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]

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

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Rubi [A]  time = 0.16, antiderivative size = 196, normalized size of antiderivative = 1.00, 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 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]))))

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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 1.12, size = 361, normalized size = 1.84 \[ \frac {4 \, A a b^{2} d e - {\left (B a b^{2} + A b^{3}\right )} d^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} e^{2} - 2 \, {\left (B b^{3} d^{2} + A a b^{2} e^{2} - {\left (B a b^{2} + A b^{3}\right )} d e\right )} x - 2 \, {\left (B a^{2} b d e - A a^{2} b e^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x^{2} + 2 \, {\left (B a b^{2} d e - A a b^{2} e^{2}\right )} x\right )} \log \left (b x + a\right ) + 2 \, {\left (B a^{2} b d e - A a^{2} b e^{2} + {\left (B b^{3} d e - A b^{3} e^{2}\right )} x^{2} + 2 \, {\left (B a b^{2} d e - A a b^{2} e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} + {\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \, {\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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

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maple [B]  time = 0.07, size = 315, normalized size = 1.61 \[ -\frac {\left (2 A \,b^{3} e^{2} x^{2} \ln \left (b x +a \right )-2 A \,b^{3} e^{2} x^{2} \ln \left (e x +d \right )-2 B \,b^{3} d e \,x^{2} \ln \left (b x +a \right )+2 B \,b^{3} d e \,x^{2} \ln \left (e x +d \right )+4 A a \,b^{2} e^{2} x \ln \left (b x +a \right )-4 A a \,b^{2} e^{2} x \ln \left (e x +d \right )-4 B a \,b^{2} d e x \ln \left (b x +a \right )+4 B a \,b^{2} d e x \ln \left (e x +d \right )+2 A \,a^{2} b \,e^{2} \ln \left (b x +a \right )-2 A \,a^{2} b \,e^{2} \ln \left (e x +d \right )-2 A a \,b^{2} e^{2} x +2 A \,b^{3} d e x -2 B \,a^{2} b d e \ln \left (b x +a \right )+2 B \,a^{2} b d e \ln \left (e x +d \right )+2 B a \,b^{2} d e x -2 B \,b^{3} d^{2} x -3 A \,a^{2} b \,e^{2}+4 A a \,b^{2} d e -A \,b^{3} d^{2}+B \,a^{3} e^{2}-B a \,b^{2} d^{2}\right ) \left (b x +a \right )}{2 \left (a e -b d \right )^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b} \]

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*b^3*e^2*x^2*ln(b*x+a)-2*A*ln(e*x+d)*x^2*b^3*e^2-2*B*b^3*d*e*x^2*ln(b*x+a)+2*B*ln(e*x+d)*x^2*b^3*d*e+
4*A*a*b^2*e^2*x*ln(b*x+a)-4*A*ln(e*x+d)*x*a*b^2*e^2-4*B*a*b^2*d*e*x*ln(b*x+a)+4*B*ln(e*x+d)*x*a*b^2*d*e+2*A*a^
2*b*e^2*ln(b*x+a)-2*A*ln(e*x+d)*a^2*b*e^2-2*A*a*b^2*e^2*x+2*A*b^3*d*e*x-2*B*a^2*b*d*e*ln(b*x+a)+2*B*ln(e*x+d)*
a^2*b*d*e+2*B*a*b^2*d*e*x-2*B*b^3*d^2*x-3*A*a^2*b*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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(((2*a*b)/e>0)', see `assume?`
for more details)Is ((2*a*b)/e    -(2*b^2*d)/e^2)    ^2    -(4*b^2       *((-(2*a*b*d)/e)        +(b^2*d^2)/e^
2+a^2))     /e^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x}{\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{\left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

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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