3.15.78 \(\int \frac {1}{(a+b x)^{5/3} (c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=66 \[ -\frac {9 d \sqrt [3]{a+b x}}{2 \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{2 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)} \]

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Rubi [A]  time = 0.01, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {9 d \sqrt [3]{a+b x}}{2 \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{2 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(5/3)*(c + d*x)^(4/3)),x]

[Out]

-3/(2*(b*c - a*d)*(a + b*x)^(2/3)*(c + d*x)^(1/3)) - (9*d*(a + b*x)^(1/3))/(2*(b*c - a*d)^2*(c + d*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)^{5/3} (c+d x)^{4/3}} \, dx &=-\frac {3}{2 (b c-a d) (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac {(3 d) \int \frac {1}{(a+b x)^{2/3} (c+d x)^{4/3}} \, dx}{2 (b c-a d)}\\ &=-\frac {3}{2 (b c-a d) (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac {9 d \sqrt [3]{a+b x}}{2 (b c-a d)^2 \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.68 \begin {gather*} -\frac {3 (2 a d+b (c+3 d x))}{2 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(5/3)*(c + d*x)^(4/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 49, normalized size = 0.74 \begin {gather*} -\frac {3 (c+d x)^{2/3} \left (\frac {2 d (a+b x)}{c+d x}+b\right )}{2 (a+b x)^{2/3} (b c-a d)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(5/3)*(c + d*x)^(4/3)),x]

[Out]

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

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fricas [B]  time = 1.21, size = 126, normalized size = 1.91 \begin {gather*} -\frac {3 \, {\left (3 \, b d x + b c + 2 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{2 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(5/3)*(d*x + c)^(4/3)), x)

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maple [A]  time = 0.00, size = 53, normalized size = 0.80 \begin {gather*} -\frac {3 \left (3 b d x +2 a d +b c \right )}{2 \left (b x +a \right )^{\frac {2}{3}} \left (d x +c \right )^{\frac {1}{3}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(5/3)/(d*x+c)^(4/3),x)

[Out]

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

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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 {5}{3}} {\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(5/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(5/3)*(d*x + c)^(4/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (a+b\,x\right )}^{5/3}\,{\left (c+d\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(5/3)*(c + d*x)^(4/3)),x)

[Out]

int(1/((a + b*x)^(5/3)*(c + d*x)^(4/3)), x)

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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 {5}{3}} \left (c + d x\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(5/3)/(d*x+c)**(4/3),x)

[Out]

Integral(1/((a + b*x)**(5/3)*(c + d*x)**(4/3)), x)

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