3.1772 \(\int \frac{A+B x}{(d+e x) (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=196 \[ -\frac{A b-a B}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{B d-A e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e (a+b x) \log (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e (a+b x) (B d-A e) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

[Out]

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

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Rubi [A]  time = 0.157076, antiderivative size = 196, 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} \[ -\frac{A b-a B}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{B d-A e}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e (a+b x) \log (a+b x) (B d-A e)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{e (a+b x) (B d-A e) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.120196, size = 132, normalized size = 0.67 \[ \frac{-(b d-a e) \left (B \left (a^2 e+a b d+2 b^2 d x\right )+A b (b (d-2 e x)-3 a e)\right )+2 b e (a+b x)^2 \log (a+b x) (A e-B d)+2 b e (a+b x)^2 (B d-A e) \log (d+e x)}{2 b (a+b x) \sqrt{(a+b x)^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.018, size = 314, normalized size = 1.6 \begin{align*}{\frac{ \left ( 2\,A\ln \left ( ex+d \right ){x}^{2}{b}^{3}{e}^{2}-2\,A\ln \left ( bx+a \right ){x}^{2}{b}^{3}{e}^{2}-2\,B\ln \left ( ex+d \right ){x}^{2}{b}^{3}de+2\,B\ln \left ( bx+a \right ){x}^{2}{b}^{3}de+4\,A\ln \left ( ex+d \right ) xa{b}^{2}{e}^{2}-4\,A\ln \left ( bx+a \right ) xa{b}^{2}{e}^{2}-4\,B\ln \left ( ex+d \right ) xa{b}^{2}de+4\,B\ln \left ( bx+a \right ) xa{b}^{2}de+2\,A\ln \left ( ex+d \right ){a}^{2}b{e}^{2}-2\,A\ln \left ( bx+a \right ){a}^{2}b{e}^{2}+2\,Axa{b}^{2}{e}^{2}-2\,Ax{b}^{3}de-2\,B\ln \left ( ex+d \right ){a}^{2}bde+2\,B\ln \left ( bx+a \right ){a}^{2}bde-2\,Bxa{b}^{2}de+2\,Bx{b}^{3}{d}^{2}+3\,Ab{a}^{2}{e}^{2}-4\,Aa{b}^{2}de+A{b}^{3}{d}^{2}-B{a}^{3}{e}^{2}+Ba{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) }{2\, \left ( ae-bd \right ) ^{3}b} \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)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

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

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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)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.4519, size = 728, normalized size = 3.71 \begin{align*} \frac{4 \, A a b^{2} d e -{\left (B a b^{2} + A b^{3}\right )} d^{2} +{\left (B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \,{\left (B b^{3} d^{2} + A a b^{2} e^{2} -{\left (B a b^{2} + A b^{3}\right )} d e\right )} x - 2 \,{\left (B a^{2} b d e - A a^{2} b e^{2} +{\left (B b^{3} d e - A b^{3} e^{2}\right )} x^{2} + 2 \,{\left (B a b^{2} d e - A a b^{2} e^{2}\right )} x\right )} \log \left (b x + a\right ) + 2 \,{\left (B a^{2} b d e - A a^{2} b e^{2} +{\left (B b^{3} d e - A b^{3} e^{2}\right )} x^{2} + 2 \,{\left (B a b^{2} d e - A a b^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \,{\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} +{\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \,{\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

sage0*x