3.1699 \(\int \frac{1}{(a+b x)^{19/4} \sqrt [4]{c+d x}} \, dx\)

Optimal. Leaf size=136 \[ \frac{512 d^3 (c+d x)^{3/4}}{1155 (a+b x)^{3/4} (b c-a d)^4}-\frac{128 d^2 (c+d x)^{3/4}}{385 (a+b x)^{7/4} (b c-a d)^3}+\frac{16 d (c+d x)^{3/4}}{55 (a+b x)^{11/4} (b c-a d)^2}-\frac{4 (c+d x)^{3/4}}{15 (a+b x)^{15/4} (b c-a d)} \]

[Out]

(-4*(c + d*x)^(3/4))/(15*(b*c - a*d)*(a + b*x)^(15/4)) + (16*d*(c + d*x)^(3/4))/(55*(b*c - a*d)^2*(a + b*x)^(1
1/4)) - (128*d^2*(c + d*x)^(3/4))/(385*(b*c - a*d)^3*(a + b*x)^(7/4)) + (512*d^3*(c + d*x)^(3/4))/(1155*(b*c -
a*d)^4*(a + b*x)^(3/4))

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Rubi [A]  time = 0.0305414, 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} \[ \frac{512 d^3 (c+d x)^{3/4}}{1155 (a+b x)^{3/4} (b c-a d)^4}-\frac{128 d^2 (c+d x)^{3/4}}{385 (a+b x)^{7/4} (b c-a d)^3}+\frac{16 d (c+d x)^{3/4}}{55 (a+b x)^{11/4} (b c-a d)^2}-\frac{4 (c+d x)^{3/4}}{15 (a+b x)^{15/4} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(19/4)*(c + d*x)^(1/4)),x]

[Out]

(-4*(c + d*x)^(3/4))/(15*(b*c - a*d)*(a + b*x)^(15/4)) + (16*d*(c + d*x)^(3/4))/(55*(b*c - a*d)^2*(a + b*x)^(1
1/4)) - (128*d^2*(c + d*x)^(3/4))/(385*(b*c - a*d)^3*(a + b*x)^(7/4)) + (512*d^3*(c + d*x)^(3/4))/(1155*(b*c -
a*d)^4*(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)^{19/4} \sqrt [4]{c+d x}} \, dx &=-\frac{4 (c+d x)^{3/4}}{15 (b c-a d) (a+b x)^{15/4}}-\frac{(4 d) \int \frac{1}{(a+b x)^{15/4} \sqrt [4]{c+d x}} \, dx}{5 (b c-a d)}\\ &=-\frac{4 (c+d x)^{3/4}}{15 (b c-a d) (a+b x)^{15/4}}+\frac{16 d (c+d x)^{3/4}}{55 (b c-a d)^2 (a+b x)^{11/4}}+\frac{\left (32 d^2\right ) \int \frac{1}{(a+b x)^{11/4} \sqrt [4]{c+d x}} \, dx}{55 (b c-a d)^2}\\ &=-\frac{4 (c+d x)^{3/4}}{15 (b c-a d) (a+b x)^{15/4}}+\frac{16 d (c+d x)^{3/4}}{55 (b c-a d)^2 (a+b x)^{11/4}}-\frac{128 d^2 (c+d x)^{3/4}}{385 (b c-a d)^3 (a+b x)^{7/4}}-\frac{\left (128 d^3\right ) \int \frac{1}{(a+b x)^{7/4} \sqrt [4]{c+d x}} \, dx}{385 (b c-a d)^3}\\ &=-\frac{4 (c+d x)^{3/4}}{15 (b c-a d) (a+b x)^{15/4}}+\frac{16 d (c+d x)^{3/4}}{55 (b c-a d)^2 (a+b x)^{11/4}}-\frac{128 d^2 (c+d x)^{3/4}}{385 (b c-a d)^3 (a+b x)^{7/4}}+\frac{512 d^3 (c+d x)^{3/4}}{1155 (b c-a d)^4 (a+b x)^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.047585, size = 118, normalized size = 0.87 \[ \frac{4 (c+d x)^{3/4} \left (165 a^2 b d^2 (4 d x-3 c)+385 a^3 d^3+15 a b^2 d \left (21 c^2-24 c d x+32 d^2 x^2\right )+b^3 \left (84 c^2 d x-77 c^3-96 c d^2 x^2+128 d^3 x^3\right )\right )}{1155 (a+b x)^{15/4} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(19/4)*(c + d*x)^(1/4)),x]

[Out]

(4*(c + d*x)^(3/4)*(385*a^3*d^3 + 165*a^2*b*d^2*(-3*c + 4*d*x) + 15*a*b^2*d*(21*c^2 - 24*c*d*x + 32*d^2*x^2) +
b^3*(-77*c^3 + 84*c^2*d*x - 96*c*d^2*x^2 + 128*d^3*x^3)))/(1155*(b*c - a*d)^4*(a + b*x)^(15/4))

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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}+1920\,a{b}^{2}{d}^{3}{x}^{2}-384\,{b}^{3}c{d}^{2}{x}^{2}+2640\,{a}^{2}b{d}^{3}x-1440\,a{b}^{2}c{d}^{2}x+336\,{b}^{3}{c}^{2}dx+1540\,{a}^{3}{d}^{3}-1980\,{a}^{2}cb{d}^{2}+1260\,a{b}^{2}{c}^{2}d-308\,{b}^{3}{c}^{3}}{1155\,{a}^{4}{d}^{4}-4620\,{a}^{3}bc{d}^{3}+6930\,{a}^{2}{c}^{2}{b}^{2}{d}^{2}-4620\,a{b}^{3}{c}^{3}d+1155\,{b}^{4}{c}^{4}} \left ( dx+c \right ) ^{{\frac{3}{4}}} \left ( bx+a \right ) ^{-{\frac{15}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(19/4)/(d*x+c)^(1/4),x)

[Out]

4/1155*(d*x+c)^(3/4)*(128*b^3*d^3*x^3+480*a*b^2*d^3*x^2-96*b^3*c*d^2*x^2+660*a^2*b*d^3*x-360*a*b^2*c*d^2*x+84*
b^3*c^2*d*x+385*a^3*d^3-495*a^2*b*c*d^2+315*a*b^2*c^2*d-77*b^3*c^3)/(b*x+a)^(15/4)/(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{19}{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)^(19/4)/(d*x+c)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(19/4)*(d*x + c)^(1/4)), x)

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Fricas [B]  time = 17.7949, size = 865, normalized size = 6.36 \begin{align*} \frac{4 \,{\left (128 \, b^{3} d^{3} x^{3} - 77 \, b^{3} c^{3} + 315 \, a b^{2} c^{2} d - 495 \, a^{2} b c d^{2} + 385 \, a^{3} d^{3} - 96 \,{\left (b^{3} c d^{2} - 5 \, a b^{2} d^{3}\right )} x^{2} + 12 \,{\left (7 \, b^{3} c^{2} d - 30 \, a b^{2} c d^{2} + 55 \, a^{2} b d^{3}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{1155 \,{\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} +{\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \,{\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \,{\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \,{\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(19/4)/(d*x+c)^(1/4),x, algorithm="fricas")

[Out]

4/1155*(128*b^3*d^3*x^3 - 77*b^3*c^3 + 315*a*b^2*c^2*d - 495*a^2*b*c*d^2 + 385*a^3*d^3 - 96*(b^3*c*d^2 - 5*a*b
^2*d^3)*x^2 + 12*(7*b^3*c^2*d - 30*a*b^2*c*d^2 + 55*a^2*b*d^3)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(a^4*b^4*c^4
- 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^
2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3
+ a^5*b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2
+ 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*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)**(19/4)/(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)^(19/4)/(d*x+c)^(1/4),x, algorithm="giac")

[Out]

Timed out