3.8.18 \(\int x^{7/2} (A+B x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=220 \[ \frac {2 b^2 x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{15 (a+b x)}+\frac {6 a b x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{13 (a+b x)}+\frac {2 a^2 x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{11 (a+b x)}+\frac {2 b^3 B x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)}+\frac {2 a^3 A x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

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Rubi [A]  time = 0.09, antiderivative size = 220, 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, 76} \begin {gather*} \frac {2 b^2 x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2} (3 a B+A b)}{15 (a+b x)}+\frac {6 a b x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+A b)}{13 (a+b x)}+\frac {2 a^2 x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2} (a B+3 A b)}{11 (a+b x)}+\frac {2 a^3 A x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 b^3 B x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(2*a^3*A*x^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*(a + b*x)) + (2*a^2*(3*A*b + a*B)*x^(11/2)*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(11*(a + b*x)) + (6*a*b*(A*b + a*B)*x^(13/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(13*(a + b*x)) + (2
*b^2*(A*b + 3*a*B)*x^(15/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(15*(a + b*x)) + (2*b^3*B*x^(17/2)*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(17*(a + b*x))

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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 x^{7/2} (A+B 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 x^{7/2} \left (a b+b^2 x\right )^3 (A+B 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 (a^3 A b^3 x^{7/2}+a^2 b^3 (3 A b+a B) x^{9/2}+3 a b^4 (A b+a B) x^{11/2}+b^5 (A b+3 a B) x^{13/2}+b^6 B x^{15/2}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {2 a^3 A x^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {2 a^2 (3 A b+a B) x^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}}{11 (a+b x)}+\frac {6 a b (A b+a B) x^{13/2} \sqrt {a^2+2 a b x+b^2 x^2}}{13 (a+b x)}+\frac {2 b^2 (A b+3 a B) x^{15/2} \sqrt {a^2+2 a b x+b^2 x^2}}{15 (a+b x)}+\frac {2 b^3 B x^{17/2} \sqrt {a^2+2 a b x+b^2 x^2}}{17 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.40 \begin {gather*} \frac {2 x^{9/2} \sqrt {(a+b x)^2} \left (1105 a^3 (11 A+9 B x)+2295 a^2 b x (13 A+11 B x)+1683 a b^2 x^2 (15 A+13 B x)+429 b^3 x^3 (17 A+15 B x)\right )}{109395 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(2*x^(9/2)*Sqrt[(a + b*x)^2]*(1105*a^3*(11*A + 9*B*x) + 2295*a^2*b*x*(13*A + 11*B*x) + 1683*a*b^2*x^2*(15*A +
13*B*x) + 429*b^3*x^3*(17*A + 15*B*x)))/(109395*(a + b*x))

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IntegrateAlgebraic [A]  time = 13.75, size = 115, normalized size = 0.52 \begin {gather*} \frac {2 \sqrt {(a+b x)^2} \left (12155 a^3 A x^{9/2}+9945 a^3 B x^{11/2}+29835 a^2 A b x^{11/2}+25245 a^2 b B x^{13/2}+25245 a A b^2 x^{13/2}+21879 a b^2 B x^{15/2}+7293 A b^3 x^{15/2}+6435 b^3 B x^{17/2}\right )}{109395 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(7/2)*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(2*Sqrt[(a + b*x)^2]*(12155*a^3*A*x^(9/2) + 29835*a^2*A*b*x^(11/2) + 9945*a^3*B*x^(11/2) + 25245*a*A*b^2*x^(13
/2) + 25245*a^2*b*B*x^(13/2) + 7293*A*b^3*x^(15/2) + 21879*a*b^2*B*x^(15/2) + 6435*b^3*B*x^(17/2)))/(109395*(a
+ b*x))

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fricas [A]  time = 0.42, size = 78, normalized size = 0.35 \begin {gather*} \frac {2}{109395} \, {\left (6435 \, B b^{3} x^{8} + 12155 \, A a^{3} x^{4} + 7293 \, {\left (3 \, B a b^{2} + A b^{3}\right )} x^{7} + 25245 \, {\left (B a^{2} b + A a b^{2}\right )} x^{6} + 9945 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{5}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

2/109395*(6435*B*b^3*x^8 + 12155*A*a^3*x^4 + 7293*(3*B*a*b^2 + A*b^3)*x^7 + 25245*(B*a^2*b + A*a*b^2)*x^6 + 99
45*(B*a^3 + 3*A*a^2*b)*x^5)*sqrt(x)

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giac [A]  time = 0.20, size = 125, normalized size = 0.57 \begin {gather*} \frac {2}{17} \, B b^{3} x^{\frac {17}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, B a b^{2} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{15} \, A b^{3} x^{\frac {15}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{13} \, B a^{2} b x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{13} \, A a b^{2} x^{\frac {13}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{11} \, B a^{3} x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {6}{11} \, A a^{2} b x^{\frac {11}{2}} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{9} \, A a^{3} x^{\frac {9}{2}} \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

2/17*B*b^3*x^(17/2)*sgn(b*x + a) + 2/5*B*a*b^2*x^(15/2)*sgn(b*x + a) + 2/15*A*b^3*x^(15/2)*sgn(b*x + a) + 6/13
*B*a^2*b*x^(13/2)*sgn(b*x + a) + 6/13*A*a*b^2*x^(13/2)*sgn(b*x + a) + 2/11*B*a^3*x^(11/2)*sgn(b*x + a) + 6/11*
A*a^2*b*x^(11/2)*sgn(b*x + a) + 2/9*A*a^3*x^(9/2)*sgn(b*x + a)

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maple [A]  time = 0.05, size = 92, normalized size = 0.42 \begin {gather*} \frac {2 \left (6435 B \,b^{3} x^{4}+7293 A \,b^{3} x^{3}+21879 B a \,b^{2} x^{3}+25245 A a \,b^{2} x^{2}+25245 B \,a^{2} b \,x^{2}+29835 A \,a^{2} b x +9945 B \,a^{3} x +12155 A \,a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{\frac {9}{2}}}{109395 \left (b x +a \right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

2/109395*x^(9/2)*(6435*B*b^3*x^4+7293*A*b^3*x^3+21879*B*a*b^2*x^3+25245*A*a*b^2*x^2+25245*B*a^2*b*x^2+29835*A*
a^2*b*x+9945*B*a^3*x+12155*A*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

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maxima [A]  time = 0.57, size = 137, normalized size = 0.62 \begin {gather*} \frac {2}{6435} \, {\left (33 \, {\left (13 \, b^{3} x^{2} + 15 \, a b^{2} x\right )} x^{\frac {11}{2}} + 90 \, {\left (11 \, a b^{2} x^{2} + 13 \, a^{2} b x\right )} x^{\frac {9}{2}} + 65 \, {\left (9 \, a^{2} b x^{2} + 11 \, a^{3} x\right )} x^{\frac {7}{2}}\right )} A + \frac {2}{36465} \, {\left (143 \, {\left (15 \, b^{3} x^{2} + 17 \, a b^{2} x\right )} x^{\frac {13}{2}} + 374 \, {\left (13 \, a b^{2} x^{2} + 15 \, a^{2} b x\right )} x^{\frac {11}{2}} + 255 \, {\left (11 \, a^{2} b x^{2} + 13 \, a^{3} x\right )} x^{\frac {9}{2}}\right )} B \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*(B*x+A)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

2/6435*(33*(13*b^3*x^2 + 15*a*b^2*x)*x^(11/2) + 90*(11*a*b^2*x^2 + 13*a^2*b*x)*x^(9/2) + 65*(9*a^2*b*x^2 + 11*
a^3*x)*x^(7/2))*A + 2/36465*(143*(15*b^3*x^2 + 17*a*b^2*x)*x^(13/2) + 374*(13*a*b^2*x^2 + 15*a^2*b*x)*x^(11/2)
+ 255*(11*a^2*b*x^2 + 13*a^3*x)*x^(9/2))*B

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{7/2}\,\left (A+B\,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(x^(7/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int(x^(7/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*(B*x+A)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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