3.1698 \(\int \frac{1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{128 d^2 (c+d x)^{3/4}}{231 (a+b x)^{3/4} (b c-a d)^3}+\frac{32 d (c+d x)^{3/4}}{77 (a+b x)^{7/4} (b c-a d)^2}-\frac{4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)} \]

[Out]

(-4*(c + d*x)^(3/4))/(11*(b*c - a*d)*(a + b*x)^(11/4)) + (32*d*(c + d*x)^(3/4))/(77*(b*c - a*d)^2*(a + b*x)^(7
/4)) - (128*d^2*(c + d*x)^(3/4))/(231*(b*c - a*d)^3*(a + b*x)^(3/4))

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Rubi [A]  time = 0.0171828, 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} \[ -\frac{128 d^2 (c+d x)^{3/4}}{231 (a+b x)^{3/4} (b c-a d)^3}+\frac{32 d (c+d x)^{3/4}}{77 (a+b x)^{7/4} (b c-a d)^2}-\frac{4 (c+d x)^{3/4}}{11 (a+b x)^{11/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(c + d*x)^(3/4))/(11*(b*c - a*d)*(a + b*x)^(11/4)) + (32*d*(c + d*x)^(3/4))/(77*(b*c - a*d)^2*(a + b*x)^(7
/4)) - (128*d^2*(c + d*x)^(3/4))/(231*(b*c - a*d)^3*(a + b*x)^(3/4))

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} \sqrt [4]{c+d x}} \, dx &=-\frac{4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}-\frac{(8 d) \int \frac{1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx}{11 (b c-a d)}\\ &=-\frac{4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}+\frac{32 d (c+d x)^{3/4}}{77 (b c-a d)^2 (a+b x)^{7/4}}+\frac{\left (32 d^2\right ) \int \frac{1}{(a+b x)^{7/4} \sqrt [4]{c+d x}} \, dx}{77 (b c-a d)^2}\\ &=-\frac{4 (c+d x)^{3/4}}{11 (b c-a d) (a+b x)^{11/4}}+\frac{32 d (c+d x)^{3/4}}{77 (b c-a d)^2 (a+b x)^{7/4}}-\frac{128 d^2 (c+d x)^{3/4}}{231 (b c-a d)^3 (a+b x)^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0310133, size = 77, normalized size = 0.76 \[ -\frac{4 (c+d x)^{3/4} \left (77 a^2 d^2+22 a b d (4 d x-3 c)+b^2 \left (21 c^2-24 c d x+32 d^2 x^2\right )\right )}{231 (a+b x)^{11/4} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(c + d*x)^(3/4)*(77*a^2*d^2 + 22*a*b*d*(-3*c + 4*d*x) + b^2*(21*c^2 - 24*c*d*x + 32*d^2*x^2)))/(231*(b*c -
a*d)^3*(a + b*x)^(11/4))

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Maple [A]  time = 0.005, size = 105, normalized size = 1. \begin{align*}{\frac{128\,{b}^{2}{d}^{2}{x}^{2}+352\,ab{d}^{2}x-96\,{b}^{2}cdx+308\,{a}^{2}{d}^{2}-264\,abcd+84\,{b}^{2}{c}^{2}}{231\,{a}^{3}{d}^{3}-693\,{a}^{2}cb{d}^{2}+693\,a{b}^{2}{c}^{2}d-231\,{b}^{3}{c}^{3}} \left ( dx+c \right ) ^{{\frac{3}{4}}} \left ( bx+a \right ) ^{-{\frac{11}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/231*(d*x+c)^(3/4)*(32*b^2*d^2*x^2+88*a*b*d^2*x-24*b^2*c*d*x+77*a^2*d^2-66*a*b*c*d+21*b^2*c^2)/(b*x+a)^(11/4)
/(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{15}{4}}{\left (d x + c\right )}^{\frac{1}{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)^(1/4),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 6.77432, size = 522, normalized size = 5.17 \begin{align*} -\frac{4 \,{\left (32 \, b^{2} d^{2} x^{2} + 21 \, b^{2} c^{2} - 66 \, a b c d + 77 \, a^{2} d^{2} - 8 \,{\left (3 \, b^{2} c d - 11 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{231 \,{\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)^(15/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

-4/231*(32*b^2*d^2*x^2 + 21*b^2*c^2 - 66*a*b*c*d + 77*a^2*d^2 - 8*(3*b^2*c*d - 11*a*b*d^2)*x)*(b*x + a)^(1/4)*
(d*x + c)^(3/4)/(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)**(15/4)/(d*x+c)**(1/4),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)^(1/4),x, algorithm="giac")

[Out]

Timed out