = 0.82 \begin {gather*} -\frac {3 \left (\frac {2 b^2 (c+d x)^{7/3}}{(a+b x)^{7/3}}+\frac {14 d^2 \sqrt [3]{c+d x}}{\sqrt [3]{a+b x}}-\frac {7 b d (c+d x)^{4/3}}{(a+b x)^{4/3}}\right )}{14 (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*((14*d^2*(c + d*x)^(1/3))/(a + b*x)^(1/3) - (7*b*d*(c + d*x)^(4/3))/(a + b*x)^(4/3) + (2*b^2*(c + d*x)^(7/
3))/(a + b*x)^(7/3)))/(14*(b*c - a*d)^3)

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fricas [B]  time = 1.11, size = 251, normalized size = 2.49 \begin {gather*} -\frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} + 2 \, b^{2} c^{2} - 7 \, a b c d + 14 \, a^{2} d^{2} - 3 \, {\left (b^{2} c d - 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{14 \, {\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} + {\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \, {\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

-3/14*(9*b^2*d^2*x^2 + 2*b^2*c^2 - 7*a*b*c*d + 14*a^2*d^2 - 3*(b^2*c*d - 7*a*b*d^2)*x)*(b*x + a)^(2/3)*(d*x +
c)^(1/3)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2
- a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 - 3
*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*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 {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(10/3)*(d*x + c)^(2/3)), 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 {1}{3}} \left (9 b^{2} x^{2} d^{2}+21 a b \,d^{2} x -3 b^{2} c d x +14 a^{2} d^{2}-7 a b c d +2 b^{2} c^{2}\right )}{14 \left (b x +a \right )^{\frac {7}{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(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x)

[Out]

3/14*(d*x+c)^(1/3)*(9*b^2*d^2*x^2+21*a*b*d^2*x-3*b^2*c*d*x+14*a^2*d^2-7*a*b*c*d+2*b^2*c^2)/(b*x+a)^(7/3)/(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 {1}{{\left (b x + a\right )}^{\frac {10}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(10/3)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(10/3)*(d*x + c)^(2/3)), x)

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mupad [B]  time = 1.51, size = 133, normalized size = 1.32 \begin {gather*} \frac {{\left (c+d\,x\right )}^{1/3}\,\left (\frac {27\,d^2\,x^2}{14\,{\left (a\,d-b\,c\right )}^3}+\frac {42\,a^2\,d^2-21\,a\,b\,c\,d+6\,b^2\,c^2}{14\,b^2\,{\left (a\,d-b\,c\right )}^3}+\frac {9\,d\,x\,\left (7\,a\,d-b\,c\right )}{14\,b\,{\left (a\,d-b\,c\right )}^3}\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(10/3)*(c + d*x)^(2/3)),x)

[Out]

((c + d*x)^(1/3)*((27*d^2*x^2)/(14*(a*d - b*c)^3) + (42*a^2*d^2 + 6*b^2*c^2 - 21*a*b*c*d)/(14*b^2*(a*d - b*c)^
3) + (9*d*x*(7*a*d - b*c))/(14*b*(a*d - b*c)^3)))/(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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + b x\right )^{\frac {10}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(10/3)/(d*x+c)**(2/3),x)

[Out]

Integral(1/((a + b*x)**(10/3)*(c + d*x)**(2/3)), x)

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