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

Optimal. Leaf size=78 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)}{5 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {770, 21, 43} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)}{5 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^2) + (e*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/(6*b^2)

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])

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 102, normalized size = 1.31 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (15 a^4 (2 d+e x)+20 a^3 b x (3 d+2 e x)+15 a^2 b^2 x^2 (4 d+3 e x)+6 a b^3 x^3 (5 d+4 e x)+b^4 x^4 (6 d+5 e x)\right )}{30 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(15*a^4*(2*d + e*x) + 20*a^3*b*x*(3*d + 2*e*x) + 15*a^2*b^2*x^2*(4*d + 3*e*x) + 6*a*b^3*x
^3*(5*d + 4*e*x) + b^4*x^4*(6*d + 5*e*x)))/(30*(a + b*x))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 1.01, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x) (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.39, size = 96, normalized size = 1.23 \begin {gather*} \frac {1}{6} \, b^{4} e x^{6} + a^{4} d x + \frac {1}{5} \, {\left (b^{4} d + 4 \, a b^{3} e\right )} x^{5} + \frac {1}{2} \, {\left (2 \, a b^{3} d + 3 \, a^{2} b^{2} e\right )} x^{4} + \frac {2}{3} \, {\left (3 \, a^{2} b^{2} d + 2 \, a^{3} b e\right )} x^{3} + \frac {1}{2} \, {\left (4 \, a^{3} b d + a^{4} e\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^4*e*x^6 + a^4*d*x + 1/5*(b^4*d + 4*a*b^3*e)*x^5 + 1/2*(2*a*b^3*d + 3*a^2*b^2*e)*x^4 + 2/3*(3*a^2*b^2*d +
2*a^3*b*e)*x^3 + 1/2*(4*a^3*b*d + a^4*e)*x^2

________________________________________________________________________________________

giac [B]  time = 0.18, size = 162, normalized size = 2.08 \begin {gather*} \frac {1}{6} \, b^{4} x^{6} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, b^{4} d x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{5} \, a b^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + a b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{3} \, a^{3} b x^{3} e \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{4} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{4} d x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^4*x^6*e*sgn(b*x + a) + 1/5*b^4*d*x^5*sgn(b*x + a) + 4/5*a*b^3*x^5*e*sgn(b*x + a) + a*b^3*d*x^4*sgn(b*x +
a) + 3/2*a^2*b^2*x^4*e*sgn(b*x + a) + 2*a^2*b^2*d*x^3*sgn(b*x + a) + 4/3*a^3*b*x^3*e*sgn(b*x + a) + 2*a^3*b*d
*x^2*sgn(b*x + a) + 1/2*a^4*x^2*e*sgn(b*x + a) + a^4*d*x*sgn(b*x + a)

________________________________________________________________________________________

maple [B]  time = 0.05, size = 114, normalized size = 1.46 \begin {gather*} \frac {\left (5 e \,b^{4} x^{5}+24 x^{4} e a \,b^{3}+6 x^{4} d \,b^{4}+45 x^{3} e \,a^{2} b^{2}+30 x^{3} d a \,b^{3}+40 x^{2} e \,a^{3} b +60 x^{2} d \,a^{2} b^{2}+15 a^{4} e x +60 a^{3} b d x +30 d \,a^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{30 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*x*(5*b^4*e*x^5+24*a*b^3*e*x^4+6*b^4*d*x^4+45*a^2*b^2*e*x^3+30*a*b^3*d*x^3+40*a^3*b*e*x^2+60*a^2*b^2*d*x^2
+15*a^4*e*x+60*a^3*b*d*x+30*a^4*d)*((b*x+a)^2)^(3/2)/(b*x+a)^3

________________________________________________________________________________________

maxima [B]  time = 0.61, size = 251, normalized size = 3.22 \begin {gather*} \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} e}{4 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (b d + a e\right )} a x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e x}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (b d + a e\right )} a^{2}}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a e}{30 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (b d + a e\right )}}{5 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d*x + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*e*x/b + 1/4*(b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*a^2*d/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*e/b^2 - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2
)*(b*d + a*e)*a*x/b + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*e*x/b - 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(b*d + a
*e)*a^2/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*e/b^2 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(b*d + a*e)/b
^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,x\right )\,\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b x\right ) \left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________