3.2104 \(\int \frac{(a+b x) (a^2+2 a b x+b^2 x^2)^{3/2}}{(d+e x)^{3/2}} \, dx\)

Optimal. Leaf size=258 \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^5 (a+b x)}-\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^5 (a+b x)}+\frac{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^5 (a+b x)}-\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x) \sqrt{d+e x}} \]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) - (8*b*(b*d - a*e)^3*Sqrt[d + e
*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) + (4*b^2*(b*d - a*e)^2*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x +
 b^2*x^2])/(e^5*(a + b*x)) - (8*b^3*(b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x
)) + (2*b^4*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(a + b*x))

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Rubi [A]  time = 0.105694, antiderivative size = 258, 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} \[ \frac{2 b^4 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{7/2}}{7 e^5 (a+b x)}-\frac{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}{5 e^5 (a+b x)}+\frac{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}{e^5 (a+b x)}-\frac{8 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}{e^5 (a+b x)}-\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^4}{e^5 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x)^(3/2),x]

[Out]

(-2*(b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)*Sqrt[d + e*x]) - (8*b*(b*d - a*e)^3*Sqrt[d + e
*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^5*(a + b*x)) + (4*b^2*(b*d - a*e)^2*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x +
 b^2*x^2])/(e^5*(a + b*x)) - (8*b^3*(b*d - a*e)*(d + e*x)^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^5*(a + b*x
)) + (2*b^4*(d + e*x)^(7/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*e^5*(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 \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{(d+e x)^{3/2}} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{(a+b x) \left (a b+b^2 x\right )^3}{(d+e x)^{3/2}} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{(a+b x)^4}{(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)^4}{e^4 (d+e x)^{3/2}}-\frac{4 b (b d-a e)^3}{e^4 \sqrt{d+e x}}+\frac{6 b^2 (b d-a e)^2 \sqrt{d+e x}}{e^4}-\frac{4 b^3 (b d-a e) (d+e x)^{3/2}}{e^4}+\frac{b^4 (d+e x)^{5/2}}{e^4}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{2 (b d-a e)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x) \sqrt{d+e x}}-\frac{8 b (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}+\frac{4 b^2 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}{e^5 (a+b x)}-\frac{8 b^3 (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^5 (a+b x)}+\frac{2 b^4 (d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}{7 e^5 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0998249, size = 169, normalized size = 0.66 \[ \frac{2 \sqrt{(a+b x)^2} \left (70 a^2 b^2 e^2 \left (-8 d^2-4 d e x+e^2 x^2\right )+140 a^3 b e^3 (2 d+e x)-35 a^4 e^4+28 a b^3 e \left (8 d^2 e x+16 d^3-2 d e^2 x^2+e^3 x^3\right )+b^4 \left (16 d^2 e^2 x^2-64 d^3 e x-128 d^4-8 d e^3 x^3+5 e^4 x^4\right )\right )}{35 e^5 (a+b x) \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[(a + b*x)^2]*(-35*a^4*e^4 + 140*a^3*b*e^3*(2*d + e*x) + 70*a^2*b^2*e^2*(-8*d^2 - 4*d*e*x + e^2*x^2) +
28*a*b^3*e*(16*d^3 + 8*d^2*e*x - 2*d*e^2*x^2 + e^3*x^3) + b^4*(-128*d^4 - 64*d^3*e*x + 16*d^2*e^2*x^2 - 8*d*e^
3*x^3 + 5*e^4*x^4)))/(35*e^5*(a + b*x)*Sqrt[d + e*x])

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Maple [A]  time = 0.007, size = 202, normalized size = 0.8 \begin{align*} -{\frac{-10\,{x}^{4}{b}^{4}{e}^{4}-56\,{x}^{3}a{b}^{3}{e}^{4}+16\,{x}^{3}{b}^{4}d{e}^{3}-140\,{x}^{2}{a}^{2}{b}^{2}{e}^{4}+112\,{x}^{2}a{b}^{3}d{e}^{3}-32\,{x}^{2}{b}^{4}{d}^{2}{e}^{2}-280\,x{a}^{3}b{e}^{4}+560\,x{a}^{2}{b}^{2}d{e}^{3}-448\,xa{b}^{3}{d}^{2}{e}^{2}+128\,x{b}^{4}{d}^{3}e+70\,{a}^{4}{e}^{4}-560\,d{e}^{3}{a}^{3}b+1120\,{a}^{2}{b}^{2}{d}^{2}{e}^{2}-896\,a{b}^{3}{d}^{3}e+256\,{b}^{4}{d}^{4}}{35\, \left ( bx+a \right ) ^{3}{e}^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{ex+d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(3/2),x)

[Out]

-2/35/(e*x+d)^(1/2)*(-5*b^4*e^4*x^4-28*a*b^3*e^4*x^3+8*b^4*d*e^3*x^3-70*a^2*b^2*e^4*x^2+56*a*b^3*d*e^3*x^2-16*
b^4*d^2*e^2*x^2-140*a^3*b*e^4*x+280*a^2*b^2*d*e^3*x-224*a*b^3*d^2*e^2*x+64*b^4*d^3*e*x+35*a^4*e^4-280*a^3*b*d*
e^3+560*a^2*b^2*d^2*e^2-448*a*b^3*d^3*e+128*b^4*d^4)*((b*x+a)^2)^(3/2)/e^5/(b*x+a)^3

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Maxima [A]  time = 1.15846, size = 381, normalized size = 1.48 \begin{align*} \frac{2 \,{\left (b^{3} e^{3} x^{3} + 16 \, b^{3} d^{3} - 40 \, a b^{2} d^{2} e + 30 \, a^{2} b d e^{2} - 5 \, a^{3} e^{3} -{\left (2 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{2} +{\left (8 \, b^{3} d^{2} e - 20 \, a b^{2} d e^{2} + 15 \, a^{2} b e^{3}\right )} x\right )} a}{5 \, \sqrt{e x + d} e^{4}} + \frac{2 \,{\left (5 \, b^{3} e^{4} x^{4} - 128 \, b^{3} d^{4} + 336 \, a b^{2} d^{3} e - 280 \, a^{2} b d^{2} e^{2} + 70 \, a^{3} d e^{3} -{\left (8 \, b^{3} d e^{3} - 21 \, a b^{2} e^{4}\right )} x^{3} +{\left (16 \, b^{3} d^{2} e^{2} - 42 \, a b^{2} d e^{3} + 35 \, a^{2} b e^{4}\right )} x^{2} -{\left (64 \, b^{3} d^{3} e - 168 \, a b^{2} d^{2} e^{2} + 140 \, a^{2} b d e^{3} - 35 \, a^{3} e^{4}\right )} x\right )} b}{35 \, \sqrt{e x + d} e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b^3*e^3*x^3 + 16*b^3*d^3 - 40*a*b^2*d^2*e + 30*a^2*b*d*e^2 - 5*a^3*e^3 - (2*b^3*d*e^2 - 5*a*b^2*e^3)*x^2
+ (8*b^3*d^2*e - 20*a*b^2*d*e^2 + 15*a^2*b*e^3)*x)*a/(sqrt(e*x + d)*e^4) + 2/35*(5*b^3*e^4*x^4 - 128*b^3*d^4 +
 336*a*b^2*d^3*e - 280*a^2*b*d^2*e^2 + 70*a^3*d*e^3 - (8*b^3*d*e^3 - 21*a*b^2*e^4)*x^3 + (16*b^3*d^2*e^2 - 42*
a*b^2*d*e^3 + 35*a^2*b*e^4)*x^2 - (64*b^3*d^3*e - 168*a*b^2*d^2*e^2 + 140*a^2*b*d*e^3 - 35*a^3*e^4)*x)*b/(sqrt
(e*x + d)*e^5)

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Fricas [A]  time = 0.988529, size = 412, normalized size = 1.6 \begin{align*} \frac{2 \,{\left (5 \, b^{4} e^{4} x^{4} - 128 \, b^{4} d^{4} + 448 \, a b^{3} d^{3} e - 560 \, a^{2} b^{2} d^{2} e^{2} + 280 \, a^{3} b d e^{3} - 35 \, a^{4} e^{4} - 4 \,{\left (2 \, b^{4} d e^{3} - 7 \, a b^{3} e^{4}\right )} x^{3} + 2 \,{\left (8 \, b^{4} d^{2} e^{2} - 28 \, a b^{3} d e^{3} + 35 \, a^{2} b^{2} e^{4}\right )} x^{2} - 4 \,{\left (16 \, b^{4} d^{3} e - 56 \, a b^{3} d^{2} e^{2} + 70 \, a^{2} b^{2} d e^{3} - 35 \, a^{3} b e^{4}\right )} x\right )} \sqrt{e x + d}}{35 \,{\left (e^{6} x + d e^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*b^4*e^4*x^4 - 128*b^4*d^4 + 448*a*b^3*d^3*e - 560*a^2*b^2*d^2*e^2 + 280*a^3*b*d*e^3 - 35*a^4*e^4 - 4*(
2*b^4*d*e^3 - 7*a*b^3*e^4)*x^3 + 2*(8*b^4*d^2*e^2 - 28*a*b^3*d*e^3 + 35*a^2*b^2*e^4)*x^2 - 4*(16*b^4*d^3*e - 5
6*a*b^3*d^2*e^2 + 70*a^2*b^2*d*e^3 - 35*a^3*b*e^4)*x)*sqrt(e*x + d)/(e^6*x + d*e^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}{\left (d + e x\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)*((a + b*x)**2)**(3/2)/(d + e*x)**(3/2), x)

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Giac [A]  time = 1.20923, size = 441, normalized size = 1.71 \begin{align*} \frac{2}{35} \,{\left (5 \,{\left (x e + d\right )}^{\frac{7}{2}} b^{4} e^{30} \mathrm{sgn}\left (b x + a\right ) - 28 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{4} d e^{30} \mathrm{sgn}\left (b x + a\right ) + 70 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{4} d^{2} e^{30} \mathrm{sgn}\left (b x + a\right ) - 140 \, \sqrt{x e + d} b^{4} d^{3} e^{30} \mathrm{sgn}\left (b x + a\right ) + 28 \,{\left (x e + d\right )}^{\frac{5}{2}} a b^{3} e^{31} \mathrm{sgn}\left (b x + a\right ) - 140 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{3} d e^{31} \mathrm{sgn}\left (b x + a\right ) + 420 \, \sqrt{x e + d} a b^{3} d^{2} e^{31} \mathrm{sgn}\left (b x + a\right ) + 70 \,{\left (x e + d\right )}^{\frac{3}{2}} a^{2} b^{2} e^{32} \mathrm{sgn}\left (b x + a\right ) - 420 \, \sqrt{x e + d} a^{2} b^{2} d e^{32} \mathrm{sgn}\left (b x + a\right ) + 140 \, \sqrt{x e + d} a^{3} b e^{33} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-35\right )} - \frac{2 \,{\left (b^{4} d^{4} \mathrm{sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm{sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm{sgn}\left (b x + a\right )\right )} e^{\left (-5\right )}}{\sqrt{x e + d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(3/2),x, algorithm="giac")

[Out]

2/35*(5*(x*e + d)^(7/2)*b^4*e^30*sgn(b*x + a) - 28*(x*e + d)^(5/2)*b^4*d*e^30*sgn(b*x + a) + 70*(x*e + d)^(3/2
)*b^4*d^2*e^30*sgn(b*x + a) - 140*sqrt(x*e + d)*b^4*d^3*e^30*sgn(b*x + a) + 28*(x*e + d)^(5/2)*a*b^3*e^31*sgn(
b*x + a) - 140*(x*e + d)^(3/2)*a*b^3*d*e^31*sgn(b*x + a) + 420*sqrt(x*e + d)*a*b^3*d^2*e^31*sgn(b*x + a) + 70*
(x*e + d)^(3/2)*a^2*b^2*e^32*sgn(b*x + a) - 420*sqrt(x*e + d)*a^2*b^2*d*e^32*sgn(b*x + a) + 140*sqrt(x*e + d)*
a^3*b*e^33*sgn(b*x + a))*e^(-35) - 2*(b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sg
n(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4*e^4*sgn(b*x + a))*e^(-5)/sqrt(x*e + d)