3.16.52 \(\int \frac {1}{(a+b x)^{7/6} (c+d x)^{17/6}} \, dx\)

Optimal. Leaf size=98 \[ -\frac {432 b d (a+b x)^{5/6}}{55 (c+d x)^{5/6} (b c-a d)^3}-\frac {72 d (a+b x)^{5/6}}{11 (c+d x)^{11/6} (b c-a d)^2}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)} \]

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Rubi [A]  time = 0.02, antiderivative size = 98, 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 {432 b d (a+b x)^{5/6}}{55 (c+d x)^{5/6} (b c-a d)^3}-\frac {72 d (a+b x)^{5/6}}{11 (c+d x)^{11/6} (b c-a d)^2}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x]

[Out]

-6/((b*c - a*d)*(a + b*x)^(1/6)*(c + d*x)^(11/6)) - (72*d*(a + b*x)^(5/6))/(11*(b*c - a*d)^2*(c + d*x)^(11/6))
- (432*b*d*(a + b*x)^(5/6))/(55*(b*c - a*d)^3*(c + d*x)^(5/6))

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)^{7/6} (c+d x)^{17/6}} \, dx &=-\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{11/6}}-\frac {(12 d) \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{17/6}} \, dx}{b c-a d}\\ &=-\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{11/6}}-\frac {72 d (a+b x)^{5/6}}{11 (b c-a d)^2 (c+d x)^{11/6}}-\frac {(72 b d) \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{11/6}} \, dx}{11 (b c-a d)^2}\\ &=-\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{11/6}}-\frac {72 d (a+b x)^{5/6}}{11 (b c-a d)^2 (c+d x)^{11/6}}-\frac {432 b d (a+b x)^{5/6}}{55 (b c-a d)^3 (c+d x)^{5/6}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 77, normalized size = 0.79 \begin {gather*} -\frac {6 \left (-5 a^2 d^2+2 a b d (11 c+6 d x)+b^2 \left (55 c^2+132 c d x+72 d^2 x^2\right )\right )}{55 \sqrt [6]{a+b x} (c+d x)^{11/6} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x]

[Out]

(-6*(-5*a^2*d^2 + 2*a*b*d*(11*c + 6*d*x) + b^2*(55*c^2 + 132*c*d*x + 72*d^2*x^2)))/(55*(b*c - a*d)^3*(a + b*x)
^(1/6)*(c + d*x)^(11/6))

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IntegrateAlgebraic [A]  time = 0.13, size = 73, normalized size = 0.74 \begin {gather*} -\frac {6 (a+b x)^{11/6} \left (\frac {55 b^2 (c+d x)^2}{(a+b x)^2}+\frac {22 b d (c+d x)}{a+b x}-5 d^2\right )}{55 (c+d x)^{11/6} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x]

[Out]

(-6*(a + b*x)^(11/6)*(-5*d^2 + (22*b*d*(c + d*x))/(a + b*x) + (55*b^2*(c + d*x)^2)/(a + b*x)^2))/(55*(b*c - a*
d)^3*(c + d*x)^(11/6))

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fricas [B]  time = 1.57, size = 273, normalized size = 2.79 \begin {gather*} -\frac {6 \, {\left (72 \, b^{2} d^{2} x^{2} + 55 \, b^{2} c^{2} + 22 \, a b c d - 5 \, a^{2} d^{2} + 12 \, {\left (11 \, b^{2} c d + a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{55 \, {\left (a b^{3} c^{5} - 3 \, a^{2} b^{2} c^{4} d + 3 \, a^{3} b c^{3} d^{2} - a^{4} c^{2} d^{3} + {\left (b^{4} c^{3} d^{2} - 3 \, a b^{3} c^{2} d^{3} + 3 \, a^{2} b^{2} c d^{4} - a^{3} b d^{5}\right )} x^{3} + {\left (2 \, b^{4} c^{4} d - 5 \, a b^{3} c^{3} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{3} + a^{3} b c d^{4} - a^{4} d^{5}\right )} x^{2} + {\left (b^{4} c^{5} - a b^{3} c^{4} d - 3 \, a^{2} b^{2} c^{3} d^{2} + 5 \, a^{3} b c^{2} d^{3} - 2 \, a^{4} c d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="fricas")

[Out]

-6/55*(72*b^2*d^2*x^2 + 55*b^2*c^2 + 22*a*b*c*d - 5*a^2*d^2 + 12*(11*b^2*c*d + a*b*d^2)*x)*(b*x + a)^(5/6)*(d*
x + c)^(1/6)/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3
*a^2*b^2*c*d^4 - a^3*b*d^5)*x^3 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*
x^2 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*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 {7}{6}} {\left (d x + c\right )}^{\frac {17}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(7/6)*(d*x + c)^(17/6)), x)

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maple [A]  time = 0.01, size = 105, normalized size = 1.07 \begin {gather*} -\frac {6 \left (-72 b^{2} x^{2} d^{2}-12 a b \,d^{2} x -132 b^{2} c d x +5 a^{2} d^{2}-22 a b c d -55 b^{2} c^{2}\right )}{55 \left (b x +a \right )^{\frac {1}{6}} \left (d x +c \right )^{\frac {11}{6}} \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)^(7/6)/(d*x+c)^(17/6),x)

[Out]

-6/55*(-72*b^2*d^2*x^2-12*a*b*d^2*x-132*b^2*c*d*x+5*a^2*d^2-22*a*b*c*d-55*b^2*c^2)/(b*x+a)^(1/6)/(d*x+c)^(11/6
)/(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 {7}{6}} {\left (d x + c\right )}^{\frac {17}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(17/6),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(7/6)*(d*x + c)^(17/6)), x)

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mupad [B]  time = 0.96, size = 132, normalized size = 1.35 \begin {gather*} \frac {{\left (c+d\,x\right )}^{1/6}\,\left (\frac {432\,b^2\,x^2}{55\,{\left (a\,d-b\,c\right )}^3}+\frac {-30\,a^2\,d^2+132\,a\,b\,c\,d+330\,b^2\,c^2}{55\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {72\,b\,x\,\left (a\,d+11\,b\,c\right )}{55\,d\,{\left (a\,d-b\,c\right )}^3}\right )}{x^2\,{\left (a+b\,x\right )}^{1/6}+\frac {c^2\,{\left (a+b\,x\right )}^{1/6}}{d^2}+\frac {2\,c\,x\,{\left (a+b\,x\right )}^{1/6}}{d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x)^(7/6)*(c + d*x)^(17/6)),x)

[Out]

((c + d*x)^(1/6)*((432*b^2*x^2)/(55*(a*d - b*c)^3) + (330*b^2*c^2 - 30*a^2*d^2 + 132*a*b*c*d)/(55*d^2*(a*d - b
*c)^3) + (72*b*x*(a*d + 11*b*c))/(55*d*(a*d - b*c)^3)))/(x^2*(a + b*x)^(1/6) + (c^2*(a + b*x)^(1/6))/d^2 + (2*
c*x*(a + b*x)^(1/6))/d)

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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)**(7/6)/(d*x+c)**(17/6),x)

[Out]

Timed out

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