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

Optimal. Leaf size=144 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x) (-a B e-A b e+2 b B d)}{e^3 (a+b x)}+\frac {b B x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 144, 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} \begin {gather*} -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) (B d-A e)}{e^3 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x) (-a B e-A b e+2 b B d)}{e^3 (a+b x)}+\frac {b B x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 96, normalized size = 0.67 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-(d+e x) \log (d+e x) (-a B e-A b e+2 b B d)+a e (B d-A e)+b \left (A d e-B d^2+B d e x+B e^2 x^2\right )\right )}{e^3 (a+b x) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 2.42, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 102, normalized size = 0.71 \begin {gather*} \frac {B b e^{2} x^{2} + B b d e x - B b d^{2} - A a e^{2} + {\left (B a + A b\right )} d e - {\left (2 \, B b d^{2} - {\left (B a + A b\right )} d e + {\left (2 \, B b d e - {\left (B a + A b\right )} e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.18, size = 123, normalized size = 0.85 \begin {gather*} B b x e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) - {\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 )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (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 (-3\right )}}{x e + d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.07, size = 158, normalized size = 1.10 \begin {gather*} \frac {\left (A b \,e^{2} x \ln \left (b e x +b d \right )+B a \,e^{2} x \ln \left (b e x +b d \right )-2 B b d e x \ln \left (b e x +b d \right )+B b \,e^{2} x^{2}+A b d e \ln \left (b e x +b d \right )+B a d e \ln \left (b e x +b d \right )+B a \,e^{2} x -2 B b \,d^{2} \ln \left (b e x +b d \right )+B b d e x -A a \,e^{2}+A b d e +2 B a d e -B b \,d^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{\left (e x +d \right ) e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/(e*x+d)^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?

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.46, size = 71, normalized size = 0.49 \begin {gather*} \frac {B b x}{e^{2}} + \frac {- A a e^{2} + A b d e + B a d e - B b d^{2}}{d e^{3} + e^{4} x} + \frac {\left (A b e + B a e - 2 B b d\right ) \log {\left (d + e x \right )}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________