3.1721 \(\int \frac{1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx\)

Optimal. Leaf size=101 \[ \frac{128 d^2 \sqrt [4]{a+b x}}{21 \sqrt [4]{c+d x} (b c-a d)^3}+\frac{32 d}{21 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac{4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)} \]

[Out]

-4/(7*(b*c - a*d)*(a + b*x)^(7/4)*(c + d*x)^(1/4)) + (32*d)/(21*(b*c - a*d)^2*(a + b*x)^(3/4)*(c + d*x)^(1/4))
+ (128*d^2*(a + b*x)^(1/4))/(21*(b*c - a*d)^3*(c + d*x)^(1/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0165107, 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} \[ \frac{128 d^2 \sqrt [4]{a+b x}}{21 \sqrt [4]{c+d x} (b c-a d)^3}+\frac{32 d}{21 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)^2}-\frac{4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]

[Out]

-4/(7*(b*c - a*d)*(a + b*x)^(7/4)*(c + d*x)^(1/4)) + (32*d)/(21*(b*c - a*d)^2*(a + b*x)^(3/4)*(c + d*x)^(1/4))
+ (128*d^2*(a + b*x)^(1/4))/(21*(b*c - a*d)^3*(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)^{11/4} (c+d x)^{5/4}} \, dx &=-\frac{4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac{(8 d) \int \frac{1}{(a+b x)^{7/4} (c+d x)^{5/4}} \, dx}{7 (b c-a d)}\\ &=-\frac{4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac{32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac{\left (32 d^2\right ) \int \frac{1}{(a+b x)^{3/4} (c+d x)^{5/4}} \, dx}{21 (b c-a d)^2}\\ &=-\frac{4}{7 (b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}+\frac{32 d}{21 (b c-a d)^2 (a+b x)^{3/4} \sqrt [4]{c+d x}}+\frac{128 d^2 \sqrt [4]{a+b x}}{21 (b c-a d)^3 \sqrt [4]{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.0314335, size = 76, normalized size = 0.75 \[ \frac{84 a^2 d^2+56 a b d (c+4 d x)+4 b^2 \left (-3 c^2+8 c d x+32 d^2 x^2\right )}{21 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]

[Out]

(84*a^2*d^2 + 56*a*b*d*(c + 4*d*x) + 4*b^2*(-3*c^2 + 8*c*d*x + 32*d^2*x^2))/(21*(b*c - a*d)^3*(a + b*x)^(7/4)*
(c + d*x)^(1/4))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 105, normalized size = 1. \begin{align*} -{\frac{128\,{b}^{2}{d}^{2}{x}^{2}+224\,ab{d}^{2}x+32\,{b}^{2}cdx+84\,{a}^{2}{d}^{2}+56\,abcd-12\,{b}^{2}{c}^{2}}{21\,{a}^{3}{d}^{3}-63\,{a}^{2}cb{d}^{2}+63\,a{b}^{2}{c}^{2}d-21\,{b}^{3}{c}^{3}} \left ( bx+a \right ) ^{-{\frac{7}{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)^(11/4)/(d*x+c)^(5/4),x)

[Out]

-4/21*(32*b^2*d^2*x^2+56*a*b*d^2*x+8*b^2*c*d*x+21*a^2*d^2+14*a*b*c*d-3*b^2*c^2)/(b*x+a)^(7/4)/(d*x+c)^(1/4)/(a
^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{11}{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)^(11/4)/(d*x+c)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(5/4)), x)

________________________________________________________________________________________

Fricas [B]  time = 4.1968, size = 558, normalized size = 5.52 \begin{align*} \frac{4 \,{\left (32 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 14 \, a b c d + 21 \, a^{2} d^{2} + 8 \,{\left (b^{2} c d + 7 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{21 \,{\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} +{\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} +{\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} +{\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="fricas")

[Out]

4/21*(32*b^2*d^2*x^2 - 3*b^2*c^2 + 14*a*b*c*d + 21*a^2*d^2 + 8*(b^2*c*d + 7*a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x +
c)^(3/4)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*
b^3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2
+ (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)

________________________________________________________________________________________

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)**(11/4)/(d*x+c)**(5/4),x)

[Out]

Timed out

________________________________________________________________________________________

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)^(11/4)/(d*x+c)^(5/4),x, algorithm="giac")

[Out]

Timed out