3.1710 \(\int \frac{1}{(a+b x)^{13/4} (c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{128 d^2 \sqrt [4]{c+d x}}{45 \sqrt [4]{a+b x} (b c-a d)^3}+\frac{32 d \sqrt [4]{c+d x}}{45 (a+b x)^{5/4} (b c-a d)^2}-\frac{4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)} \]

[Out]

(-4*(c + d*x)^(1/4))/(9*(b*c - a*d)*(a + b*x)^(9/4)) + (32*d*(c + d*x)^(1/4))/(45*(b*c - a*d)^2*(a + b*x)^(5/4
)) - (128*d^2*(c + d*x)^(1/4))/(45*(b*c - a*d)^3*(a + b*x)^(1/4))

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Rubi [A]  time = 0.0174494, antiderivative size = 101, normalized size of antiderivative = 1., 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} \[ -\frac{128 d^2 \sqrt [4]{c+d x}}{45 \sqrt [4]{a+b x} (b c-a d)^3}+\frac{32 d \sqrt [4]{c+d x}}{45 (a+b x)^{5/4} (b c-a d)^2}-\frac{4 \sqrt [4]{c+d x}}{9 (a+b x)^{9/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4))/(9*(b*c - a*d)*(a + b*x)^(9/4)) + (32*d*(c + d*x)^(1/4))/(45*(b*c - a*d)^2*(a + b*x)^(5/4
)) - (128*d^2*(c + d*x)^(1/4))/(45*(b*c - a*d)^3*(a + b*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)^{13/4} (c+d x)^{3/4}} \, dx &=-\frac{4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}-\frac{(8 d) \int \frac{1}{(a+b x)^{9/4} (c+d x)^{3/4}} \, dx}{9 (b c-a d)}\\ &=-\frac{4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac{32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}+\frac{\left (32 d^2\right ) \int \frac{1}{(a+b x)^{5/4} (c+d x)^{3/4}} \, dx}{45 (b c-a d)^2}\\ &=-\frac{4 \sqrt [4]{c+d x}}{9 (b c-a d) (a+b x)^{9/4}}+\frac{32 d \sqrt [4]{c+d x}}{45 (b c-a d)^2 (a+b x)^{5/4}}-\frac{128 d^2 \sqrt [4]{c+d x}}{45 (b c-a d)^3 \sqrt [4]{a+b x}}\\ \end{align*}

Mathematica [A]  time = 0.0316454, size = 75, normalized size = 0.74 \[ -\frac{4 \sqrt [4]{c+d x} \left (45 a^2 d^2-18 a b d (c-4 d x)+b^2 \left (5 c^2-8 c d x+32 d^2 x^2\right )\right )}{45 (a+b x)^{9/4} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(13/4)*(c + d*x)^(3/4)),x]

[Out]

(-4*(c + d*x)^(1/4)*(45*a^2*d^2 - 18*a*b*d*(c - 4*d*x) + b^2*(5*c^2 - 8*c*d*x + 32*d^2*x^2)))/(45*(b*c - a*d)^
3*(a + b*x)^(9/4))

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Maple [A]  time = 0.005, size = 105, normalized size = 1. \begin{align*}{\frac{128\,{b}^{2}{d}^{2}{x}^{2}+288\,ab{d}^{2}x-32\,{b}^{2}cdx+180\,{a}^{2}{d}^{2}-72\,abcd+20\,{b}^{2}{c}^{2}}{45\,{a}^{3}{d}^{3}-135\,{a}^{2}cb{d}^{2}+135\,a{b}^{2}{c}^{2}d-45\,{b}^{3}{c}^{3}}\sqrt [4]{dx+c} \left ( bx+a \right ) ^{-{\frac{9}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x)

[Out]

4/45*(d*x+c)^(1/4)*(32*b^2*d^2*x^2+72*a*b*d^2*x-8*b^2*c*d*x+45*a^2*d^2-18*a*b*c*d+5*b^2*c^2)/(b*x+a)^(9/4)/(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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{13}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(13/4)*(d*x + c)^(3/4)), x)

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Fricas [B]  time = 2.36675, size = 516, normalized size = 5.11 \begin{align*} -\frac{4 \,{\left (32 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 18 \, a b c d + 45 \, a^{2} d^{2} - 8 \,{\left (b^{2} c d - 9 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{3}{4}}{\left (d x + c\right )}^{\frac{1}{4}}}{45 \,{\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

-4/45*(32*b^2*d^2*x^2 + 5*b^2*c^2 - 18*a*b*c*d + 45*a^2*d^2 - 8*(b^2*c*d - 9*a*b*d^2)*x)*(b*x + a)^(3/4)*(d*x
+ c)^(1/4)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d
^2 - a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 -
3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*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)**(13/4)/(d*x+c)**(3/4),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)^(13/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

Timed out