3.1839 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{d+e x}} \, dx\)

Optimal. Leaf size=162 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e) (B d-A e)}{e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)} \]

[Out]

(2*(b*d - a*e)*(B*d - A*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)) - (2*(2*b*B*d - A*b*e
- a*B*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(5/2)*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x))

________________________________________________________________________________________

Rubi [A]  time = 0.0881532, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.057, Rules used = {770, 77} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (-a B e-A b e+2 b B d)}{3 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e) (B d-A e)}{e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2}}{5 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)*(B*d - A*e)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a + b*x)) - (2*(2*b*B*d - A*b*e
- a*B*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(5/2)*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x))

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) \sqrt{a^2+2 a b x+b^2 x^2}}{\sqrt{d+e x}} \, 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)}{\sqrt{d+e x}} \, 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 \sqrt{d+e x}}+\frac{b (-2 b B d+A b e+a B e) \sqrt{d+e x}}{e^2}+\frac{b^2 B (d+e x)^{3/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e) (B d-A e) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac{2 (2 b B d-A b e-a B e) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}+\frac{2 b B (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0703842, size = 86, normalized size = 0.53 \[ \frac{2 \sqrt{(a+b x)^2} \sqrt{d+e x} \left (5 a e (3 A e-2 B d+B e x)+5 A b e (e x-2 d)+b B \left (8 d^2-4 d e x+3 e^2 x^2\right )\right )}{15 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*Sqrt[d + e*x]*(5*A*b*e*(-2*d + e*x) + 5*a*e*(-2*B*d + 3*A*e + B*e*x) + b*B*(8*d^2 - 4*d*e
*x + 3*e^2*x^2)))/(15*e^3*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 89, normalized size = 0.6 \begin{align*}{\frac{6\,B{x}^{2}b{e}^{2}+10\,Axb{e}^{2}+10\,aB{e}^{2}x-8\,Bxbde+30\,aA{e}^{2}-20\,Abde-20\,aBde+16\,Bb{d}^{2}}{15\, \left ( bx+a \right ){e}^{3}}\sqrt{ex+d}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(e*x+d)^(1/2)*(3*B*b*e^2*x^2+5*A*b*e^2*x+5*B*a*e^2*x-4*B*b*d*e*x+15*A*a*e^2-10*A*b*d*e-10*B*a*d*e+8*B*b*d
^2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

________________________________________________________________________________________

Maxima [A]  time = 1.06632, size = 161, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (b e^{2} x^{2} - 2 \, b d^{2} + 3 \, a d e -{\left (b d e - 3 \, a e^{2}\right )} x\right )} A}{3 \, \sqrt{e x + d} e^{2}} + \frac{2 \,{\left (3 \, b e^{3} x^{3} + 8 \, b d^{3} - 10 \, a d^{2} e -{\left (b d e^{2} - 5 \, a e^{3}\right )} x^{2} +{\left (4 \, b d^{2} e - 5 \, a d e^{2}\right )} x\right )} B}{15 \, \sqrt{e x + d} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*e^2*x^2 - 2*b*d^2 + 3*a*d*e - (b*d*e - 3*a*e^2)*x)*A/(sqrt(e*x + d)*e^2) + 2/15*(3*b*e^3*x^3 + 8*b*d^3
- 10*a*d^2*e - (b*d*e^2 - 5*a*e^3)*x^2 + (4*b*d^2*e - 5*a*d*e^2)*x)*B/(sqrt(e*x + d)*e^3)

________________________________________________________________________________________

Fricas [A]  time = 1.27167, size = 165, normalized size = 1.02 \begin{align*} \frac{2 \,{\left (3 \, B b e^{2} x^{2} + 8 \, B b d^{2} + 15 \, A a e^{2} - 10 \,{\left (B a + A b\right )} d e -{\left (4 \, B b d e - 5 \,{\left (B a + A b\right )} e^{2}\right )} x\right )} \sqrt{e x + d}}{15 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*B*b*e^2*x^2 + 8*B*b*d^2 + 15*A*a*e^2 - 10*(B*a + A*b)*d*e - (4*B*b*d*e - 5*(B*a + A*b)*e^2)*x)*sqrt(e*
x + d)/e^3

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)*sqrt((a + b*x)**2)/sqrt(d + e*x), x)

________________________________________________________________________________________

Giac [A]  time = 1.12098, size = 180, normalized size = 1.11 \begin{align*} \frac{2}{15} \,{\left (5 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} B a e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left ({\left (x e + d\right )}^{\frac{3}{2}} - 3 \, \sqrt{x e + d} d\right )} A b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) +{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} d + 15 \, \sqrt{x e + d} d^{2}\right )} B b e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 15 \, \sqrt{x e + d} A a \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-1\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d)*B*a*e^(-1)*sgn(b*x + a) + 5*((x*e + d)^(3/2) - 3*sqrt(x*e + d)*d
)*A*b*e^(-1)*sgn(b*x + a) + (3*(x*e + d)^(5/2) - 10*(x*e + d)^(3/2)*d + 15*sqrt(x*e + d)*d^2)*B*b*e^(-2)*sgn(b
*x + a) + 15*sqrt(x*e + d)*A*a*sgn(b*x + a))*e^(-1)