3.1768 \(\int \frac{(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=249 \[ \frac{e^2 x (a+b x) (-3 a B e+A b e+3 b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)^3}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^3 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.222905, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ \frac{e^2 x (a+b x) (-3 a B e+A b e+3 b B d)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^2 (-4 a B e+3 A b e+b B d)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (b d-a e)^3}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^3 x^2 (a+b x)}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

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

Mathematica [A]  time = 0.165044, size = 256, normalized size = 1.03 \[ \frac{-A b \left (a^2 b e^2 (4 e x-9 d)+5 a^3 e^3+a b^2 e \left (3 d^2-12 d e x-4 e^2 x^2\right )+b^3 \left (6 d^2 e x+d^3-2 e^3 x^3\right )\right )+B \left (a^2 b^2 e \left (9 d^2-12 d e x-11 e^2 x^2\right )+a^3 b e^2 (2 e x-15 d)+7 a^4 e^3-a b^3 \left (-12 d^2 e x+d^3-12 d e^2 x^2+4 e^3 x^3\right )+b^4 x \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )\right )+6 e (a+b x)^2 (b d-a e) \log (a+b x) (-2 a B e+A b e+b B d)}{2 b^5 (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(A*b*(5*a^3*e^3 + a^2*b*e^2*(-9*d + 4*e*x) + a*b^2*e*(3*d^2 - 12*d*e*x - 4*e^2*x^2) + b^3*(d^3 + 6*d^2*e*x -
2*e^3*x^3))) + B*(7*a^4*e^3 + a^3*b*e^2*(-15*d + 2*e*x) + a^2*b^2*e*(9*d^2 - 12*d*e*x - 11*e^2*x^2) + b^4*x*(
-2*d^3 + 6*d*e^2*x^2 + e^3*x^3) - a*b^3*(d^3 - 12*d^2*e*x - 12*d*e^2*x^2 + 4*e^3*x^3)) + 6*e*(b*d - a*e)*(b*B*
d + A*b*e - 2*a*B*e)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.017, size = 556, normalized size = 2.2 \begin{align*} -{\frac{ \left ( -24\,B\ln \left ( bx+a \right ) x{a}^{3}b{e}^{3}-12\,B{x}^{2}a{b}^{3}d{e}^{2}+3\,A{d}^{2}a{b}^{3}e-9\,{b}^{2}B{a}^{2}{d}^{2}e-9\,Ad{a}^{2}{b}^{2}{e}^{2}+6\,Ax{b}^{4}{d}^{2}e-2\,Bx{a}^{3}b{e}^{3}+6\,A\ln \left ( bx+a \right ){a}^{3}b{e}^{3}+4\,Ax{a}^{2}{b}^{2}{e}^{3}-4\,A{x}^{2}a{b}^{3}{e}^{3}+11\,B{x}^{2}{a}^{2}{b}^{2}{e}^{3}+4\,B{x}^{3}a{b}^{3}{e}^{3}-6\,B{x}^{3}{b}^{4}d{e}^{2}+A{d}^{3}{b}^{4}-7\,B{e}^{3}{a}^{4}+15\,B{a}^{3}bd{e}^{2}-12\,A\ln \left ( bx+a \right ) xa{b}^{3}d{e}^{2}+36\,B\ln \left ( bx+a \right ) x{a}^{2}{b}^{2}d{e}^{2}-12\,B\ln \left ( bx+a \right ) xa{b}^{3}{d}^{2}e+18\,B\ln \left ( bx+a \right ){x}^{2}a{b}^{3}d{e}^{2}+12\,A\ln \left ( bx+a \right ) x{a}^{2}{b}^{2}{e}^{3}+6\,A\ln \left ( bx+a \right ){x}^{2}a{b}^{3}{e}^{3}-6\,A\ln \left ( bx+a \right ){x}^{2}{b}^{4}d{e}^{2}-6\,B\ln \left ( bx+a \right ){x}^{2}{b}^{4}{d}^{2}e-12\,B\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{3}+Ba{b}^{3}{d}^{3}-B{x}^{4}{b}^{4}{e}^{3}-2\,A{x}^{3}{b}^{4}{e}^{3}-12\,B\ln \left ( bx+a \right ){a}^{4}{e}^{3}+2\,Bx{b}^{4}{d}^{3}-12\,Axa{b}^{3}d{e}^{2}+18\,B\ln \left ( bx+a \right ){a}^{3}bd{e}^{2}-6\,B\ln \left ( bx+a \right ){a}^{2}{b}^{2}{d}^{2}e+12\,Bx{a}^{2}{b}^{2}d{e}^{2}-12\,Bxa{b}^{3}{d}^{2}e-6\,A\ln \left ( bx+a \right ){a}^{2}{b}^{2}d{e}^{2}+5\,A{a}^{3}b{e}^{3} \right ) \left ( bx+a \right ) }{2\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-24*B*ln(b*x+a)*x*a^3*b*e^3-12*B*x^2*a*b^3*d*e^2+3*A*d^2*a*b^3*e-9*b^2*B*a^2*d^2*e-9*A*d*a^2*b^2*e^2+6*A
*x*b^4*d^2*e-2*B*x*a^3*b*e^3+6*A*ln(b*x+a)*a^3*b*e^3+4*A*x*a^2*b^2*e^3-4*A*x^2*a*b^3*e^3+11*B*x^2*a^2*b^2*e^3+
4*B*x^3*a*b^3*e^3-6*B*x^3*b^4*d*e^2+A*d^3*b^4-7*B*e^3*a^4+15*B*a^3*b*d*e^2-12*A*ln(b*x+a)*x*a*b^3*d*e^2+36*B*l
n(b*x+a)*x*a^2*b^2*d*e^2-12*B*ln(b*x+a)*x*a*b^3*d^2*e+18*B*ln(b*x+a)*x^2*a*b^3*d*e^2+12*A*ln(b*x+a)*x*a^2*b^2*
e^3+6*A*ln(b*x+a)*x^2*a*b^3*e^3-6*A*ln(b*x+a)*x^2*b^4*d*e^2-6*B*ln(b*x+a)*x^2*b^4*d^2*e-12*B*ln(b*x+a)*x^2*a^2
*b^2*e^3+B*a*b^3*d^3-B*x^4*b^4*e^3-2*A*x^3*b^4*e^3-12*B*ln(b*x+a)*a^4*e^3+2*B*x*b^4*d^3-12*A*x*a*b^3*d*e^2+18*
B*ln(b*x+a)*a^3*b*d*e^2-6*B*ln(b*x+a)*a^2*b^2*d^2*e+12*B*x*a^2*b^2*d*e^2-12*B*x*a*b^3*d^2*e-6*A*ln(b*x+a)*a^2*
b^2*d*e^2+5*A*a^3*b*e^3)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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Maxima [B]  time = 0.976826, size = 792, normalized size = 3.18 \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)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*B*e^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 6*B*
a^2*e^3*log(x + a/b)/((b^2)^(3/2)*b^2) + 9*B*a^4*e^3/((b^2)^(7/2)*(x + a/b)^2) + 12*B*a^3*e^3*x/((b^2)^(5/2)*b
*(x + a/b)^2) - 5*B*a^3*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) - 1/2*A*d^3/((b^2)^(3/2)*(x + a/b)^2) + 5/2*B*
a^4*e^3/((b^2)^(3/2)*b^4*(x + a/b)^2) + (3*B*d*e^2 + A*e^3)*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 3*(B*d^2
*e + A*d*e^2)*log(x + a/b)/(b^2)^(3/2) - 3*(3*B*d*e^2 + A*e^3)*a*log(x + a/b)/((b^2)^(3/2)*b) - 9/2*(3*B*d*e^2
+ A*e^3)*a^3*b/((b^2)^(7/2)*(x + a/b)^2) + 9/2*(B*d^2*e + A*d*e^2)*a^2*b^2/((b^2)^(7/2)*(x + a/b)^2) - 6*(3*B
*d*e^2 + A*e^3)*a^2*x/((b^2)^(5/2)*(x + a/b)^2) + 6*(B*d^2*e + A*d*e^2)*a*b*x/((b^2)^(5/2)*(x + a/b)^2) + 2*(3
*B*d*e^2 + A*e^3)*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (B*d^3 + 3*A*d^2*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)
*b^2) - (3*B*d*e^2 + A*e^3)*a^3/((b^2)^(3/2)*b^3*(x + a/b)^2) + 1/2*(B*d^3 + 3*A*d^2*e)*a/((b^2)^(3/2)*b*(x +
a/b)^2)

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Fricas [B]  time = 1.33452, size = 875, normalized size = 3.51 \begin{align*} \frac{B b^{4} e^{3} x^{4} -{\left (B a b^{3} + A b^{4}\right )} d^{3} + 3 \,{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d^{2} e - 3 \,{\left (5 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} d e^{2} +{\left (7 \, B a^{4} - 5 \, A a^{3} b\right )} e^{3} + 2 \,{\left (3 \, B b^{4} d e^{2} -{\left (2 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} +{\left (12 \, B a b^{3} d e^{2} -{\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} e^{3}\right )} x^{2} - 2 \,{\left (B b^{4} d^{3} - 3 \,{\left (2 \, B a b^{3} - A b^{4}\right )} d^{2} e + 6 \,{\left (B a^{2} b^{2} - A a b^{3}\right )} d e^{2} -{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} e^{3}\right )} x + 6 \,{\left (B a^{2} b^{2} d^{2} e -{\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} d e^{2} +{\left (2 \, B a^{4} - A a^{3} b\right )} e^{3} +{\left (B b^{4} d^{2} e -{\left (3 \, B a b^{3} - A b^{4}\right )} d e^{2} +{\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 2 \,{\left (B a b^{3} d^{2} e -{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} d e^{2} +{\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \,{\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^4*e^3*x^4 - (B*a*b^3 + A*b^4)*d^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*d^2*e - 3*(5*B*a^3*b - 3*A*a^2*b^2)*d*e
^2 + (7*B*a^4 - 5*A*a^3*b)*e^3 + 2*(3*B*b^4*d*e^2 - (2*B*a*b^3 - A*b^4)*e^3)*x^3 + (12*B*a*b^3*d*e^2 - (11*B*a
^2*b^2 - 4*A*a*b^3)*e^3)*x^2 - 2*(B*b^4*d^3 - 3*(2*B*a*b^3 - A*b^4)*d^2*e + 6*(B*a^2*b^2 - A*a*b^3)*d*e^2 - (B
*a^3*b - 2*A*a^2*b^2)*e^3)*x + 6*(B*a^2*b^2*d^2*e - (3*B*a^3*b - A*a^2*b^2)*d*e^2 + (2*B*a^4 - A*a^3*b)*e^3 +
(B*b^4*d^2*e - (3*B*a*b^3 - A*b^4)*d*e^2 + (2*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 2*(B*a*b^3*d^2*e - (3*B*a^2*b^2
- A*a*b^3)*d*e^2 + (2*B*a^3*b - A*a^2*b^2)*e^3)*x)*log(b*x + a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x