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]

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

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Rubi [A]  time = 0.25, antiderivative size = 266, 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} \[ -\frac {A b-a B}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {a B e-2 A b e+b B d}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {e (a+b x) (B d-A e)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^3}-\frac {e (a+b x) \log (a+b x) (a B e-3 A b e+2 b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {e (a+b x) \log (d+e x) (a B e-3 A b e+2 b B d)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4} \]

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 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)^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*}

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Mathematica [A]  time = 0.18, 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))

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fricas [B]  time = 0.76, size = 803, normalized size = 3.02 \[ -\frac {2 \, A a^{3} e^{3} + {\left (B a b^{2} + A b^{3}\right )} d^{3} + 2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d^{2} e - {\left (5 \, B a^{3} - 3 \, A a^{2} b\right )} d e^{2} + 2 \, {\left (2 \, B b^{3} d^{2} e - {\left (B a b^{2} + 3 \, A b^{3}\right )} d e^{2} - {\left (B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} + {\left (2 \, B b^{3} d^{3} + {\left (5 \, B a b^{2} - 3 \, A b^{3}\right )} d^{2} e - 2 \, {\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} d e^{2} - 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} e^{3}\right )} x + 2 \, {\left (2 \, B a^{2} b d^{2} e + {\left (B a^{3} - 3 \, A a^{2} b\right )} d e^{2} + {\left (2 \, B b^{3} d e^{2} + {\left (B a b^{2} - 3 \, A b^{3}\right )} e^{3}\right )} x^{3} + {\left (2 \, B b^{3} d^{2} e + {\left (5 \, B a b^{2} - 3 \, A b^{3}\right )} d e^{2} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} + {\left (4 \, B a b^{2} d^{2} e + 2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} e^{3}\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left (2 \, B a^{2} b d^{2} e + {\left (B a^{3} - 3 \, A a^{2} b\right )} d e^{2} + {\left (2 \, B b^{3} d e^{2} + {\left (B a b^{2} - 3 \, A b^{3}\right )} e^{3}\right )} x^{3} + {\left (2 \, B b^{3} d^{2} e + {\left (5 \, B a b^{2} - 3 \, A b^{3}\right )} d e^{2} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} e^{3}\right )} x^{2} + {\left (4 \, B a b^{2} d^{2} e + 2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} d e^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (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} + {\left (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}\right )} x^{3} + {\left (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}\right )} x^{2} + {\left (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}\right )} x\right )}} \]

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)

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giac [B]  time = 0.53, size = 708, normalized size = 2.66 \[ \frac {{\left (2 \, B b d e^{2} + B a e^{3} - 3 \, A b e^{3}\right )} \log \left ({\left | -b + \frac {b d}{x e + d} - \frac {a e}{x e + d} \right |}\right )}{b^{4} d^{4} e \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 4 \, a b^{3} d^{3} e^{2} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 6 \, a^{2} b^{2} d^{2} e^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 4 \, a^{3} b d e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + a^{4} e^{5} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} + \frac {\frac {B d e^{4}}{x e + d} - \frac {A e^{5}}{x e + d}}{b^{3} d^{3} e^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 3 \, a b^{2} d^{2} e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 3 \, a^{2} b d e^{5} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - a^{3} e^{6} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} + \frac {2 \, B b^{3} d e + 3 \, B a b^{2} e^{2} - 5 \, A b^{3} e^{2} - \frac {2 \, {\left (B b^{3} d^{2} e^{2} + B a b^{2} d e^{3} - 3 \, A b^{3} d e^{3} - 2 \, B a^{2} b e^{4} + 3 \, A a b^{2} e^{4}\right )} e^{\left (-1\right )}}{x e + d}}{2 \, {\left (b d - a e\right )}^{4} {\left (b - \frac {b d}{x e + d} + \frac {a e}{x e + d}\right )}^{2} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} \]

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]

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

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maple [B]  time = 0.07, size = 828, normalized size = 3.11 \[ \frac {\left (6 A \,b^{3} e^{3} x^{3} \ln \left (b x +a \right )-6 A \,b^{3} e^{3} x^{3} \ln \left (e x +d \right )-2 B a \,b^{2} e^{3} x^{3} \ln \left (b x +a \right )+2 B a \,b^{2} e^{3} x^{3} \ln \left (e x +d \right )-4 B \,b^{3} d \,e^{2} x^{3} \ln \left (b x +a \right )+4 B \,b^{3} d \,e^{2} x^{3} \ln \left (e x +d \right )+12 A a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-12 A a \,b^{2} e^{3} x^{2} \ln \left (e x +d \right )+6 A \,b^{3} d \,e^{2} x^{2} \ln \left (b x +a \right )-6 A \,b^{3} d \,e^{2} x^{2} \ln \left (e x +d \right )-4 B \,a^{2} b \,e^{3} x^{2} \ln \left (b x +a \right )+4 B \,a^{2} b \,e^{3} x^{2} \ln \left (e x +d \right )-10 B a \,b^{2} d \,e^{2} x^{2} \ln \left (b x +a \right )+10 B a \,b^{2} d \,e^{2} x^{2} \ln \left (e x +d \right )-4 B \,b^{3} d^{2} e \,x^{2} \ln \left (b x +a \right )+4 B \,b^{3} d^{2} e \,x^{2} \ln \left (e x +d \right )+6 A \,a^{2} b \,e^{3} x \ln \left (b x +a \right )-6 A \,a^{2} b \,e^{3} x \ln \left (e x +d \right )+12 A a \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-12 A a \,b^{2} d \,e^{2} x \ln \left (e x +d \right )-6 A a \,b^{2} e^{3} x^{2}+6 A \,b^{3} d \,e^{2} x^{2}-2 B \,a^{3} e^{3} x \ln \left (b x +a \right )+2 B \,a^{3} e^{3} x \ln \left (e x +d \right )-8 B \,a^{2} b d \,e^{2} x \ln \left (b x +a \right )+8 B \,a^{2} b d \,e^{2} x \ln \left (e x +d \right )+2 B \,a^{2} b \,e^{3} x^{2}-8 B a \,b^{2} d^{2} e x \ln \left (b x +a \right )+8 B a \,b^{2} d^{2} e x \ln \left (e x +d \right )+2 B a \,b^{2} d \,e^{2} x^{2}-4 B \,b^{3} d^{2} e \,x^{2}+6 A \,a^{2} b d \,e^{2} \ln \left (b x +a \right )-6 A \,a^{2} b d \,e^{2} \ln \left (e x +d \right )-9 A \,a^{2} b \,e^{3} x +6 A a \,b^{2} d \,e^{2} x +3 A \,b^{3} d^{2} e x -2 B \,a^{3} d \,e^{2} \ln \left (b x +a \right )+2 B \,a^{3} d \,e^{2} \ln \left (e x +d \right )+3 B \,a^{3} e^{3} x -4 B \,a^{2} b \,d^{2} e \ln \left (b x +a \right )+4 B \,a^{2} b \,d^{2} e \ln \left (e x +d \right )+4 B \,a^{2} b d \,e^{2} x -5 B a \,b^{2} d^{2} e x -2 B \,b^{3} d^{3} x -2 A \,a^{3} e^{3}-3 A \,a^{2} b d \,e^{2}+6 A a \,b^{2} d^{2} e -A \,b^{3} d^{3}+5 B \,a^{3} d \,e^{2}-4 B \,a^{2} b \,d^{2} e -B a \,b^{2} d^{3}\right ) \left (b x +a \right )}{2 \left (e x +d \right ) \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \]

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*(-A*b^3*d^3-2*A*a^3*e^3-3*A*a^2*b*d*e^2+6*A*a*b^2*d^2*e+4*B*ln(e*x+d)*x^3*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*a^2*b*d^2*e+3*B*a^3*e^3*x-2*B*b^3*d^3*x+5*B*a^3*d*e^2-B*a*b^2*d^3-12*A*
ln(e*x+d)*x^2*a*b^2*e^3-6*A*ln(e*x+d)*x^2*b^3*d*e^2-4*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+4*
B*ln(e*x+d)*a^2*b*d^2*e-2*B*ln(b*x+a)*x^3*a*b^2*e^3+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-6*
A*ln(e*x+d)*x^3*b^3*e^3-2*B*ln(b*x+a)*x*a^3*e^3+2*B*ln(e*x+d)*x*a^3*e^3-2*B*ln(b*x+a)*a^3*d*e^2+3*A*b^3*d^2*e*
x+6*A*x^2*b^3*d*e^2+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-6*A*x^2*a*b^2*e^3+2*B*ln(e*x+d)*a^3*d*
e^2+6*A*ln(b*x+a)*x^3*b^3*e^3-5*B*a*b^2*d^2*e*x+4*B*a^2*b*d*e^2*x+6*A*a*b^2*d*e^2*x+2*B*x^2*a*b^2*d*e^2-4*B*ln
(b*x+a)*x^3*b^3*d*e^2+2*B*ln(e*x+d)*x^3*a*b^2*e^3-8*B*ln(b*x+a)*x*a*b^2*d^2*e+8*B*ln(e*x+d)*x*a^2*b*d*e^2+8*B*
ln(e*x+d)*x*a*b^2*d^2*e+12*A*ln(b*x+a)*x*a*b^2*d*e^2-12*A*ln(e*x+d)*x*a*b^2*d*e^2-8*B*ln(b*x+a)*x*a^2*b*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+6*A*ln(b*x+a)*x*a^2*b*e^3-6*A*ln(e*x+d)*x*a^2*b*
e^3+6*A*ln(b*x+a)*a^2*b*d*e^2-6*A*ln(e*x+d)*a^2*b*d*e^2-4*B*ln(b*x+a)*a^2*b*d^2*e)*(b*x+a)/(e*x+d)/(a*e-b*d)^4
/((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)^2/(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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*x)/((d + e*x)^2*(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 \[ \int \frac {A + B x}{\left (d + e x\right )^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**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)

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