3.1715 \(\int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=151 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) (d+e x)}+\frac {b B \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 151, 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} \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{2 e^3 (a+b x) (d+e x)^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (-a B e-A b e+2 b B d)}{e^3 (a+b x) (d+e x)}+\frac {b B \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 89, normalized size = 0.59 \[ -\frac {\sqrt {(a+b x)^2} \left (a e (A e+B (d+2 e x))+b (A e (d+2 e x)-B d (3 d+4 e x))-2 b B (d+e x)^2 \log (d+e x)\right )}{2 e^3 (a+b x) (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[(a + b*x)^2]*(a*e*(A*e + B*(d + 2*e*x)) + b*(A*e*(d + 2*e*x) - B*d*(3*d + 4*e*x)) - 2*b*B*(d + e*x)
^2*Log[d + e*x]))/(e^3*(a + b*x)*(d + e*x)^2)

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fricas [A]  time = 0.56, size = 105, normalized size = 0.70 \[ \frac {3 \, B b d^{2} - A a e^{2} - {\left (B a + A b\right )} d e + 2 \, {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x + 2 \, {\left (B b e^{2} x^{2} + 2 \, B b d e x + B b d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 127, normalized size = 0.84 \[ B b e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (2 \, {\left (2 \, B b d \mathrm {sgn}\left (b x + a\right ) - B a e \mathrm {sgn}\left (b x + a\right ) - A b e \mathrm {sgn}\left (b x + a\right )\right )} x + {\left (3 \, B b d^{2} \mathrm {sgn}\left (b x + a\right ) - B a d e \mathrm {sgn}\left (b x + a\right ) - A b d e \mathrm {sgn}\left (b x + a\right ) - A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.09, size = 117, normalized size = 0.77 \[ -\frac {\left (-2 B b \,e^{2} x^{2} \ln \left (b e x +b d \right )-4 B b d e x \ln \left (b e x +b d \right )+2 A b \,e^{2} x +2 B a \,e^{2} x -2 B b \,d^{2} \ln \left (b e x +b d \right )-4 B b d e x +A a \,e^{2}+A b d e +B a d e -3 B b \,d^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{2 \left (e x +d \right )^{2} e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*csgn(b*x+a)*(-2*B*ln(b*e*x+b*d)*x^2*b*e^2-4*B*b*d*e*x*ln(b*e*x+b*d)+2*A*x*b*e^2-2*B*b*d^2*ln(b*e*x+b*d)+2
*B*a*e^2*x-4*B*b*d*e*x+A*a*e^2+A*b*d*e+B*a*d*e-3*B*b*d^2)/e^3/(e*x+d)^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)*((b*x+a)^2)^(1/2)/(e*x+d)^3,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 = -0.01 \[ \int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\left (A+B\,x\right )}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.84, size = 94, normalized size = 0.62 \[ \frac {B b \log {\left (d + e x \right )}}{e^{3}} + \frac {- A a e^{2} - A b d e - B a d e + 3 B b d^{2} + x \left (- 2 A b e^{2} - 2 B a e^{2} + 4 B b d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/(e*x+d)**3,x)

[Out]

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

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