3.1722 \(\int \frac{1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{512 d^3 \sqrt [4]{a+b x}}{77 \sqrt [4]{c+d x} (b c-a d)^4}-\frac{128 d^2}{77 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^3}+\frac{48 d}{77 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac{4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)} \]

[Out]

-4/(11*(b*c - a*d)*(a + b*x)^(11/4)*(c + d*x)^(1/4)) + (48*d)/(77*(b*c - a*d)^2*(a + b*x)^(7/4)*(c + d*x)^(1/4
)) - (128*d^2)/(77*(b*c - a*d)^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)) - (512*d^3*(a + b*x)^(1/4))/(77*(b*c - a*d)^
4*(c + d*x)^(1/4))

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Rubi [A]  time = 0.027462, antiderivative size = 136, normalized size of antiderivative = 1., 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} \[ -\frac{512 d^3 \sqrt [4]{a+b x}}{77 \sqrt [4]{c+d x} (b c-a d)^4}-\frac{128 d^2}{77 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^3}+\frac{48 d}{77 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac{4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]

[Out]

-4/(11*(b*c - a*d)*(a + b*x)^(11/4)*(c + d*x)^(1/4)) + (48*d)/(77*(b*c - a*d)^2*(a + b*x)^(7/4)*(c + d*x)^(1/4
)) - (128*d^2)/(77*(b*c - a*d)^3*(a + b*x)^(3/4)*(c + d*x)^(1/4)) - (512*d^3*(a + b*x)^(1/4))/(77*(b*c - a*d)^
4*(c + d*x)^(1/4))

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])

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]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx &=-\frac{4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}-\frac{(12 d) \int \frac{1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx}{11 (b c-a d)}\\ &=-\frac{4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac{48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac{\left (96 d^2\right ) \int \frac{1}{(a+b x)^{7/4} (c+d x)^{5/4}} \, dx}{77 (b c-a d)^2}\\ &=-\frac{4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac{48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac{128 d^2}{77 (b c-a d)^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}-\frac{\left (128 d^3\right ) \int \frac{1}{(a+b x)^{3/4} (c+d x)^{5/4}} \, dx}{77 (b c-a d)^3}\\ &=-\frac{4}{11 (b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}+\frac{48 d}{77 (b c-a d)^2 (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac{128 d^2}{77 (b c-a d)^3 (a+b x)^{3/4} \sqrt [4]{c+d x}}-\frac{512 d^3 \sqrt [4]{a+b x}}{77 (b c-a d)^4 \sqrt [4]{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.0422317, size = 116, normalized size = 0.85 \[ -\frac{4 \left (77 a^2 b d^2 (c+4 d x)+77 a^3 d^3+11 a b^2 d \left (-3 c^2+8 c d x+32 d^2 x^2\right )+b^3 \left (-12 c^2 d x+7 c^3+32 c d^2 x^2+128 d^3 x^3\right )\right )}{77 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]

[Out]

(-4*(77*a^3*d^3 + 77*a^2*b*d^2*(c + 4*d*x) + 11*a*b^2*d*(-3*c^2 + 8*c*d*x + 32*d^2*x^2) + b^3*(7*c^3 - 12*c^2*
d*x + 32*c*d^2*x^2 + 128*d^3*x^3)))/(77*(b*c - a*d)^4*(a + b*x)^(11/4)*(c + d*x)^(1/4))

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Maple [A]  time = 0.008, size = 171, normalized size = 1.3 \begin{align*} -{\frac{512\,{x}^{3}{b}^{3}{d}^{3}+1408\,a{b}^{2}{d}^{3}{x}^{2}+128\,{b}^{3}c{d}^{2}{x}^{2}+1232\,{a}^{2}b{d}^{3}x+352\,a{b}^{2}c{d}^{2}x-48\,{b}^{3}{c}^{2}dx+308\,{a}^{3}{d}^{3}+308\,{a}^{2}cb{d}^{2}-132\,a{b}^{2}{c}^{2}d+28\,{b}^{3}{c}^{3}}{77\,{a}^{4}{d}^{4}-308\,{a}^{3}bc{d}^{3}+462\,{a}^{2}{c}^{2}{b}^{2}{d}^{2}-308\,a{b}^{3}{c}^{3}d+77\,{b}^{4}{c}^{4}} \left ( bx+a \right ) ^{-{\frac{11}{4}}}{\frac{1}{\sqrt [4]{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x)

[Out]

-4/77*(128*b^3*d^3*x^3+352*a*b^2*d^3*x^2+32*b^3*c*d^2*x^2+308*a^2*b*d^3*x+88*a*b^2*c*d^2*x-12*b^3*c^2*d*x+77*a
^3*d^3+77*a^2*b*c*d^2-33*a*b^2*c^2*d+7*b^3*c^3)/(b*x+a)^(11/4)/(d*x+c)^(1/4)/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{15}{4}}{\left (d x + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(15/4)*(d*x + c)^(5/4)), x)

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Fricas [B]  time = 7.96956, size = 934, normalized size = 6.87 \begin{align*} -\frac{4 \,{\left (128 \, b^{3} d^{3} x^{3} + 7 \, b^{3} c^{3} - 33 \, a b^{2} c^{2} d + 77 \, a^{2} b c d^{2} + 77 \, a^{3} d^{3} + 32 \,{\left (b^{3} c d^{2} + 11 \, a b^{2} d^{3}\right )} x^{2} - 4 \,{\left (3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} - 77 \, a^{2} b d^{3}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{77 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

-4/77*(128*b^3*d^3*x^3 + 7*b^3*c^3 - 33*a*b^2*c^2*d + 77*a^2*b*c*d^2 + 77*a^3*d^3 + 32*(b^3*c*d^2 + 11*a*b^2*d
^3)*x^2 - 4*(3*b^3*c^2*d - 22*a*b^2*c*d^2 - 77*a^2*b*d^3)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(15/4)/(d*x+c)**(5/4),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

Timed out