3.16.17 \(\int \frac {\sqrt [6]{a+b x}}{(c+d x)^{25/6}} \, dx\)

Optimal. Leaf size=101 \[ \frac {432 b^2 (a+b x)^{7/6}}{1729 (c+d x)^{7/6} (b c-a d)^3}+\frac {72 b (a+b x)^{7/6}}{247 (c+d x)^{13/6} (b c-a d)^2}+\frac {6 (a+b x)^{7/6}}{19 (c+d x)^{19/6} (b c-a d)} \]

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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 {432 b^2 (a+b x)^{7/6}}{1729 (c+d x)^{7/6} (b c-a d)^3}+\frac {72 b (a+b x)^{7/6}}{247 (c+d x)^{13/6} (b c-a d)^2}+\frac {6 (a+b x)^{7/6}}{19 (c+d x)^{19/6} (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(1/6)/(c + d*x)^(25/6),x]

[Out]

(6*(a + b*x)^(7/6))/(19*(b*c - a*d)*(c + d*x)^(19/6)) + (72*b*(a + b*x)^(7/6))/(247*(b*c - a*d)^2*(c + d*x)^(1
3/6)) + (432*b^2*(a + b*x)^(7/6))/(1729*(b*c - a*d)^3*(c + d*x)^(7/6))

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 {\sqrt [6]{a+b x}}{(c+d x)^{25/6}} \, dx &=\frac {6 (a+b x)^{7/6}}{19 (b c-a d) (c+d x)^{19/6}}+\frac {(12 b) \int \frac {\sqrt [6]{a+b x}}{(c+d x)^{19/6}} \, dx}{19 (b c-a d)}\\ &=\frac {6 (a+b x)^{7/6}}{19 (b c-a d) (c+d x)^{19/6}}+\frac {72 b (a+b x)^{7/6}}{247 (b c-a d)^2 (c+d x)^{13/6}}+\frac {\left (72 b^2\right ) \int \frac {\sqrt [6]{a+b x}}{(c+d x)^{13/6}} \, dx}{247 (b c-a d)^2}\\ &=\frac {6 (a+b x)^{7/6}}{19 (b c-a d) (c+d x)^{19/6}}+\frac {72 b (a+b x)^{7/6}}{247 (b c-a d)^2 (c+d x)^{13/6}}+\frac {432 b^2 (a+b x)^{7/6}}{1729 (b c-a d)^3 (c+d x)^{7/6}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 0.76 \begin {gather*} \frac {6 (a+b x)^{7/6} \left (91 a^2 d^2-14 a b d (19 c+6 d x)+b^2 \left (247 c^2+228 c d x+72 d^2 x^2\right )\right )}{1729 (c+d x)^{19/6} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(1/6)/(c + d*x)^(25/6),x]

[Out]

(6*(a + b*x)^(7/6)*(91*a^2*d^2 - 14*a*b*d*(19*c + 6*d*x) + b^2*(247*c^2 + 228*c*d*x + 72*d^2*x^2)))/(1729*(b*c
- a*d)^3*(c + d*x)^(19/6))

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IntegrateAlgebraic [A]  time = 0.17, size = 73, normalized size = 0.72 \begin {gather*} \frac {6 (a+b x)^{7/6} \left (\frac {91 d^2 (a+b x)^2}{(c+d x)^2}-\frac {266 b d (a+b x)}{c+d x}+247 b^2\right )}{1729 (c+d x)^{7/6} (b c-a d)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(1/6)/(c + d*x)^(25/6),x]

[Out]

(6*(a + b*x)^(7/6)*(247*b^2 + (91*d^2*(a + b*x)^2)/(c + d*x)^2 - (266*b*d*(a + b*x))/(c + d*x)))/(1729*(b*c -
a*d)^3*(c + d*x)^(7/6))

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fricas [B]  time = 0.75, size = 338, normalized size = 3.35 \begin {gather*} \frac {6 \, {\left (72 \, b^{3} d^{2} x^{3} + 247 \, a b^{2} c^{2} - 266 \, a^{2} b c d + 91 \, a^{3} d^{2} + 12 \, {\left (19 \, b^{3} c d - a b^{2} d^{2}\right )} x^{2} + {\left (247 \, b^{3} c^{2} - 38 \, a b^{2} c d + 7 \, a^{2} b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{6}} {\left (d x + c\right )}^{\frac {5}{6}}}{1729 \, {\left (b^{3} c^{7} - 3 \, a b^{2} c^{6} d + 3 \, a^{2} b c^{5} d^{2} - a^{3} c^{4} d^{3} + {\left (b^{3} c^{3} d^{4} - 3 \, a b^{2} c^{2} d^{5} + 3 \, a^{2} b c d^{6} - a^{3} d^{7}\right )} x^{4} + 4 \, {\left (b^{3} c^{4} d^{3} - 3 \, a b^{2} c^{3} d^{4} + 3 \, a^{2} b c^{2} d^{5} - a^{3} c d^{6}\right )} x^{3} + 6 \, {\left (b^{3} c^{5} d^{2} - 3 \, a b^{2} c^{4} d^{3} + 3 \, a^{2} b c^{3} d^{4} - a^{3} c^{2} d^{5}\right )} x^{2} + 4 \, {\left (b^{3} c^{6} d - 3 \, a b^{2} c^{5} d^{2} + 3 \, a^{2} b c^{4} d^{3} - a^{3} c^{3} d^{4}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/6)/(d*x+c)^(25/6),x, algorithm="fricas")

[Out]

6/1729*(72*b^3*d^2*x^3 + 247*a*b^2*c^2 - 266*a^2*b*c*d + 91*a^3*d^2 + 12*(19*b^3*c*d - a*b^2*d^2)*x^2 + (247*b
^3*c^2 - 38*a*b^2*c*d + 7*a^2*b*d^2)*x)*(b*x + a)^(1/6)*(d*x + c)^(5/6)/(b^3*c^7 - 3*a*b^2*c^6*d + 3*a^2*b*c^5
*d^2 - a^3*c^4*d^3 + (b^3*c^3*d^4 - 3*a*b^2*c^2*d^5 + 3*a^2*b*c*d^6 - a^3*d^7)*x^4 + 4*(b^3*c^4*d^3 - 3*a*b^2*
c^3*d^4 + 3*a^2*b*c^2*d^5 - a^3*c*d^6)*x^3 + 6*(b^3*c^5*d^2 - 3*a*b^2*c^4*d^3 + 3*a^2*b*c^3*d^4 - a^3*c^2*d^5)
*x^2 + 4*(b^3*c^6*d - 3*a*b^2*c^5*d^2 + 3*a^2*b*c^4*d^3 - a^3*c^3*d^4)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )}^{\frac {1}{6}}}{{\left (d x + c\right )}^{\frac {25}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/6)/(d*x+c)^(25/6),x, algorithm="giac")

[Out]

integrate((b*x + a)^(1/6)/(d*x + c)^(25/6), x)

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maple [A]  time = 0.01, size = 105, normalized size = 1.04 \begin {gather*} -\frac {6 \left (b x +a \right )^{\frac {7}{6}} \left (72 b^{2} x^{2} d^{2}-84 a b \,d^{2} x +228 b^{2} c d x +91 a^{2} d^{2}-266 a b c d +247 b^{2} c^{2}\right )}{1729 \left (d x +c \right )^{\frac {19}{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((b*x+a)^(1/6)/(d*x+c)^(25/6),x)

[Out]

-6/1729*(b*x+a)^(7/6)*(72*b^2*d^2*x^2-84*a*b*d^2*x+228*b^2*c*d*x+91*a^2*d^2-266*a*b*c*d+247*b^2*c^2)/(d*x+c)^(
19/6)/(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 {{\left (b x + a\right )}^{\frac {1}{6}}}{{\left (d x + c\right )}^{\frac {25}{6}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(1/6)/(d*x+c)^(25/6),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(1/6)/(d*x + c)^(25/6), x)

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mupad [B]  time = 0.95, size = 213, normalized size = 2.11 \begin {gather*} -\frac {{\left (c+d\,x\right )}^{5/6}\,\left (\frac {{\left (a+b\,x\right )}^{1/6}\,\left (546\,a^3\,d^2-1596\,a^2\,b\,c\,d+1482\,a\,b^2\,c^2\right )}{1729\,d^4\,{\left (a\,d-b\,c\right )}^3}+\frac {432\,b^3\,x^3\,{\left (a+b\,x\right )}^{1/6}}{1729\,d^2\,{\left (a\,d-b\,c\right )}^3}+\frac {x\,{\left (a+b\,x\right )}^{1/6}\,\left (42\,a^2\,b\,d^2-228\,a\,b^2\,c\,d+1482\,b^3\,c^2\right )}{1729\,d^4\,{\left (a\,d-b\,c\right )}^3}-\frac {72\,b^2\,x^2\,\left (a\,d-19\,b\,c\right )\,{\left (a+b\,x\right )}^{1/6}}{1729\,d^3\,{\left (a\,d-b\,c\right )}^3}\right )}{x^4+\frac {c^4}{d^4}+\frac {4\,c\,x^3}{d}+\frac {4\,c^3\,x}{d^3}+\frac {6\,c^2\,x^2}{d^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(1/6)/(c + d*x)^(25/6),x)

[Out]

-((c + d*x)^(5/6)*(((a + b*x)^(1/6)*(546*a^3*d^2 + 1482*a*b^2*c^2 - 1596*a^2*b*c*d))/(1729*d^4*(a*d - b*c)^3)
+ (432*b^3*x^3*(a + b*x)^(1/6))/(1729*d^2*(a*d - b*c)^3) + (x*(a + b*x)^(1/6)*(1482*b^3*c^2 + 42*a^2*b*d^2 - 2
28*a*b^2*c*d))/(1729*d^4*(a*d - b*c)^3) - (72*b^2*x^2*(a*d - 19*b*c)*(a + b*x)^(1/6))/(1729*d^3*(a*d - b*c)^3)
))/(x^4 + c^4/d^4 + (4*c*x^3)/d + (4*c^3*x)/d^3 + (6*c^2*x^2)/d^2)

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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((b*x+a)**(1/6)/(d*x+c)**(25/6),x)

[Out]

Timed out

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