3.1777 \(\int \frac{(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=227 \[ -\frac{(d+e x)^4 (A b-a B)}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 B e^2 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^3 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.1745, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 78, 43} \[ -\frac{(d+e x)^4 (A b-a B)}{4 b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{3 B e^2 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^3 (a+b x) \log (a+b x)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*B*e^2*(b*d - a*e))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (B*(b*d - a*e)^3)/(3*b^5*(a + b*x)^2*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) - (3*B*e*(b*d - a*e)^2)/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((A*b - a*B)*(d +
e*x)^4)/(4*b*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*e^3*(a + b*x)*Log[a + b*x])/(b^5*Sqr
t[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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{(A+B x) (d+e x)^3}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 B \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^3}{\left (a b+b^2 x\right )^4} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 B \left (a b+b^2 x\right )\right ) \int \left (\frac{(b d-a e)^3}{b^7 (a+b x)^4}+\frac{3 e (b d-a e)^2}{b^7 (a+b x)^3}+\frac{3 e^2 (b d-a e)}{b^7 (a+b x)^2}+\frac{e^3}{b^7 (a+b x)}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 B e^2 (b d-a e)}{b^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{B e^3 (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 = 0.153164, size = 239, normalized size = 1.05 \[ \frac{-3 A b \left (a^2 b e^2 (d+4 e x)+a^3 e^3+a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (4 d^2 e x+d^3+6 d e^2 x^2+4 e^3 x^3\right )\right )+B \left (-3 a^2 b^2 e \left (d^2+12 d e x-36 e^2 x^2\right )+a^3 b e^2 (88 e x-9 d)+25 a^4 e^3-a b^3 \left (12 d^2 e x+d^3+54 d e^2 x^2-48 e^3 x^3\right )-2 b^4 d x \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )+12 B e^3 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(25*a^4*e^3 + a^3*b*e^2*(-9*d + 88*e*x) - 3*a^2*b^2*e*(d^2 + 12*d*e*x - 36*e^2*x^2) - 2*b^4*d*x*(2*d^2 + 9*
d*e*x + 18*e^2*x^2) - a*b^3*(d^3 + 12*d^2*e*x + 54*d*e^2*x^2 - 48*e^3*x^3)) - 3*A*b*(a^3*e^3 + a^2*b*e^2*(d +
4*e*x) + a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + b^3*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) + 12*B*e^3*(a
+ b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.014, size = 385, normalized size = 1.7 \begin{align*} -{\frac{ \left ( -48\,B\ln \left ( bx+a \right ){x}^{3}a{b}^{3}{e}^{3}-48\,B\ln \left ( bx+a \right ) x{a}^{3}b{e}^{3}+54\,B{x}^{2}a{b}^{3}d{e}^{2}+3\,A{d}^{2}a{b}^{3}e+3\,{b}^{2}B{a}^{2}{d}^{2}e+3\,Ad{a}^{2}{b}^{2}{e}^{2}+12\,Ax{b}^{4}{d}^{2}e-88\,Bx{a}^{3}b{e}^{3}+18\,B{x}^{2}{b}^{4}{d}^{2}e+12\,Ax{a}^{2}{b}^{2}{e}^{3}+18\,A{x}^{2}a{b}^{3}{e}^{3}+18\,A{x}^{2}{b}^{4}d{e}^{2}-108\,B{x}^{2}{a}^{2}{b}^{2}{e}^{3}-48\,B{x}^{3}a{b}^{3}{e}^{3}+36\,B{x}^{3}{b}^{4}d{e}^{2}+3\,A{d}^{3}{b}^{4}-25\,B{e}^{3}{a}^{4}+9\,B{a}^{3}bd{e}^{2}-72\,B\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}{e}^{3}+Ba{b}^{3}{d}^{3}+12\,A{x}^{3}{b}^{4}{e}^{3}-12\,B\ln \left ( bx+a \right ){a}^{4}{e}^{3}+4\,Bx{b}^{4}{d}^{3}+12\,Axa{b}^{3}d{e}^{2}+36\,Bx{a}^{2}{b}^{2}d{e}^{2}+12\,Bxa{b}^{3}{d}^{2}e+3\,A{a}^{3}b{e}^{3}-12\,B\ln \left ( bx+a \right ){x}^{4}{b}^{4}{e}^{3} \right ) \left ( bx+a \right ) }{12\,{b}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{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)^(5/2),x)

[Out]

-1/12*(-48*B*ln(b*x+a)*x^3*a*b^3*e^3-48*B*ln(b*x+a)*x*a^3*b*e^3+54*B*x^2*a*b^3*d*e^2+3*A*d^2*a*b^3*e+3*b^2*B*a
^2*d^2*e+3*A*d*a^2*b^2*e^2+12*A*x*b^4*d^2*e-88*B*x*a^3*b*e^3+18*B*x^2*b^4*d^2*e+12*A*x*a^2*b^2*e^3+18*A*x^2*a*
b^3*e^3+18*A*x^2*b^4*d*e^2-108*B*x^2*a^2*b^2*e^3-48*B*x^3*a*b^3*e^3+36*B*x^3*b^4*d*e^2+3*A*d^3*b^4-25*B*e^3*a^
4+9*B*a^3*b*d*e^2-72*B*ln(b*x+a)*x^2*a^2*b^2*e^3+B*a*b^3*d^3+12*A*x^3*b^4*e^3-12*B*ln(b*x+a)*a^4*e^3+4*B*x*b^4
*d^3+12*A*x*a*b^3*d*e^2+36*B*x*a^2*b^2*d*e^2+12*B*x*a*b^3*d^2*e+3*A*a^3*b*e^3-12*B*ln(b*x+a)*x^4*b^4*e^3)*(b*x
+a)/b^5/((b*x+a)^2)^(5/2)

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Maxima [B]  time = 1.08935, size = 849, normalized size = 3.74 \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)^(5/2),x, algorithm="maxima")

[Out]

1/12*B*e^3*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*
a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 1/4*B*d*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^
2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^2/((b^2)^(7/2)*(x + a/b)^3)
+ 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(x + a/b)^4)) - 1/12*A*e^3*(12*x^2/((b^2*x^2 + 2*a*
b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^
2/((b^2)^(7/2)*(x + a/b)^3) + 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(x + a/b)^4)) - 1/12*B*
d^3*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/((b^2)^(5/2)*b*(x + a/b)^4)) - 1/4*A*d^2*e*(4/((b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*b^2) - 3*a/((b^2)^(5/2)*b*(x + a/b)^4)) - 1/4*B*d^2*e*(3*a^2*b^2/((b^2)^(9/2)*(x + a/b)^4)
- 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/4*A*d*e^2*(3*a^2*b^2/((b^2)^(9/2)*(x + a
/b)^4) - 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/4*A*d^3/((b^2)^(5/2)*(x + a/b)^4)

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Fricas [B]  time = 1.44778, size = 733, normalized size = 3.23 \begin{align*} -\frac{{\left (B a b^{3} + 3 \, A b^{4}\right )} d^{3} + 3 \,{\left (B a^{2} b^{2} + A a b^{3}\right )} d^{2} e + 3 \,{\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} d e^{2} -{\left (25 \, B a^{4} - 3 \, A a^{3} b\right )} e^{3} + 12 \,{\left (3 \, B b^{4} d e^{2} -{\left (4 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 18 \,{\left (B b^{4} d^{2} e +{\left (3 \, B a b^{3} + A b^{4}\right )} d e^{2} -{\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 4 \,{\left (B b^{4} d^{3} + 3 \,{\left (B a b^{3} + A b^{4}\right )} d^{2} e + 3 \,{\left (3 \, B a^{2} b^{2} + A a b^{3}\right )} d e^{2} -{\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} e^{3}\right )} x - 12 \,{\left (B b^{4} e^{3} x^{4} + 4 \, B a b^{3} e^{3} x^{3} + 6 \, B a^{2} b^{2} e^{3} x^{2} + 4 \, B a^{3} b e^{3} x + B a^{4} e^{3}\right )} \log \left (b x + a\right )}{12 \,{\left (b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*((B*a*b^3 + 3*A*b^4)*d^3 + 3*(B*a^2*b^2 + A*a*b^3)*d^2*e + 3*(3*B*a^3*b + A*a^2*b^2)*d*e^2 - (25*B*a^4 -
3*A*a^3*b)*e^3 + 12*(3*B*b^4*d*e^2 - (4*B*a*b^3 - A*b^4)*e^3)*x^3 + 18*(B*b^4*d^2*e + (3*B*a*b^3 + A*b^4)*d*e
^2 - (6*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 4*(B*b^4*d^3 + 3*(B*a*b^3 + A*b^4)*d^2*e + 3*(3*B*a^2*b^2 + A*a*b^3)*d
*e^2 - (22*B*a^3*b - 3*A*a^2*b^2)*e^3)*x - 12*(B*b^4*e^3*x^4 + 4*B*a*b^3*e^3*x^3 + 6*B*a^2*b^2*e^3*x^2 + 4*B*a
^3*b*e^3*x + B*a^4*e^3)*log(b*x + a))/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*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{5}{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)**(5/2),x)

[Out]

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

[Out]

sage0*x