3.1740 \(\int \frac{(a+b x)^3 (A+B x)}{(d+e x)^{5/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac{6 b \sqrt{d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac{2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt{d+e x}}-\frac{2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac{2 b^3 B (d+e x)^{5/2}}{5 e^5} \]

[Out]

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

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Rubi [A]  time = 0.0666049, antiderivative size = 169, 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} \[ -\frac{2 b^2 (d+e x)^{3/2} (-3 a B e-A b e+4 b B d)}{3 e^5}+\frac{6 b \sqrt{d+e x} (b d-a e) (-a B e-A b e+2 b B d)}{e^5}+\frac{2 (b d-a e)^2 (-a B e-3 A b e+4 b B d)}{e^5 \sqrt{d+e x}}-\frac{2 (b d-a e)^3 (B d-A e)}{3 e^5 (d+e x)^{3/2}}+\frac{2 b^3 B (d+e x)^{5/2}}{5 e^5} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.137473, size = 145, normalized size = 0.86 \[ \frac{2 \left (-5 b^2 (d+e x)^3 (-3 a B e-A b e+4 b B d)+45 b (d+e x)^2 (b d-a e) (-a B e-A b e+2 b B d)+15 (d+e x) (b d-a e)^2 (-a B e-3 A b e+4 b B d)-5 (b d-a e)^3 (B d-A e)+3 b^3 B (d+e x)^4\right )}{15 e^5 (d+e x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^3*(A + B*x))/(d + e*x)^(5/2),x]

[Out]

(2*(-5*(b*d - a*e)^3*(B*d - A*e) + 15*(b*d - a*e)^2*(4*b*B*d - 3*A*b*e - a*B*e)*(d + e*x) + 45*b*(b*d - a*e)*(
2*b*B*d - A*b*e - a*B*e)*(d + e*x)^2 - 5*b^2*(4*b*B*d - A*b*e - 3*a*B*e)*(d + e*x)^3 + 3*b^3*B*(d + e*x)^4))/(
15*e^5*(d + e*x)^(3/2))

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Maple [A]  time = 0.006, size = 301, normalized size = 1.8 \begin{align*} -{\frac{-6\,{b}^{3}B{x}^{4}{e}^{4}-10\,A{b}^{3}{e}^{4}{x}^{3}-30\,Ba{b}^{2}{e}^{4}{x}^{3}+16\,B{b}^{3}d{e}^{3}{x}^{3}-90\,Aa{b}^{2}{e}^{4}{x}^{2}+60\,A{b}^{3}d{e}^{3}{x}^{2}-90\,B{a}^{2}b{e}^{4}{x}^{2}+180\,Ba{b}^{2}d{e}^{3}{x}^{2}-96\,B{b}^{3}{d}^{2}{e}^{2}{x}^{2}+90\,A{a}^{2}b{e}^{4}x-360\,Aa{b}^{2}d{e}^{3}x+240\,A{b}^{3}{d}^{2}{e}^{2}x+30\,B{a}^{3}{e}^{4}x-360\,B{a}^{2}bd{e}^{3}x+720\,Ba{b}^{2}{d}^{2}{e}^{2}x-384\,B{b}^{3}{d}^{3}ex+10\,{a}^{3}A{e}^{4}+60\,A{a}^{2}bd{e}^{3}-240\,Aa{b}^{2}{d}^{2}{e}^{2}+160\,A{b}^{3}{d}^{3}e+20\,B{a}^{3}d{e}^{3}-240\,B{a}^{2}b{d}^{2}{e}^{2}+480\,Ba{b}^{2}{d}^{3}e-256\,B{b}^{3}{d}^{4}}{15\,{e}^{5}} \left ( ex+d \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x)

[Out]

-2/15/(e*x+d)^(3/2)*(-3*B*b^3*e^4*x^4-5*A*b^3*e^4*x^3-15*B*a*b^2*e^4*x^3+8*B*b^3*d*e^3*x^3-45*A*a*b^2*e^4*x^2+
30*A*b^3*d*e^3*x^2-45*B*a^2*b*e^4*x^2+90*B*a*b^2*d*e^3*x^2-48*B*b^3*d^2*e^2*x^2+45*A*a^2*b*e^4*x-180*A*a*b^2*d
*e^3*x+120*A*b^3*d^2*e^2*x+15*B*a^3*e^4*x-180*B*a^2*b*d*e^3*x+360*B*a*b^2*d^2*e^2*x-192*B*b^3*d^3*e*x+5*A*a^3*
e^4+30*A*a^2*b*d*e^3-120*A*a*b^2*d^2*e^2+80*A*b^3*d^3*e+10*B*a^3*d*e^3-120*B*a^2*b*d^2*e^2+240*B*a*b^2*d^3*e-1
28*B*b^3*d^4)/e^5

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Maxima [A]  time = 1.02987, size = 366, normalized size = 2.17 \begin{align*} \frac{2 \,{\left (\frac{3 \,{\left (e x + d\right )}^{\frac{5}{2}} B b^{3} - 5 \,{\left (4 \, B b^{3} d -{\left (3 \, B a b^{2} + A b^{3}\right )} e\right )}{\left (e x + d\right )}^{\frac{3}{2}} + 45 \,{\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 )} \sqrt{e x + d}}{e^{4}} - \frac{5 \,{\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} - 3 \,{\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 )}\right )}}{{\left (e x + d\right )}^{\frac{3}{2}} e^{4}}\right )}}{15 \, e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="maxima")

[Out]

2/15*((3*(e*x + d)^(5/2)*B*b^3 - 5*(4*B*b^3*d - (3*B*a*b^2 + A*b^3)*e)*(e*x + d)^(3/2) + 45*(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)*sqrt(e*x + d))/e^4 - 5*(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 - 3*(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))/((e*x + d)^(3/2)*e^4))/e

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Fricas [A]  time = 1.5329, size = 605, normalized size = 3.58 \begin{align*} \frac{2 \,{\left (3 \, B b^{3} e^{4} x^{4} + 128 \, B b^{3} d^{4} - 5 \, A a^{3} e^{4} - 80 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{3} e + 120 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} e^{2} - 10 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e^{3} -{\left (8 \, B b^{3} d e^{3} - 5 \,{\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} - 10 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e^{3} + 15 \,{\left (B a^{2} b + A a b^{2}\right )} e^{4}\right )} x^{2} + 3 \,{\left (64 \, B b^{3} d^{3} e - 40 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} e^{2} + 60 \,{\left (B a^{2} b + A a b^{2}\right )} d e^{3} - 5 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{4}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (e^{7} x^{2} + 2 \, d e^{6} x + d^{2} e^{5}\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)^(5/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b^3*e^4*x^4 + 128*B*b^3*d^4 - 5*A*a^3*e^4 - 80*(3*B*a*b^2 + A*b^3)*d^3*e + 120*(B*a^2*b + A*a*b^2)*d
^2*e^2 - 10*(B*a^3 + 3*A*a^2*b)*d*e^3 - (8*B*b^3*d*e^3 - 5*(3*B*a*b^2 + A*b^3)*e^4)*x^3 + 3*(16*B*b^3*d^2*e^2
- 10*(3*B*a*b^2 + A*b^3)*d*e^3 + 15*(B*a^2*b + A*a*b^2)*e^4)*x^2 + 3*(64*B*b^3*d^3*e - 40*(3*B*a*b^2 + A*b^3)*
d^2*e^2 + 60*(B*a^2*b + A*a*b^2)*d*e^3 - 5*(B*a^3 + 3*A*a^2*b)*e^4)*x)*sqrt(e*x + d)/(e^7*x^2 + 2*d*e^6*x + d^
2*e^5)

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Sympy [A]  time = 58.7335, size = 199, normalized size = 1.18 \begin{align*} \frac{2 B b^{3} \left (d + e x\right )^{\frac{5}{2}}}{5 e^{5}} + \frac{\left (d + e x\right )^{\frac{3}{2}} \left (2 A b^{3} e + 6 B a b^{2} e - 8 B b^{3} d\right )}{3 e^{5}} + \frac{\sqrt{d + e x} \left (6 A a b^{2} e^{2} - 6 A b^{3} d e + 6 B a^{2} b e^{2} - 18 B a b^{2} d e + 12 B b^{3} d^{2}\right )}{e^{5}} - \frac{2 \left (a e - b d\right )^{2} \left (3 A b e + B a e - 4 B b d\right )}{e^{5} \sqrt{d + e x}} + \frac{2 \left (- A e + B d\right ) \left (a e - b d\right )^{3}}{3 e^{5} \left (d + e x\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3*(B*x+A)/(e*x+d)**(5/2),x)

[Out]

2*B*b**3*(d + e*x)**(5/2)/(5*e**5) + (d + e*x)**(3/2)*(2*A*b**3*e + 6*B*a*b**2*e - 8*B*b**3*d)/(3*e**5) + sqrt
(d + e*x)*(6*A*a*b**2*e**2 - 6*A*b**3*d*e + 6*B*a**2*b*e**2 - 18*B*a*b**2*d*e + 12*B*b**3*d**2)/e**5 - 2*(a*e
- b*d)**2*(3*A*b*e + B*a*e - 4*B*b*d)/(e**5*sqrt(d + e*x)) + 2*(-A*e + B*d)*(a*e - b*d)**3/(3*e**5*(d + e*x)**
(3/2))

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Giac [B]  time = 1.48557, size = 493, normalized size = 2.92 \begin{align*} \frac{2}{15} \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} B b^{3} e^{20} - 20 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{3} d e^{20} + 90 \, \sqrt{x e + d} B b^{3} d^{2} e^{20} + 15 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b^{2} e^{21} + 5 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{3} e^{21} - 135 \, \sqrt{x e + d} B a b^{2} d e^{21} - 45 \, \sqrt{x e + d} A b^{3} d e^{21} + 45 \, \sqrt{x e + d} B a^{2} b e^{22} + 45 \, \sqrt{x e + d} A a b^{2} e^{22}\right )} e^{\left (-25\right )} + \frac{2 \,{\left (12 \,{\left (x e + d\right )} B b^{3} d^{3} - B b^{3} d^{4} - 27 \,{\left (x e + d\right )} B a b^{2} d^{2} e - 9 \,{\left (x e + d\right )} A b^{3} d^{2} e + 3 \, B a b^{2} d^{3} e + A b^{3} d^{3} e + 18 \,{\left (x e + d\right )} B a^{2} b d e^{2} + 18 \,{\left (x e + d\right )} A a b^{2} d e^{2} - 3 \, B a^{2} b d^{2} e^{2} - 3 \, A a b^{2} d^{2} e^{2} - 3 \,{\left (x e + d\right )} B a^{3} e^{3} - 9 \,{\left (x e + d\right )} A a^{2} b e^{3} + B a^{3} d e^{3} + 3 \, A a^{2} b d e^{3} - A a^{3} e^{4}\right )} e^{\left (-5\right )}}{3 \,{\left (x e + d\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3*(B*x+A)/(e*x+d)^(5/2),x, algorithm="giac")

[Out]

2/15*(3*(x*e + d)^(5/2)*B*b^3*e^20 - 20*(x*e + d)^(3/2)*B*b^3*d*e^20 + 90*sqrt(x*e + d)*B*b^3*d^2*e^20 + 15*(x
*e + d)^(3/2)*B*a*b^2*e^21 + 5*(x*e + d)^(3/2)*A*b^3*e^21 - 135*sqrt(x*e + d)*B*a*b^2*d*e^21 - 45*sqrt(x*e + d
)*A*b^3*d*e^21 + 45*sqrt(x*e + d)*B*a^2*b*e^22 + 45*sqrt(x*e + d)*A*a*b^2*e^22)*e^(-25) + 2/3*(12*(x*e + d)*B*
b^3*d^3 - B*b^3*d^4 - 27*(x*e + d)*B*a*b^2*d^2*e - 9*(x*e + d)*A*b^3*d^2*e + 3*B*a*b^2*d^3*e + A*b^3*d^3*e + 1
8*(x*e + d)*B*a^2*b*d*e^2 + 18*(x*e + d)*A*a*b^2*d*e^2 - 3*B*a^2*b*d^2*e^2 - 3*A*a*b^2*d^2*e^2 - 3*(x*e + d)*B
*a^3*e^3 - 9*(x*e + d)*A*a^2*b*e^3 + B*a^3*d*e^3 + 3*A*a^2*b*d*e^3 - A*a^3*e^4)*e^(-5)/(x*e + d)^(3/2)