3.17.20 \(\int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx\) [1620]

Optimal. Leaf size=101 \[ -\frac {3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac {9 d}{5 (b c-a d)^2 (a+b x)^{2/3} \sqrt [3]{c+d x}}+\frac {27 d^2 \sqrt [3]{a+b x}}{5 (b c-a d)^3 \sqrt [3]{c+d x}} \]

[Out]

-3/5/(-a*d+b*c)/(b*x+a)^(5/3)/(d*x+c)^(1/3)+9/5*d/(-a*d+b*c)^2/(b*x+a)^(2/3)/(d*x+c)^(1/3)+27/5*d^2*(b*x+a)^(1
/3)/(-a*d+b*c)^3/(d*x+c)^(1/3)

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Rubi [A]
time = 0.01, 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 = {47, 37} \begin {gather*} \frac {27 d^2 \sqrt [3]{a+b x}}{5 \sqrt [3]{c+d x} (b c-a d)^3}+\frac {9 d}{5 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]

[Out]

-3/(5*(b*c - a*d)*(a + b*x)^(5/3)*(c + d*x)^(1/3)) + (9*d)/(5*(b*c - a*d)^2*(a + b*x)^(2/3)*(c + d*x)^(1/3)) +
(27*d^2*(a + b*x)^(1/3))/(5*(b*c - a*d)^3*(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 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*} \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx &=-\frac {3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac {(6 d) \int \frac {1}{(a+b x)^{5/3} (c+d x)^{4/3}} \, dx}{5 (b c-a d)}\\ &=-\frac {3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac {9 d}{5 (b c-a d)^2 (a+b x)^{2/3} \sqrt [3]{c+d x}}+\frac {\left (9 d^2\right ) \int \frac {1}{(a+b x)^{2/3} (c+d x)^{4/3}} \, dx}{5 (b c-a d)^2}\\ &=-\frac {3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac {9 d}{5 (b c-a d)^2 (a+b x)^{2/3} \sqrt [3]{c+d x}}+\frac {27 d^2 \sqrt [3]{a+b x}}{5 (b c-a d)^3 \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 71, normalized size = 0.70 \begin {gather*} -\frac {3 (c+d x)^{5/3} \left (b^2-\frac {5 d^2 (a+b x)^2}{(c+d x)^2}-\frac {5 b d (a+b x)}{c+d x}\right )}{5 (b c-a d)^3 (a+b x)^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]

[Out]

(-3*(c + d*x)^(5/3)*(b^2 - (5*d^2*(a + b*x)^2)/(c + d*x)^2 - (5*b*d*(a + b*x))/(c + d*x)))/(5*(b*c - a*d)^3*(a
+ b*x)^(5/3))

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Maple [A]
time = 0.17, size = 105, normalized size = 1.04

method	result	size
gosper	\(-\frac {3 \left (9 b^{2} x^{2} d^{2}+15 a b \,d^{2} x +3 b^{2} c d x +5 a^{2} d^{2}+5 a b c d -b^{2} c^{2}\right )}{5 \left (b x +a \right )^{\frac {5}{3}} \left (d x +c \right )^{\frac {1}{3}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x,method=_RETURNVERBOSE)

[Out]

-3/5*(9*b^2*d^2*x^2+15*a*b*d^2*x+3*b^2*c*d*x+5*a^2*d^2+5*a*b*c*d-b^2*c^2)/(b*x+a)^(5/3)/(d*x+c)^(1/3)/(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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (83) = 166\).
time = 0.99, size = 273, normalized size = 2.70 \begin {gather*} \frac {3 \, {\left (9 \, b^{2} d^{2} x^{2} - b^{2} c^{2} + 5 \, a b c d + 5 \, a^{2} d^{2} + 3 \, {\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{5 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

3/5*(9*b^2*d^2*x^2 - b^2*c^2 + 5*a*b*c*d + 5*a^2*d^2 + 3*(b^2*c*d + 5*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2
/3)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*
d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*
a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*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 {8}{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)**(8/3)/(d*x+c)**(4/3),x)

[Out]

Integral(1/((a + b*x)**(8/3)*(c + d*x)**(4/3)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), 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 )}^{8/3}\,{\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)^(8/3)*(c + d*x)^(4/3)),x)

[Out]

int(1/((a + b*x)^(8/3)*(c + d*x)^(4/3)), x)

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