3.1837 \(\int (A+B x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=164 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (-a B e-A b e+2 b B d)}{7 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e) (B d-A e)}{5 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)} \]

[Out]

(2*(b*d - a*e)*(B*d - A*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(9/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(9*e^3*(a + b*x))

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Rubi [A]  time = 0.0842699, 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} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (-a B e-A b e+2 b B d)}{7 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e) (B d-A e)}{5 e^3 (a+b x)}+\frac{2 b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(d + e*x)^(3/2)*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)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) - (2*(2*b*B*d - A*
b*e - a*B*e)*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^3*(a + b*x)) + (2*b*B*(d + e*x)^(9/2)*Sqrt[a^
2 + 2*a*b*x + b^2*x^2])/(9*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) (d+e x)^{3/2} \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) (d+e x)^{3/2} \, 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) (d+e x)^{3/2}}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^{5/2}}{e^2}+\frac{b^2 B (d+e x)^{7/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e) (B d-A 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 (2 b B d-A b e-a B e) (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^3 (a+b x)}+\frac{2 b B (d+e x)^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}{9 e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0753875, size = 88, normalized size = 0.54 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{5/2} \left (9 a e (7 A e-2 B d+5 B e x)+9 A b e (5 e x-2 d)+b B \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(d + e*x)^(5/2)*(9*A*b*e*(-2*d + 5*e*x) + 9*a*e*(-2*B*d + 7*A*e + 5*B*e*x) + b*B*(8*d^2 -
20*d*e*x + 35*e^2*x^2)))/(315*e^3*(a + b*x))

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Maple [A]  time = 0.005, size = 89, normalized size = 0.5 \begin{align*}{\frac{70\,B{x}^{2}b{e}^{2}+90\,Axb{e}^{2}+90\,aB{e}^{2}x-40\,Bxbde+126\,aA{e}^{2}-36\,Abde-36\,aBde+16\,Bb{d}^{2}}{315\,{e}^{3} \left ( bx+a \right ) } \left ( ex+d \right ) ^{{\frac{5}{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)^(3/2)*((b*x+a)^2)^(1/2),x)

[Out]

2/315*(e*x+d)^(5/2)*(35*B*b*e^2*x^2+45*A*b*e^2*x+45*B*a*e^2*x-20*B*b*d*e*x+63*A*a*e^2-18*A*b*d*e-18*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 = 1.0386, size = 225, normalized size = 1.37 \begin{align*} \frac{2 \,{\left (5 \, b e^{3} x^{3} - 2 \, b d^{3} + 7 \, a d^{2} e +{\left (8 \, b d e^{2} + 7 \, a e^{3}\right )} x^{2} +{\left (b d^{2} e + 14 \, a d e^{2}\right )} x\right )} \sqrt{e x + d} A}{35 \, e^{2}} + \frac{2 \,{\left (35 \, b e^{4} x^{4} + 8 \, b d^{4} - 18 \, a d^{3} e + 5 \,{\left (10 \, b d e^{3} + 9 \, a e^{4}\right )} x^{3} + 3 \,{\left (b d^{2} e^{2} + 24 \, a d e^{3}\right )} x^{2} -{\left (4 \, b d^{3} e - 9 \, a d^{2} e^{2}\right )} x\right )} \sqrt{e x + d} B}{315 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b*e^3*x^3 - 2*b*d^3 + 7*a*d^2*e + (8*b*d*e^2 + 7*a*e^3)*x^2 + (b*d^2*e + 14*a*d*e^2)*x)*sqrt(e*x + d)*
A/e^2 + 2/315*(35*b*e^4*x^4 + 8*b*d^4 - 18*a*d^3*e + 5*(10*b*d*e^3 + 9*a*e^4)*x^3 + 3*(b*d^2*e^2 + 24*a*d*e^3)
*x^2 - (4*b*d^3*e - 9*a*d^2*e^2)*x)*sqrt(e*x + d)*B/e^3

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Fricas [A]  time = 1.30477, size = 342, normalized size = 2.09 \begin{align*} \frac{2 \,{\left (35 \, B b e^{4} x^{4} + 8 \, B b d^{4} + 63 \, A a d^{2} e^{2} - 18 \,{\left (B a + A b\right )} d^{3} e + 5 \,{\left (10 \, B b d e^{3} + 9 \,{\left (B a + A b\right )} e^{4}\right )} x^{3} + 3 \,{\left (B b d^{2} e^{2} + 21 \, A a e^{4} + 24 \,{\left (B a + A b\right )} d e^{3}\right )} x^{2} -{\left (4 \, B b d^{3} e - 126 \, A a d e^{3} - 9 \,{\left (B a + A b\right )} d^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{315 \, e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(35*B*b*e^4*x^4 + 8*B*b*d^4 + 63*A*a*d^2*e^2 - 18*(B*a + A*b)*d^3*e + 5*(10*B*b*d*e^3 + 9*(B*a + A*b)*e^
4)*x^3 + 3*(B*b*d^2*e^2 + 21*A*a*e^4 + 24*(B*a + A*b)*d*e^3)*x^2 - (4*B*b*d^3*e - 126*A*a*d*e^3 - 9*(B*a + A*b
)*d^2*e^2)*x)*sqrt(e*x + d)/e^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.18844, size = 441, normalized size = 2.69 \begin{align*} \frac{2}{315} \,{\left (21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} B a d e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} A b d e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 3 \,{\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 d e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} A a d \mathrm{sgn}\left (b x + a\right ) + 3 \,{\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 a e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) + 3 \,{\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 )} A b e^{\left (-1\right )} \mathrm{sgn}\left (b x + a\right ) +{\left (35 \,{\left (x e + d\right )}^{\frac{9}{2}} - 135 \,{\left (x e + d\right )}^{\frac{7}{2}} d + 189 \,{\left (x e + d\right )}^{\frac{5}{2}} d^{2} - 105 \,{\left (x e + d\right )}^{\frac{3}{2}} d^{3}\right )} B b e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 21 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} - 5 \,{\left (x e + d\right )}^{\frac{3}{2}} d\right )} 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)^(3/2)*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

2/315*(21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*B*a*d*e^(-1)*sgn(b*x + a) + 21*(3*(x*e + d)^(5/2) - 5*(x*e
+ d)^(3/2)*d)*A*b*d*e^(-1)*sgn(b*x + a) + 3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*(x*e + d)^(3/2)*d
^2)*B*b*d*e^(-2)*sgn(b*x + a) + 105*(x*e + d)^(3/2)*A*a*d*sgn(b*x + a) + 3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^
(5/2)*d + 35*(x*e + d)^(3/2)*d^2)*B*a*e^(-1)*sgn(b*x + a) + 3*(15*(x*e + d)^(7/2) - 42*(x*e + d)^(5/2)*d + 35*
(x*e + d)^(3/2)*d^2)*A*b*e^(-1)*sgn(b*x + a) + (35*(x*e + d)^(9/2) - 135*(x*e + d)^(7/2)*d + 189*(x*e + d)^(5/
2)*d^2 - 105*(x*e + d)^(3/2)*d^3)*B*b*e^(-2)*sgn(b*x + a) + 21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*A*a*s
gn(b*x + a))*e^(-1)