3.1621 \(\int \frac {1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=136 \[ -\frac {243 d^3 \sqrt [3]{a+b x}}{40 \sqrt [3]{c+d x} (b c-a d)^4}-\frac {81 d^2}{40 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^3}+\frac {27 d}{40 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)} \]

[Out]

-3/8/(-a*d+b*c)/(b*x+a)^(8/3)/(d*x+c)^(1/3)+27/40*d/(-a*d+b*c)^2/(b*x+a)^(5/3)/(d*x+c)^(1/3)-81/40*d^2/(-a*d+b
*c)^3/(b*x+a)^(2/3)/(d*x+c)^(1/3)-243/40*d^3*(b*x+a)^(1/3)/(-a*d+b*c)^4/(d*x+c)^(1/3)

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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} \[ -\frac {243 d^3 \sqrt [3]{a+b x}}{40 \sqrt [3]{c+d x} (b c-a d)^4}-\frac {81 d^2}{40 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^3}+\frac {27 d}{40 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac {3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]

[Out]

-3/(8*(b*c - a*d)*(a + b*x)^(8/3)*(c + d*x)^(1/3)) + (27*d)/(40*(b*c - a*d)^2*(a + b*x)^(5/3)*(c + d*x)^(1/3))
- (81*d^2)/(40*(b*c - a*d)^3*(a + b*x)^(2/3)*(c + d*x)^(1/3)) - (243*d^3*(a + b*x)^(1/3))/(40*(b*c - a*d)^4*(
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)^{11/3} (c+d x)^{4/3}} \, dx &=-\frac {3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}-\frac {(9 d) \int \frac {1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx}{8 (b c-a d)}\\ &=-\frac {3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac {27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac {\left (27 d^2\right ) \int \frac {1}{(a+b x)^{5/3} (c+d x)^{4/3}} \, dx}{20 (b c-a d)^2}\\ &=-\frac {3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac {27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac {81 d^2}{40 (b c-a d)^3 (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac {\left (81 d^3\right ) \int \frac {1}{(a+b x)^{2/3} (c+d x)^{4/3}} \, dx}{40 (b c-a d)^3}\\ &=-\frac {3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac {27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac {81 d^2}{40 (b c-a d)^3 (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac {243 d^3 \sqrt [3]{a+b x}}{40 (b c-a d)^4 \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 116, normalized size = 0.85 \[ -\frac {3 \left (40 a^3 d^3+60 a^2 b d^2 (c+3 d x)+24 a b^2 d \left (-c^2+3 c d x+9 d^2 x^2\right )+b^3 \left (5 c^3-9 c^2 d x+27 c d^2 x^2+81 d^3 x^3\right )\right )}{40 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]

[Out]

(-3*(40*a^3*d^3 + 60*a^2*b*d^2*(c + 3*d*x) + 24*a*b^2*d*(-c^2 + 3*c*d*x + 9*d^2*x^2) + b^3*(5*c^3 - 9*c^2*d*x
+ 27*c*d^2*x^2 + 81*d^3*x^3)))/(40*(b*c - a*d)^4*(a + b*x)^(8/3)*(c + d*x)^(1/3))

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fricas [B]  time = 0.53, size = 456, normalized size = 3.35 \[ -\frac {3 \, {\left (81 \, b^{3} d^{3} x^{3} + 5 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 60 \, a^{2} b c d^{2} + 40 \, a^{3} d^{3} + 27 \, {\left (b^{3} c d^{2} + 8 \, a b^{2} d^{3}\right )} x^{2} - 9 \, {\left (b^{3} c^{2} d - 8 \, a b^{2} c d^{2} - 20 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{40 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

-3/40*(81*b^3*d^3*x^3 + 5*b^3*c^3 - 24*a*b^2*c^2*d + 60*a^2*b*c*d^2 + 40*a^3*d^3 + 27*(b^3*c*d^2 + 8*a*b^2*d^3
)*x^2 - 9*(b^3*c^2*d - 8*a*b^2*c*d^2 - 20*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^3*b^4*c^5 - 4*a^4*b
^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3
- 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^
4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*
a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 -
a^6*b*c*d^4 + a^7*d^5)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)

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maple [A]  time = 0.01, size = 171, normalized size = 1.26 \[ -\frac {3 \left (81 b^{3} d^{3} x^{3}+216 a \,b^{2} d^{3} x^{2}+27 b^{3} c \,d^{2} x^{2}+180 a^{2} b \,d^{3} x +72 a \,b^{2} c \,d^{2} x -9 b^{3} c^{2} d x +40 a^{3} d^{3}+60 a^{2} b c \,d^{2}-24 a \,b^{2} c^{2} d +5 b^{3} c^{3}\right )}{40 \left (b x +a \right )^{\frac {8}{3}} \left (d x +c \right )^{\frac {1}{3}} \left (a^{4} d^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +b^{4} c^{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x)

[Out]

-3/40*(81*b^3*d^3*x^3+216*a*b^2*d^3*x^2+27*b^3*c*d^2*x^2+180*a^2*b*d^3*x+72*a*b^2*c*d^2*x-9*b^3*c^2*d*x+40*a^3
*d^3+60*a^2*b*c*d^2-24*a*b^2*c^2*d+5*b^3*c^3)/(b*x+a)^(8/3)/(d*x+c)^(1/3)/(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 \[ \int \frac {1}{{\left (b x + a\right )}^{\frac {11}{3}} {\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+b\,x\right )}^{11/3}\,{\left (c+d\,x\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x)

[Out]

int(1/((a + b*x)^(11/3)*(c + d*x)^(4/3)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b x\right )^{\frac {11}{3}} \left (c + d x\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(11/3)/(d*x+c)**(4/3),x)

[Out]

Integral(1/((a + b*x)**(11/3)*(c + d*x)**(4/3)), x)

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