3.18.71 \(\int (a+b x) (d+e x) (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=78 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e)}{7 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^2} \]

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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)^6 (b d-a e)}{7 b^2}+\frac {e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 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)^(5/2),x]

[Out]

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

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Mathematica [A]  time = 0.05, size = 140, normalized size = 1.79 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (28 a^6 (2 d+e x)+56 a^5 b x (3 d+2 e x)+70 a^4 b^2 x^2 (4 d+3 e x)+56 a^3 b^3 x^3 (5 d+4 e x)+28 a^2 b^4 x^4 (6 d+5 e x)+8 a b^5 x^5 (7 d+6 e x)+b^6 x^6 (8 d+7 e x)\right )}{56 (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)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(28*a^6*(2*d + e*x) + 56*a^5*b*x*(3*d + 2*e*x) + 70*a^4*b^2*x^2*(4*d + 3*e*x) + 56*a^3*b^
3*x^3*(5*d + 4*e*x) + 28*a^2*b^4*x^4*(6*d + 5*e*x) + 8*a*b^5*x^5*(7*d + 6*e*x) + b^6*x^6*(8*d + 7*e*x)))/(56*(
a + b*x))

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IntegrateAlgebraic [F]  time = 1.32, 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 )^{5/2} \, 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)^(5/2),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)*(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2), x]

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fricas [B]  time = 0.40, size = 142, normalized size = 1.82 \begin {gather*} \frac {1}{8} \, b^{6} e x^{8} + a^{6} d x + \frac {1}{7} \, {\left (b^{6} d + 6 \, a b^{5} e\right )} x^{7} + \frac {1}{2} \, {\left (2 \, a b^{5} d + 5 \, a^{2} b^{4} e\right )} x^{6} + {\left (3 \, a^{2} b^{4} d + 4 \, a^{3} b^{3} e\right )} x^{5} + \frac {5}{4} \, {\left (4 \, a^{3} b^{3} d + 3 \, a^{4} b^{2} e\right )} x^{4} + {\left (5 \, a^{4} b^{2} d + 2 \, a^{5} b e\right )} x^{3} + \frac {1}{2} \, {\left (6 \, a^{5} b d + a^{6} 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)^(5/2),x, algorithm="fricas")

[Out]

1/8*b^6*e*x^8 + a^6*d*x + 1/7*(b^6*d + 6*a*b^5*e)*x^7 + 1/2*(2*a*b^5*d + 5*a^2*b^4*e)*x^6 + (3*a^2*b^4*d + 4*a
^3*b^3*e)*x^5 + 5/4*(4*a^3*b^3*d + 3*a^4*b^2*e)*x^4 + (5*a^4*b^2*d + 2*a^5*b*e)*x^3 + 1/2*(6*a^5*b*d + a^6*e)*
x^2

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giac [B]  time = 0.17, size = 236, normalized size = 3.03 \begin {gather*} \frac {1}{8} \, b^{6} x^{8} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{7} \, b^{6} d x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{7} \, a b^{5} x^{7} e \mathrm {sgn}\left (b x + a\right ) + a b^{5} d x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{2} b^{4} x^{6} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{2} b^{4} d x^{5} \mathrm {sgn}\left (b x + a\right ) + 4 \, a^{3} b^{3} x^{5} e \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{3} b^{3} d x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {15}{4} \, a^{4} b^{2} x^{4} e \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} d x^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{5} b x^{3} e \mathrm {sgn}\left (b x + a\right ) + 3 \, a^{5} b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{6} x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{6} 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)^(5/2),x, algorithm="giac")

[Out]

1/8*b^6*x^8*e*sgn(b*x + a) + 1/7*b^6*d*x^7*sgn(b*x + a) + 6/7*a*b^5*x^7*e*sgn(b*x + a) + a*b^5*d*x^6*sgn(b*x +
a) + 5/2*a^2*b^4*x^6*e*sgn(b*x + a) + 3*a^2*b^4*d*x^5*sgn(b*x + a) + 4*a^3*b^3*x^5*e*sgn(b*x + a) + 5*a^3*b^3
*d*x^4*sgn(b*x + a) + 15/4*a^4*b^2*x^4*e*sgn(b*x + a) + 5*a^4*b^2*d*x^3*sgn(b*x + a) + 2*a^5*b*x^3*e*sgn(b*x +
a) + 3*a^5*b*d*x^2*sgn(b*x + a) + 1/2*a^6*x^2*e*sgn(b*x + a) + a^6*d*x*sgn(b*x + a)

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maple [B]  time = 0.05, size = 162, normalized size = 2.08 \begin {gather*} \frac {\left (7 e \,b^{6} x^{7}+48 x^{6} e a \,b^{5}+8 x^{6} d \,b^{6}+140 x^{5} e \,a^{2} b^{4}+56 x^{5} d a \,b^{5}+224 a^{3} b^{3} e \,x^{4}+168 a^{2} b^{4} d \,x^{4}+210 x^{3} e \,a^{4} b^{2}+280 x^{3} d \,a^{3} b^{3}+112 a^{5} b e \,x^{2}+280 a^{4} b^{2} d \,x^{2}+28 x e \,a^{6}+168 x d \,a^{5} b +56 d \,a^{6}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} x}{56 \left (b x +a \right )^{5}} \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)^(5/2),x)

[Out]

1/56*x*(7*b^6*e*x^7+48*a*b^5*e*x^6+8*b^6*d*x^6+140*a^2*b^4*e*x^5+56*a*b^5*d*x^5+224*a^3*b^3*e*x^4+168*a^2*b^4*
d*x^4+210*a^4*b^2*e*x^3+280*a^3*b^3*d*x^3+112*a^5*b*e*x^2+280*a^4*b^2*d*x^2+28*a^6*e*x+168*a^5*b*d*x+56*a^6*d)
*((b*x+a)^2)^(5/2)/(b*x+a)^5

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maxima [B]  time = 0.63, size = 251, normalized size = 3.22 \begin {gather*} \frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a d x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} e x}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} d}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3} e}{6 \, b^{2}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (b d + a e\right )} a x}{6 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} e x}{8 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (b d + a e\right )} a^{2}}{6 \, b^{2}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a e}{56 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} {\left (b d + a e\right )}}{7 \, 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)^(5/2),x, algorithm="maxima")

[Out]

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*d*x + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*e*x/b + 1/6*(b^2*x^2 + 2*a
*b*x + a^2)^(5/2)*a^2*d/b + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*e/b^2 - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2
)*(b*d + a*e)*a*x/b + 1/8*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*e*x/b - 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(b*d + a
*e)*a^2/b^2 - 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a*e/b^2 + 1/7*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*(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 )}^{5/2} \,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)^(5/2),x)

[Out]

int((a + b*x)*(d + e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), 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 {5}{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)**(5/2),x)

[Out]

Integral((a + b*x)*(d + e*x)*((a + b*x)**2)**(5/2), x)

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