3.1838 \(\int (A+B x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \]

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(7/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x))

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Rubi [A]  time = 0.0844633, antiderivative size = 164, 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} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(7/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(7*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 (A+B x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \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) \sqrt{d+e x}}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^{3/2}}{e^2}+\frac{b^2 B (d+e x)^{5/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e) (B d-A 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 (2 b B d-A b e-a B e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac{2 b B (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0687802, size = 88, normalized size = 0.54 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{3/2} \left (7 a e (5 A e-2 B d+3 B e x)+7 A b e (3 e x-2 d)+b B \left (8 d^2-12 d e x+15 e^2 x^2\right )\right )}{105 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(3/2)*(7*A*b*e*(-2*d + 3*e*x) + 7*a*e*(-2*B*d + 5*A*e + 3*B*e*x) + b*B*(8*d^2 -
12*d*e*x + 15*e^2*x^2)))/(105*e^3*(a + b*x))

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Maple [A]  time = 0.004, size = 89, normalized size = 0.5 \begin{align*}{\frac{30\,B{x}^{2}b{e}^{2}+42\,Axb{e}^{2}+42\,aB{e}^{2}x-24\,Bxbde+70\,aA{e}^{2}-28\,Abde-28\,aBde+16\,Bb{d}^{2}}{105\,{e}^{3} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(e*x+d)^(3/2)*(15*B*b*e^2*x^2+21*A*b*e^2*x+21*B*a*e^2*x-12*B*b*d*e*x+35*A*a*e^2-14*A*b*d*e-14*B*a*d*e+8*
B*b*d^2)*((b*x+a)^2)^(1/2)/e^3/(b*x+a)

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Maxima [A]  time = 1.02186, size = 162, normalized size = 0.99 \begin{align*} \frac{2 \,{\left (3 \, b e^{2} x^{2} - 2 \, b d^{2} + 5 \, a d e +{\left (b d e + 5 \, a e^{2}\right )} x\right )} \sqrt{e x + d} A}{15 \, e^{2}} + \frac{2 \,{\left (15 \, b e^{3} x^{3} + 8 \, b d^{3} - 14 \, a d^{2} e + 3 \,{\left (b d e^{2} + 7 \, a e^{3}\right )} x^{2} -{\left (4 \, b d^{2} e - 7 \, a d e^{2}\right )} x\right )} \sqrt{e x + d} B}{105 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*b*e^2*x^2 - 2*b*d^2 + 5*a*d*e + (b*d*e + 5*a*e^2)*x)*sqrt(e*x + d)*A/e^2 + 2/105*(15*b*e^3*x^3 + 8*b*d
^3 - 14*a*d^2*e + 3*(b*d*e^2 + 7*a*e^3)*x^2 - (4*b*d^2*e - 7*a*d*e^2)*x)*sqrt(e*x + d)*B/e^3

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Fricas [A]  time = 1.28489, size = 250, normalized size = 1.52 \begin{align*} \frac{2 \,{\left (15 \, B b e^{3} x^{3} + 8 \, B b d^{3} + 35 \, A a d e^{2} - 14 \,{\left (B a + A b\right )} d^{2} e + 3 \,{\left (B b d e^{2} + 7 \,{\left (B a + A b\right )} e^{3}\right )} x^{2} -{\left (4 \, B b d^{2} e - 35 \, A a e^{3} - 7 \,{\left (B a + A b\right )} d e^{2}\right )} x\right )} \sqrt{e x + d}}{105 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(15*B*b*e^3*x^3 + 8*B*b*d^3 + 35*A*a*d*e^2 - 14*(B*a + A*b)*d^2*e + 3*(B*b*d*e^2 + 7*(B*a + A*b)*e^3)*x^
2 - (4*B*b*d^2*e - 35*A*a*e^3 - 7*(B*a + A*b)*d*e^2)*x)*sqrt(e*x + d)/e^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14764, size = 185, normalized size = 1.13 \begin{align*} \frac{2}{105} \,{\left (7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} B a e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 7 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} A b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) +{\left (15 \,{\left (x e + d\right )}^{\frac{7}{2}} - 42 \,{\left (x e + d\right )}^{\frac{5}{2}} d + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{2}\right )} B b e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 35 \,{\left (x e + d\right )}^{\frac{3}{2}} 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)*(e*x+d)^(1/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/105*(7*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*B*a*e^(-1)*sgn(b*x + a) + 7*(3*(x*e + d)^(5/2) - 5*(x*e + d
)^(3/2)*d)*A*b*e^(-1)*sgn(b*x + a) + (15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*B*b*
e^(-2)*sgn(b*x + a) + 35*(x*e + d)^(3/2)*A*a*sgn(b*x + a))*e^(-1)