3.1738 \(\int \frac{(a+b x)^3 (A+B x)}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=171 \[ -\frac{2 b^2 (d+e x)^{7/2} (-3 a B e-A b e+4 b B d)}{7 e^5}+\frac{6 b (d+e x)^{5/2} (b d-a e) (-a B e-A b e+2 b B d)}{5 e^5}-\frac{2 (d+e x)^{3/2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{3 e^5}+\frac{2 \sqrt{d+e x} (b d-a e)^3 (B d-A e)}{e^5}+\frac{2 b^3 B (d+e x)^{9/2}}{9 e^5} \]

[Out]

(2*(b*d - a*e)^3*(B*d - A*e)*Sqrt[d + e*x])/e^5 - (2*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x)^(3/2)
)/(3*e^5) + (6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(5/2))/(5*e^5) - (2*b^2*(4*b*B*d - A*b*e - 3*
a*B*e)*(d + e*x)^(7/2))/(7*e^5) + (2*b^3*B*(d + e*x)^(9/2))/(9*e^5)

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Rubi [A]  time = 0.0682577, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {77} \[ -\frac{2 b^2 (d+e x)^{7/2} (-3 a B e-A b e+4 b B d)}{7 e^5}+\frac{6 b (d+e x)^{5/2} (b d-a e) (-a B e-A b e+2 b B d)}{5 e^5}-\frac{2 (d+e x)^{3/2} (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{3 e^5}+\frac{2 \sqrt{d+e x} (b d-a e)^3 (B d-A e)}{e^5}+\frac{2 b^3 B (d+e x)^{9/2}}{9 e^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/Sqrt[d + e*x],x]

[Out]

(2*(b*d - a*e)^3*(B*d - A*e)*Sqrt[d + e*x])/e^5 - (2*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x)^(3/2)
)/(3*e^5) + (6*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(5/2))/(5*e^5) - (2*b^2*(4*b*B*d - A*b*e - 3*
a*B*e)*(d + e*x)^(7/2))/(7*e^5) + (2*b^3*B*(d + e*x)^(9/2))/(9*e^5)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{(a+b x)^3 (A+B x)}{\sqrt{d+e x}} \, dx &=\int \left (\frac{(-b d+a e)^3 (-B d+A e)}{e^4 \sqrt{d+e x}}+\frac{(-b d+a e)^2 (-4 b B d+3 A b e+a B e) \sqrt{d+e x}}{e^4}-\frac{3 b (b d-a e) (-2 b B d+A b e+a B e) (d+e x)^{3/2}}{e^4}+\frac{b^2 (-4 b B d+A b e+3 a B e) (d+e x)^{5/2}}{e^4}+\frac{b^3 B (d+e x)^{7/2}}{e^4}\right ) \, dx\\ &=\frac{2 (b d-a e)^3 (B d-A e) \sqrt{d+e x}}{e^5}-\frac{2 (b d-a e)^2 (4 b B d-3 A b e-a B e) (d+e x)^{3/2}}{3 e^5}+\frac{6 b (b d-a e) (2 b B d-A b e-a B e) (d+e x)^{5/2}}{5 e^5}-\frac{2 b^2 (4 b B d-A b e-3 a B e) (d+e x)^{7/2}}{7 e^5}+\frac{2 b^3 B (d+e x)^{9/2}}{9 e^5}\\ \end{align*}

Mathematica [A]  time = 0.0783036, size = 145, normalized size = 0.85 \[ \frac{2 \sqrt{d+e x} \left (-45 b^2 (d+e x)^3 (-3 a B e-A b e+4 b B d)+189 b (d+e x)^2 (b d-a e) (-a B e-A b e+2 b B d)-105 (d+e x) (b d-a e)^2 (-a B e-3 A b e+4 b B d)+315 (b d-a e)^3 (B d-A e)+35 b^3 B (d+e x)^4\right )}{315 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/Sqrt[d + e*x],x]

[Out]

(2*Sqrt[d + e*x]*(315*(b*d - a*e)^3*(B*d - A*e) - 105*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x) + 18
9*b*(b*d - a*e)*(2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2 - 45*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^3 + 35*b^
3*B*(d + e*x)^4))/(315*e^5)

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Maple [A]  time = 0.007, size = 301, normalized size = 1.8 \begin{align*}{\frac{70\,{b}^{3}B{x}^{4}{e}^{4}+90\,A{b}^{3}{e}^{4}{x}^{3}+270\,Ba{b}^{2}{e}^{4}{x}^{3}-80\,B{b}^{3}d{e}^{3}{x}^{3}+378\,Aa{b}^{2}{e}^{4}{x}^{2}-108\,A{b}^{3}d{e}^{3}{x}^{2}+378\,B{a}^{2}b{e}^{4}{x}^{2}-324\,Ba{b}^{2}d{e}^{3}{x}^{2}+96\,B{b}^{3}{d}^{2}{e}^{2}{x}^{2}+630\,A{a}^{2}b{e}^{4}x-504\,Aa{b}^{2}d{e}^{3}x+144\,A{b}^{3}{d}^{2}{e}^{2}x+210\,B{a}^{3}{e}^{4}x-504\,B{a}^{2}bd{e}^{3}x+432\,Ba{b}^{2}{d}^{2}{e}^{2}x-128\,B{b}^{3}{d}^{3}ex+630\,{a}^{3}A{e}^{4}-1260\,A{a}^{2}bd{e}^{3}+1008\,Aa{b}^{2}{d}^{2}{e}^{2}-288\,A{b}^{3}{d}^{3}e-420\,B{a}^{3}d{e}^{3}+1008\,B{a}^{2}b{d}^{2}{e}^{2}-864\,Ba{b}^{2}{d}^{3}e+256\,B{b}^{3}{d}^{4}}{315\,{e}^{5}}\sqrt{ex+d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^(1/2),x)

[Out]

2/315*(e*x+d)^(1/2)*(35*B*b^3*e^4*x^4+45*A*b^3*e^4*x^3+135*B*a*b^2*e^4*x^3-40*B*b^3*d*e^3*x^3+189*A*a*b^2*e^4*
x^2-54*A*b^3*d*e^3*x^2+189*B*a^2*b*e^4*x^2-162*B*a*b^2*d*e^3*x^2+48*B*b^3*d^2*e^2*x^2+315*A*a^2*b*e^4*x-252*A*
a*b^2*d*e^3*x+72*A*b^3*d^2*e^2*x+105*B*a^3*e^4*x-252*B*a^2*b*d*e^3*x+216*B*a*b^2*d^2*e^2*x-64*B*b^3*d^3*e*x+31
5*A*a^3*e^4-630*A*a^2*b*d*e^3+504*A*a*b^2*d^2*e^2-144*A*b^3*d^3*e-210*B*a^3*d*e^3+504*B*a^2*b*d^2*e^2-432*B*a*
b^2*d^3*e+128*B*b^3*d^4)/e^5

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Maxima [A]  time = 1.04201, size = 358, normalized size = 2.09 \begin{align*} \frac{2 \,{\left (35 \,{\left (e x + d\right )}^{\frac{9}{2}} B b^{3} - 45 \,{\left (4 \, B b^{3} d -{\left (3 \, B a b^{2} + A b^{3}\right )} e\right )}{\left (e x + d\right )}^{\frac{7}{2}} + 189 \,{\left (2 \, B b^{3} d^{2} -{\left (3 \, B a b^{2} + A b^{3}\right )} d e +{\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )}{\left (e x + d\right )}^{\frac{5}{2}} - 105 \,{\left (4 \, B b^{3} d^{3} - 3 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e + 6 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{2} -{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{3}\right )}{\left (e x + d\right )}^{\frac{3}{2}} + 315 \,{\left (B b^{3} d^{4} + A a^{3} e^{4} -{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 3 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} -{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3}\right )} \sqrt{e x + d}\right )}}{315 \, e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

2/315*(35*(e*x + d)^(9/2)*B*b^3 - 45*(4*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*(e*x + d)^(7/2) + 189*(2*B*b^3*d^2 -
(3*B*a*b^2 + A*b^3)*d*e + (B*a^2*b + A*a*b^2)*e^2)*(e*x + d)^(5/2) - 105*(4*B*b^3*d^3 - 3*(3*B*a*b^2 + A*b^3)*
d^2*e + 6*(B*a^2*b + A*a*b^2)*d*e^2 - (B*a^3 + 3*A*a^2*b)*e^3)*(e*x + d)^(3/2) + 315*(B*b^3*d^4 + A*a^3*e^4 -
(3*B*a*b^2 + A*b^3)*d^3*e + 3*(B*a^2*b + A*a*b^2)*d^2*e^2 - (B*a^3 + 3*A*a^2*b)*d*e^3)*sqrt(e*x + d))/e^5

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Fricas [A]  time = 1.35619, size = 579, normalized size = 3.39 \begin{align*} \frac{2 \,{\left (35 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} + 315 \, A a^{3} e^{4} - 144 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 504 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 210 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} - 5 \,{\left (8 \, B b^{3} d e^{3} - 9 \,{\left (3 \, B a b^{2} + A b^{3}\right )} e^{4}\right )} x^{3} + 3 \,{\left (16 \, B b^{3} d^{2} e^{2} - 18 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 63 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} -{\left (64 \, B b^{3} d^{3} e - 72 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 252 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 105 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

2/315*(35*B*b^3*e^4*x^4 + 128*B*b^3*d^4 + 315*A*a^3*e^4 - 144*(3*B*a*b^2 + A*b^3)*d^3*e + 504*(B*a^2*b + A*a*b
^2)*d^2*e^2 - 210*(B*a^3 + 3*A*a^2*b)*d*e^3 - 5*(8*B*b^3*d*e^3 - 9*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*
d^2*e^2 - 18*(3*B*a*b^2 + A*b^3)*d*e^3 + 63*(B*a^2*b + A*a*b^2)*e^4)*x^2 - (64*B*b^3*d^3*e - 72*(3*B*a*b^2 + A
*b^3)*d^2*e^2 + 252*(B*a^2*b + A*a*b^2)*d*e^3 - 105*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/e^5

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Sympy [A]  time = 66.7763, size = 916, normalized size = 5.36 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**(1/2),x)

[Out]

Piecewise((-(2*A*a**3*d/sqrt(d + e*x) + 2*A*a**3*(-d/sqrt(d + e*x) - sqrt(d + e*x)) + 6*A*a**2*b*d*(-d/sqrt(d
+ e*x) - sqrt(d + e*x))/e + 6*A*a**2*b*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e + 6*A*a
*b**2*d*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e**2 + 6*A*a*b**2*(-d**3/sqrt(d + e*x) -
3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**2 + 2*A*b**3*d*(-d**3/sqrt(d + e*x) - 3*d*
*2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**3 + 2*A*b**3*(d**4/sqrt(d + e*x) + 4*d**3*sqrt(
d + e*x) - 2*d**2*(d + e*x)**(3/2) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7)/e**3 + 2*B*a**3*d*(-d/sqrt(d
+ e*x) - sqrt(d + e*x))/e + 2*B*a**3*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e + 6*B*a*
*2*b*d*(d**2/sqrt(d + e*x) + 2*d*sqrt(d + e*x) - (d + e*x)**(3/2)/3)/e**2 + 6*B*a**2*b*(-d**3/sqrt(d + e*x) -
3*d**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**2 + 6*B*a*b**2*d*(-d**3/sqrt(d + e*x) - 3*d
**2*sqrt(d + e*x) + d*(d + e*x)**(3/2) - (d + e*x)**(5/2)/5)/e**3 + 6*B*a*b**2*(d**4/sqrt(d + e*x) + 4*d**3*sq
rt(d + e*x) - 2*d**2*(d + e*x)**(3/2) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7)/e**3 + 2*B*b**3*d*(d**4/s
qrt(d + e*x) + 4*d**3*sqrt(d + e*x) - 2*d**2*(d + e*x)**(3/2) + 4*d*(d + e*x)**(5/2)/5 - (d + e*x)**(7/2)/7)/e
**4 + 2*B*b**3*(-d**5/sqrt(d + e*x) - 5*d**4*sqrt(d + e*x) + 10*d**3*(d + e*x)**(3/2)/3 - 2*d**2*(d + e*x)**(5
/2) + 5*d*(d + e*x)**(7/2)/7 - (d + e*x)**(9/2)/9)/e**4)/e, Ne(e, 0)), ((A*a**3*x + B*b**3*x**5/5 + x**4*(A*b*
*3 + 3*B*a*b**2)/4 + x**3*(3*A*a*b**2 + 3*B*a**2*b)/3 + x**2*(3*A*a**2*b + B*a**3)/2)/sqrt(d), True))

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Giac [B]  time = 1.37284, size = 466, normalized size = 2.73 \begin{align*} \frac{2}{315} \,{\left (105 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} B a^{3} e^{\left (-1\right )} + 315 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} A a^{2} b e^{\left (-1\right )} + 63 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} d + 15 \, \sqrt{x e + d} d^{2}\right )} B a^{2} b e^{\left (-2\right )} + 63 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} d + 15 \, \sqrt{x e + d} d^{2}\right )} A a b^{2} e^{\left (-2\right )} + 27 \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2} - 35 \, \sqrt{x e + d} d^{3}\right )} B a b^{2} e^{\left (-3\right )} + 9 \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} - 21 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2} - 35 \, \sqrt{x e + d} d^{3}\right )} A b^{3} e^{\left (-3\right )} +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 180 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 378 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 420 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3} + 315 \, \sqrt{x e + d} d^{4}\right )} B b^{3} e^{\left (-4\right )} + 315 \, \sqrt{x e + d} A a^{3}\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

2/315*(105*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*B*a^3*e^(-1) + 315*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*A*a^
2*b*e^(-1) + 63*(3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*B*a^2*b*e^(-2) + 63*(3*(x*e
+ d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*A*a*b^2*e^(-2) + 27*(5*(x*e + d)^(7/2) - 21*(x*e + d
)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*B*a*b^2*e^(-3) + 9*(5*(x*e + d)^(7/2) - 21*(x*e + d
)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2 - 35*sqrt(x*e + d)*d^3)*A*b^3*e^(-3) + (35*(x*e + d)^(9/2) - 180*(x*e + d)^
(7/2)*d + 378*(x*e + d)^(5/2)*d^2 - 420*(x*e + d)^(3/2)*d^3 + 315*sqrt(x*e + d)*d^4)*B*b^3*e^(-4) + 315*sqrt(x
*e + d)*A*a^3)*e^(-1)