3.2094 \(\int (a+b x) (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=152 \[ \frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)}-\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}{5 e^3 (a+b x)} \]

[Out]

(2*(b*d - a*e)^2*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) - (4*b*(b*d - a*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^2*(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 = 0.067063, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {770, 21, 43} \[ \frac{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{9/2}}{9 e^3 (a+b x)}-\frac{4 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2} (b d-a e)}{7 e^3 (a+b x)}+\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)^2}{5 e^3 (a+b x)} \]

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)^2*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) - (4*b*(b*d - a*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^2*(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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 (a+b x) \left (a b+b^2 x\right ) (d+e x)^{3/2} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^2 (d+e x)^{3/2} \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(-b d+a e)^2 (d+e x)^{3/2}}{e^2}-\frac{2 b (b d-a e) (d+e x)^{5/2}}{e^2}+\frac{b^2 (d+e x)^{7/2}}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{2 (b d-a e)^2 (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}-\frac{4 b (b d-a 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^2 (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 = 0.0491592, size = 79, normalized size = 0.52 \[ \frac{2 \sqrt{(a+b x)^2} (d+e x)^{5/2} \left (63 a^2 e^2+18 a b e (5 e x-2 d)+b^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )\right )}{315 e^3 (a+b x)} \]

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)*(63*a^2*e^2 + 18*a*b*e*(-2*d + 5*e*x) + b^2*(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 = 0.004, size = 79, normalized size = 0.5 \begin{align*}{\frac{70\,{x}^{2}{b}^{2}{e}^{2}+180\,xab{e}^{2}-40\,x{b}^{2}de+126\,{a}^{2}{e}^{2}-72\,abde+16\,{b}^{2}{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^2*e^2*x^2+90*a*b*e^2*x-20*b^2*d*e*x+63*a^2*e^2-36*a*b*d*e+8*b^2*d^2)*((b*x+a)^2)^(1/
2)/e^3/(b*x+a)

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Maxima [A]  time = 1.12236, size = 225, normalized size = 1.48 \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 = 0.997398, size = 300, normalized size = 1.97 \begin{align*} \frac{2 \,{\left (35 \, b^{2} e^{4} x^{4} + 8 \, b^{2} d^{4} - 36 \, a b d^{3} e + 63 \, a^{2} d^{2} e^{2} + 10 \,{\left (5 \, b^{2} d e^{3} + 9 \, a b e^{4}\right )} x^{3} + 3 \,{\left (b^{2} d^{2} e^{2} + 48 \, a b d e^{3} + 21 \, a^{2} e^{4}\right )} x^{2} - 2 \,{\left (2 \, b^{2} d^{3} e - 9 \, a b d^{2} e^{2} - 63 \, a^{2} d e^{3}\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^2*e^4*x^4 + 8*b^2*d^4 - 36*a*b*d^3*e + 63*a^2*d^2*e^2 + 10*(5*b^2*d*e^3 + 9*a*b*e^4)*x^3 + 3*(b^2*
d^2*e^2 + 48*a*b*d*e^3 + 21*a^2*e^4)*x^2 - 2*(2*b^2*d^3*e - 9*a*b*d^2*e^2 - 63*a^2*d*e^3)*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 = 1.19973, size = 336, normalized size = 2.21 \begin{align*} \frac{2}{315} \,{\left (42 \,{\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^{2} d e^{\left (-2\right )} \mathrm{sgn}\left (b x + a\right ) + 105 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} d \mathrm{sgn}\left (b x + a\right ) + 6 \,{\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^{2} 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^{2} \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*(42*(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^2*d*e^(-2)*sgn(b*x + a) + 105*(x*e + d)^(3/2)*a^2*d*sgn(b*x + a) +
6*(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^2*e^(-2)*sgn(b*x + a) +
21*(3*(x*e + d)^(5/2) - 5*(x*e + d)^(3/2)*d)*a^2*sgn(b*x + a))*e^(-1)