3.18.31 \(\int \frac {(a+b x)^2 (A+B x)}{(d+e x)^{3/2}} \, dx\) [1731]

Optimal. Leaf size=124 \[ \frac {2 (b d-a e)^2 (B d-A e)}{e^4 \sqrt {d+e x}}+\frac {2 (b d-a e) (3 b B d-2 A b e-a B e) \sqrt {d+e x}}{e^4}-\frac {2 b (3 b B d-A b e-2 a B e) (d+e x)^{3/2}}{3 e^4}+\frac {2 b^2 B (d+e x)^{5/2}}{5 e^4} \]

[Out]

-2/3*b*(-A*b*e-2*B*a*e+3*B*b*d)*(e*x+d)^(3/2)/e^4+2/5*b^2*B*(e*x+d)^(5/2)/e^4+2*(-a*e+b*d)^2*(-A*e+B*d)/e^4/(e
*x+d)^(1/2)+2*(-a*e+b*d)*(-2*A*b*e-B*a*e+3*B*b*d)*(e*x+d)^(1/2)/e^4

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Rubi [A]
time = 0.03, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 = {78} \begin {gather*} -\frac {2 b (d+e x)^{3/2} (-2 a B e-A b e+3 b B d)}{3 e^4}+\frac {2 \sqrt {d+e x} (b d-a e) (-a B e-2 A b e+3 b B d)}{e^4}+\frac {2 (b d-a e)^2 (B d-A e)}{e^4 \sqrt {d+e x}}+\frac {2 b^2 B (d+e x)^{5/2}}{5 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

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

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Mathematica [A]
time = 0.12, size = 135, normalized size = 1.09 \begin {gather*} \frac {30 a^2 e^2 (2 B d-A e+B e x)+20 a b e \left (3 A e (2 d+e x)+B \left (-8 d^2-4 d e x+e^2 x^2\right )\right )+2 b^2 \left (5 A e \left (-8 d^2-4 d e x+e^2 x^2\right )+3 B \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right )\right )}{15 e^4 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(30*a^2*e^2*(2*B*d - A*e + B*e*x) + 20*a*b*e*(3*A*e*(2*d + e*x) + B*(-8*d^2 - 4*d*e*x + e^2*x^2)) + 2*b^2*(5*A
*e*(-8*d^2 - 4*d*e*x + e^2*x^2) + 3*B*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3)))/(15*e^4*Sqrt[d + e*x])

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Maple [A]
time = 0.09, size = 200, normalized size = 1.61

method	result	size
risch	\(\frac {2 \left (3 b^{2} B \,x^{2} e^{2}+5 b^{2} A x \,e^{2}+10 a b B x \,e^{2}-9 B \,b^{2} d e x +30 A a b \,e^{2}-25 A \,b^{2} d e +15 B \,a^{2} e^{2}-50 B a b d e +33 b^{2} B \,d^{2}\right ) \sqrt {e x +d}}{15 e^{4}}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{e^{4} \sqrt {e x +d}}\)	\(163\)
gosper	\(-\frac {2 \left (-3 b^{2} B \,x^{3} e^{3}-5 A \,b^{2} e^{3} x^{2}-10 B a b \,e^{3} x^{2}+6 B \,b^{2} d \,e^{2} x^{2}-30 A a b \,e^{3} x +20 A \,b^{2} d \,e^{2} x -15 B \,a^{2} e^{3} x +40 B a b d \,e^{2} x -24 B \,b^{2} d^{2} e x +15 a^{2} A \,e^{3}-60 A a b d \,e^{2}+40 A \,b^{2} d^{2} e -30 B \,a^{2} d \,e^{2}+80 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\)	\(169\)
trager	\(-\frac {2 \left (-3 b^{2} B \,x^{3} e^{3}-5 A \,b^{2} e^{3} x^{2}-10 B a b \,e^{3} x^{2}+6 B \,b^{2} d \,e^{2} x^{2}-30 A a b \,e^{3} x +20 A \,b^{2} d \,e^{2} x -15 B \,a^{2} e^{3} x +40 B a b d \,e^{2} x -24 B \,b^{2} d^{2} e x +15 a^{2} A \,e^{3}-60 A a b d \,e^{2}+40 A \,b^{2} d^{2} e -30 B \,a^{2} d \,e^{2}+80 B a b \,d^{2} e -48 b^{2} B \,d^{3}\right )}{15 \sqrt {e x +d}\, e^{4}}\)	\(169\)
derivativedivides	\(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}+4 A a b \,e^{2} \sqrt {e x +d}-4 A \,b^{2} d e \sqrt {e x +d}+2 B \,a^{2} e^{2} \sqrt {e x +d}-8 B a b d e \sqrt {e x +d}+6 B \,b^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(200\)
default	\(\frac {\frac {2 b^{2} B \left (e x +d \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{2} e \left (e x +d \right )^{\frac {3}{2}}}{3}+\frac {4 B a b e \left (e x +d \right )^{\frac {3}{2}}}{3}-2 B \,b^{2} d \left (e x +d \right )^{\frac {3}{2}}+4 A a b \,e^{2} \sqrt {e x +d}-4 A \,b^{2} d e \sqrt {e x +d}+2 B \,a^{2} e^{2} \sqrt {e x +d}-8 B a b d e \sqrt {e x +d}+6 B \,b^{2} d^{2} \sqrt {e x +d}-\frac {2 \left (a^{2} A \,e^{3}-2 A a b d \,e^{2}+A \,b^{2} d^{2} e -B \,a^{2} d \,e^{2}+2 B a b \,d^{2} e -b^{2} B \,d^{3}\right )}{\sqrt {e x +d}}}{e^{4}}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/e^4*(1/5*b^2*B*(e*x+d)^(5/2)+1/3*A*b^2*e*(e*x+d)^(3/2)+2/3*B*a*b*e*(e*x+d)^(3/2)-B*b^2*d*(e*x+d)^(3/2)+2*A*a
*b*e^2*(e*x+d)^(1/2)-2*A*b^2*d*e*(e*x+d)^(1/2)+B*a^2*e^2*(e*x+d)^(1/2)-4*B*a*b*d*e*(e*x+d)^(1/2)+3*B*b^2*d^2*(
e*x+d)^(1/2)-(A*a^2*e^3-2*A*a*b*d*e^2+A*b^2*d^2*e-B*a^2*d*e^2+2*B*a*b*d^2*e-B*b^2*d^3)/(e*x+d)^(1/2))

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Maxima [A]
time = 0.45, size = 174, normalized size = 1.40 \begin {gather*} \frac {2}{15} \, {\left ({\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{2} - 5 \, {\left (3 \, B b^{2} d - 2 \, B a b e - A b^{2} e\right )} {\left (x e + d\right )}^{\frac {3}{2}} + 15 \, {\left (3 \, B b^{2} d^{2} + B a^{2} e^{2} + 2 \, A a b e^{2} - 2 \, {\left (2 \, B a b e + A b^{2} e\right )} d\right )} \sqrt {x e + d}\right )} e^{\left (-3\right )} + \frac {15 \, {\left (B b^{2} d^{3} - A a^{2} e^{3} - {\left (2 \, B a b e + A b^{2} e\right )} d^{2} + {\left (B a^{2} e^{2} + 2 \, A a b e^{2}\right )} d\right )} e^{\left (-3\right )}}{\sqrt {x e + d}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*((3*(x*e + d)^(5/2)*B*b^2 - 5*(3*B*b^2*d - 2*B*a*b*e - A*b^2*e)*(x*e + d)^(3/2) + 15*(3*B*b^2*d^2 + B*a^2
*e^2 + 2*A*a*b*e^2 - 2*(2*B*a*b*e + A*b^2*e)*d)*sqrt(x*e + d))*e^(-3) + 15*(B*b^2*d^3 - A*a^2*e^3 - (2*B*a*b*e
+ A*b^2*e)*d^2 + (B*a^2*e^2 + 2*A*a*b*e^2)*d)*e^(-3)/sqrt(x*e + d))*e^(-1)

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Fricas [A]
time = 1.00, size = 155, normalized size = 1.25 \begin {gather*} \frac {2 \, {\left (48 \, B b^{2} d^{3} + {\left (3 \, B b^{2} x^{3} - 15 \, A a^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )} e^{3} - 2 \, {\left (3 \, B b^{2} d x^{2} + 10 \, {\left (2 \, B a b + A b^{2}\right )} d x - 15 \, {\left (B a^{2} + 2 \, A a b\right )} d\right )} e^{2} + 8 \, {\left (3 \, B b^{2} d^{2} x - 5 \, {\left (2 \, B a b + A b^{2}\right )} d^{2}\right )} e\right )} \sqrt {x e + d}}{15 \, {\left (x e^{5} + d e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(48*B*b^2*d^3 + (3*B*b^2*x^3 - 15*A*a^2 + 5*(2*B*a*b + A*b^2)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)*e^3 - 2*(3*B*
b^2*d*x^2 + 10*(2*B*a*b + A*b^2)*d*x - 15*(B*a^2 + 2*A*a*b)*d)*e^2 + 8*(3*B*b^2*d^2*x - 5*(2*B*a*b + A*b^2)*d^
2)*e)*sqrt(x*e + d)/(x*e^5 + d*e^4)

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Sympy [A]
time = 16.31, size = 150, normalized size = 1.21 \begin {gather*} \frac {2 B b^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} + \frac {\left (d + e x\right )^{\frac {3}{2}} \cdot \left (2 A b^{2} e + 4 B a b e - 6 B b^{2} d\right )}{3 e^{4}} + \frac {\sqrt {d + e x} \left (4 A a b e^{2} - 4 A b^{2} d e + 2 B a^{2} e^{2} - 8 B a b d e + 6 B b^{2} d^{2}\right )}{e^{4}} + \frac {2 \left (- A e + B d\right ) \left (a e - b d\right )^{2}}{e^{4} \sqrt {d + e x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.65, size = 219, normalized size = 1.77 \begin {gather*} \frac {2}{15} \, {\left (3 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{2} e^{16} - 15 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{2} d e^{16} + 45 \, \sqrt {x e + d} B b^{2} d^{2} e^{16} + 10 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b e^{17} + 5 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{2} e^{17} - 60 \, \sqrt {x e + d} B a b d e^{17} - 30 \, \sqrt {x e + d} A b^{2} d e^{17} + 15 \, \sqrt {x e + d} B a^{2} e^{18} + 30 \, \sqrt {x e + d} A a b e^{18}\right )} e^{\left (-20\right )} + \frac {2 \, {\left (B b^{2} d^{3} - 2 \, B a b d^{2} e - A b^{2} d^{2} e + B a^{2} d e^{2} + 2 \, A a b d e^{2} - A a^{2} e^{3}\right )} e^{\left (-4\right )}}{\sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*(x*e + d)^(5/2)*B*b^2*e^16 - 15*(x*e + d)^(3/2)*B*b^2*d*e^16 + 45*sqrt(x*e + d)*B*b^2*d^2*e^16 + 10*(x
*e + d)^(3/2)*B*a*b*e^17 + 5*(x*e + d)^(3/2)*A*b^2*e^17 - 60*sqrt(x*e + d)*B*a*b*d*e^17 - 30*sqrt(x*e + d)*A*b
^2*d*e^17 + 15*sqrt(x*e + d)*B*a^2*e^18 + 30*sqrt(x*e + d)*A*a*b*e^18)*e^(-20) + 2*(B*b^2*d^3 - 2*B*a*b*d^2*e
- A*b^2*d^2*e + B*a^2*d*e^2 + 2*A*a*b*d*e^2 - A*a^2*e^3)*e^(-4)/sqrt(x*e + d)

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Mupad [B]
time = 0.08, size = 154, normalized size = 1.24 \begin {gather*} \frac {{\left (d+e\,x\right )}^{3/2}\,\left (2\,A\,b^2\,e-6\,B\,b^2\,d+4\,B\,a\,b\,e\right )}{3\,e^4}-\frac {-2\,B\,a^2\,d\,e^2+2\,A\,a^2\,e^3+4\,B\,a\,b\,d^2\,e-4\,A\,a\,b\,d\,e^2-2\,B\,b^2\,d^3+2\,A\,b^2\,d^2\,e}{e^4\,\sqrt {d+e\,x}}+\frac {2\,B\,b^2\,{\left (d+e\,x\right )}^{5/2}}{5\,e^4}+\frac {2\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (2\,A\,b\,e+B\,a\,e-3\,B\,b\,d\right )}{e^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d + e*x)^(3/2)*(2*A*b^2*e - 6*B*b^2*d + 4*B*a*b*e))/(3*e^4) - (2*A*a^2*e^3 - 2*B*b^2*d^3 + 2*A*b^2*d^2*e - 2
*B*a^2*d*e^2 - 4*A*a*b*d*e^2 + 4*B*a*b*d^2*e)/(e^4*(d + e*x)^(1/2)) + (2*B*b^2*(d + e*x)^(5/2))/(5*e^4) + (2*(
a*e - b*d)*(d + e*x)^(1/2)*(2*A*b*e + B*a*e - 3*B*b*d))/e^4

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