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

Optimal. Leaf size=135 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac {B e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

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Rubi [A]  time = 0.14, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {770, 77} \begin {gather*} \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac {B e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \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]

((A*b - a*B)*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^3) + ((b*B*d + A*b*e - 2*a*B*e)*(a +
b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^3) + (B*e*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3)

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]))))

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

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Mathematica [A]  time = 0.09, size = 144, normalized size = 1.07 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (10 a^3 (3 A (2 d+e x)+B x (3 d+2 e x))+15 a^2 b x (A (6 d+4 e x)+B x (4 d+3 e x))+3 a b^2 x^2 (5 A (4 d+3 e x)+3 B x (5 d+4 e x))+b^3 x^3 (3 A (5 d+4 e x)+2 B x (6 d+5 e x))\right )}{60 (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]*(10*a^3*(3*A*(2*d + e*x) + B*x*(3*d + 2*e*x)) + 3*a*b^2*x^2*(5*A*(4*d + 3*e*x) + 3*B*x*(5
*d + 4*e*x)) + 15*a^2*b*x*(B*x*(4*d + 3*e*x) + A*(6*d + 4*e*x)) + b^3*x^3*(3*A*(5*d + 4*e*x) + 2*B*x*(6*d + 5*
e*x))))/(60*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.95, 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]

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fricas [A]  time = 0.41, size = 146, normalized size = 1.08 \begin {gather*} \frac {1}{6} \, B b^{3} e x^{6} + A a^{3} d x + \frac {1}{5} \, {\left (B b^{3} d + {\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} x^{5} + \frac {1}{4} \, {\left ({\left (3 \, B a b^{2} + A b^{3}\right )} d + 3 \, {\left (B a^{2} b + A a b^{2}\right )} e\right )} x^{4} + \frac {1}{3} \, {\left (3 \, {\left (B a^{2} b + A a b^{2}\right )} d + {\left (B a^{3} + 3 \, A a^{2} b\right )} e\right )} x^{3} + \frac {1}{2} \, {\left (A a^{3} e + {\left (B a^{3} + 3 \, A a^{2} b\right )} d\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*b^3*e*x^6 + A*a^3*d*x + 1/5*(B*b^3*d + (3*B*a*b^2 + A*b^3)*e)*x^5 + 1/4*((3*B*a*b^2 + A*b^3)*d + 3*(B*a^
2*b + A*a*b^2)*e)*x^4 + 1/3*(3*(B*a^2*b + A*a*b^2)*d + (B*a^3 + 3*A*a^2*b)*e)*x^3 + 1/2*(A*a^3*e + (B*a^3 + 3*
A*a^2*b)*d)*x^2

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giac [B]  time = 0.17, size = 267, normalized size = 1.98 \begin {gather*} \frac {1}{6} \, B b^{3} x^{6} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, B b^{3} d x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, B a b^{2} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, A b^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a b^{2} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, A b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, B a^{2} b x^{4} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, A a b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + B a^{2} b d x^{3} \mathrm {sgn}\left (b x + a\right ) + A a b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, B a^{3} x^{3} e \mathrm {sgn}\left (b x + a\right ) + A a^{2} b x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a^{3} d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, A a^{2} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A a^{3} x^{2} e \mathrm {sgn}\left (b x + a\right ) + A a^{3} 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*b^3*x^6*e*sgn(b*x + a) + 1/5*B*b^3*d*x^5*sgn(b*x + a) + 3/5*B*a*b^2*x^5*e*sgn(b*x + a) + 1/5*A*b^3*x^5*e
*sgn(b*x + a) + 3/4*B*a*b^2*d*x^4*sgn(b*x + a) + 1/4*A*b^3*d*x^4*sgn(b*x + a) + 3/4*B*a^2*b*x^4*e*sgn(b*x + a)
+ 3/4*A*a*b^2*x^4*e*sgn(b*x + a) + B*a^2*b*d*x^3*sgn(b*x + a) + A*a*b^2*d*x^3*sgn(b*x + a) + 1/3*B*a^3*x^3*e*
sgn(b*x + a) + A*a^2*b*x^3*e*sgn(b*x + a) + 1/2*B*a^3*d*x^2*sgn(b*x + a) + 3/2*A*a^2*b*d*x^2*sgn(b*x + a) + 1/
2*A*a^3*x^2*e*sgn(b*x + a) + A*a^3*d*x*sgn(b*x + a)

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maple [A]  time = 0.05, size = 180, normalized size = 1.33 \begin {gather*} \frac {\left (10 B \,b^{3} e \,x^{5}+12 x^{4} A \,b^{3} e +36 x^{4} B e a \,b^{2}+12 x^{4} B \,b^{3} d +45 x^{3} A a \,b^{2} e +15 x^{3} A d \,b^{3}+45 x^{3} B e \,a^{2} b +45 x^{3} B a \,b^{2} d +60 x^{2} A \,a^{2} b e +60 x^{2} A d a \,b^{2}+20 x^{2} B e \,a^{3}+60 x^{2} B \,a^{2} b d +30 x A \,a^{3} e +90 x A d \,a^{2} b +30 x B \,a^{3} d +60 A d \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{60 \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]

1/60*x*(10*B*b^3*e*x^5+12*A*b^3*e*x^4+36*B*a*b^2*e*x^4+12*B*b^3*d*x^4+45*A*a*b^2*e*x^3+15*A*b^3*d*x^3+45*B*a^2
*b*e*x^3+45*B*a*b^2*d*x^3+60*A*a^2*b*e*x^2+60*A*a*b^2*d*x^2+20*B*a^3*e*x^2+60*B*a^2*b*d*x^2+30*A*a^3*e*x+90*A*
a^2*b*d*x+30*B*a^3*d*x+60*A*a^3*d)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [B]  time = 0.54, size = 254, normalized size = 1.88 \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}} B a^{2} e x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} A a d}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B a^{3} e}{4 \, b^{3}} - \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}} B e x}{6 \, 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^{2}}{4 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} B a e}{30 \, b^{3}} + \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)*B*a^2*e*x/b^2 + 1/4*(b^2*x^2 +
2*a*b*x + a^2)^(3/2)*A*a*d/b + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B*a^3*e/b^3 - 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)*B*e*x/b^2 - 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)*B*a*e/b^3 + 1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(
B*d + A*e)/b^2

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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)

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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)

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