3.1773 \(\int \frac{A+B x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=266 \[ -\frac{A b-a B}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{a B e-2 A b e+b B d}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{e (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}-\frac{e (a+b x) \log (a+b x) (a B e-3 A b e+2 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{e (a+b x) \log (d+e x) (a B e-3 A b e+2 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.24596, antiderivative size = 266, 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} \[ -\frac{A b-a B}{2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{a B e-2 A b e+b B d}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac{e (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}-\frac{e (a+b x) \log (a+b x) (a B e-3 A b e+2 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac{e (a+b x) \log (d+e x) (a B e-3 A b e+2 b B d)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.167993, size = 174, normalized size = 0.65 \[ \frac{(a+b x) \left (-2 (a+b x) (b d-a e) (a B e-2 A b e+b B d)+\frac{2 e (a+b x)^2 (b d-a e) (A e-B d)}{d+e x}-2 e (a+b x)^2 \log (a+b x) (a B e-3 A b e+2 b B d)+2 e (a+b x)^2 \log (d+e x) (a B e-3 A b e+2 b B d)+(a B-A b) (b d-a e)^2\right )}{2 \left ((a+b x)^2\right )^{3/2} (b d-a e)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.022, size = 826, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*A*ln(e*x+d)*x*a*b^2*d*e^2+10*B*ln(b*x+a)*x^2*a*b^2*d*e^2-10*B*ln(e*x+d)*x^2*a*b^2*d*e^2-12*A*ln(b*x+a
)*x*a*b^2*d*e^2-8*B*ln(e*x+d)*x*a^2*b*d*e^2-4*B*ln(e*x+d)*x^3*b^3*d*e^2+2*B*ln(b*x+a)*x^3*a*b^2*e^3-4*B*a^2*b*
d*e^2*x+5*B*a*b^2*d^2*e*x-6*A*a*b^2*d*e^2*x-2*B*x^2*a*b^2*d*e^2-6*A*ln(b*x+a)*x*a^2*b*e^3+6*A*ln(e*x+d)*a^2*b*
d*e^2-6*A*ln(b*x+a)*a^2*b*d*e^2-4*B*ln(e*x+d)*a^2*b*d^2*e+4*B*ln(b*x+a)*a^2*b*d^2*e+4*B*ln(b*x+a)*x^3*b^3*d*e^
2+12*A*ln(e*x+d)*x^2*a*b^2*e^3+6*A*ln(e*x+d)*x^2*b^3*d*e^2-12*A*ln(b*x+a)*x^2*a*b^2*e^3-6*A*ln(b*x+a)*x^2*b^3*
d*e^2-4*B*ln(e*x+d)*x^2*a^2*b*e^3-4*B*ln(e*x+d)*x^2*b^3*d^2*e+4*B*ln(b*x+a)*x^2*a^2*b*e^3+4*B*ln(b*x+a)*x^2*b^
3*d^2*e+6*A*ln(e*x+d)*x*a^2*b*e^3-2*B*ln(e*x+d)*x^3*a*b^2*e^3-5*B*a^3*d*e^2+B*a*b^2*d^3-3*B*e^3*a^3*x+2*B*b^3*
d^3*x+2*A*a^3*e^3+A*b^3*d^3-2*B*x^2*a^2*b*e^3+4*B*x^2*b^3*d^2*e+9*A*a^2*b*e^3*x-3*A*b^3*d^2*e*x-6*A*x^2*b^3*d*
e^2+6*A*x^2*a*b^2*e^3-6*A*ln(b*x+a)*x^3*b^3*e^3-2*B*ln(e*x+d)*x*a^3*e^3+2*B*ln(b*x+a)*x*a^3*e^3-2*B*ln(e*x+d)*
a^3*d*e^2+2*B*ln(b*x+a)*a^3*d*e^2+6*A*ln(e*x+d)*x^3*b^3*e^3+4*B*a^2*b*d^2*e-6*A*a*b^2*d^2*e+3*A*a^2*b*d*e^2-8*
B*ln(e*x+d)*x*a*b^2*d^2*e+8*B*ln(b*x+a)*x*a^2*b*d*e^2+8*B*ln(b*x+a)*x*a*b^2*d^2*e)*(b*x+a)/(e*x+d)/(a*e-b*d)^4
/((b*x+a)^2)^(3/2)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.43023, size = 1652, normalized size = 6.21 \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)^2/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

-1/2*(2*A*a^3*e^3 + (B*a*b^2 + A*b^3)*d^3 + 2*(2*B*a^2*b - 3*A*a*b^2)*d^2*e - (5*B*a^3 - 3*A*a^2*b)*d*e^2 + 2*
(2*B*b^3*d^2*e - (B*a*b^2 + 3*A*b^3)*d*e^2 - (B*a^2*b - 3*A*a*b^2)*e^3)*x^2 + (2*B*b^3*d^3 + (5*B*a*b^2 - 3*A*
b^3)*d^2*e - 2*(2*B*a^2*b + 3*A*a*b^2)*d*e^2 - 3*(B*a^3 - 3*A*a^2*b)*e^3)*x + 2*(2*B*a^2*b*d^2*e + (B*a^3 - 3*
A*a^2*b)*d*e^2 + (2*B*b^3*d*e^2 + (B*a*b^2 - 3*A*b^3)*e^3)*x^3 + (2*B*b^3*d^2*e + (5*B*a*b^2 - 3*A*b^3)*d*e^2
+ 2*(B*a^2*b - 3*A*a*b^2)*e^3)*x^2 + (4*B*a*b^2*d^2*e + 2*(2*B*a^2*b - 3*A*a*b^2)*d*e^2 + (B*a^3 - 3*A*a^2*b)*
e^3)*x)*log(b*x + a) - 2*(2*B*a^2*b*d^2*e + (B*a^3 - 3*A*a^2*b)*d*e^2 + (2*B*b^3*d*e^2 + (B*a*b^2 - 3*A*b^3)*e
^3)*x^3 + (2*B*b^3*d^2*e + (5*B*a*b^2 - 3*A*b^3)*d*e^2 + 2*(B*a^2*b - 3*A*a*b^2)*e^3)*x^2 + (4*B*a*b^2*d^2*e +
2*(2*B*a^2*b - 3*A*a*b^2)*d*e^2 + (B*a^3 - 3*A*a^2*b)*e^3)*x)*log(e*x + d))/(a^2*b^4*d^5 - 4*a^3*b^3*d^4*e +
6*a^4*b^2*d^3*e^2 - 4*a^5*b*d^2*e^3 + a^6*d*e^4 + (b^6*d^4*e - 4*a*b^5*d^3*e^2 + 6*a^2*b^4*d^2*e^3 - 4*a^3*b^3
*d*e^4 + a^4*b^2*e^5)*x^3 + (b^6*d^5 - 2*a*b^5*d^4*e - 2*a^2*b^4*d^3*e^2 + 8*a^3*b^3*d^2*e^3 - 7*a^4*b^2*d*e^4
+ 2*a^5*b*e^5)*x^2 + (2*a*b^5*d^5 - 7*a^2*b^4*d^4*e + 8*a^3*b^3*d^3*e^2 - 2*a^4*b^2*d^2*e^3 - 2*a^5*b*d*e^4 +
a^6*e^5)*x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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