3.3.3 \(\int x^{5/2} (A+B x) (b x+c x^2)^{3/2} \, dx\)

Optimal. Leaf size=207 \[ -\frac {256 b^4 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{45045 c^6 x^{5/2}}+\frac {128 b^3 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{9009 c^5 x^{3/2}}-\frac {32 b^2 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{1287 c^4 \sqrt {x}}+\frac {16 b \sqrt {x} \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{429 c^3}-\frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c} \]

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Rubi [A]  time = 0.15, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {794, 656, 648} \begin {gather*} -\frac {256 b^4 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{45045 c^6 x^{5/2}}+\frac {128 b^3 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{9009 c^5 x^{3/2}}-\frac {32 b^2 \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{1287 c^4 \sqrt {x}}+\frac {16 b \sqrt {x} \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{429 c^3}-\frac {2 x^{3/2} \left (b x+c x^2\right )^{5/2} (2 b B-3 A c)}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(-256*b^4*(2*b*B - 3*A*c)*(b*x + c*x^2)^(5/2))/(45045*c^6*x^(5/2)) + (128*b^3*(2*b*B - 3*A*c)*(b*x + c*x^2)^(5
/2))/(9009*c^5*x^(3/2)) - (32*b^2*(2*b*B - 3*A*c)*(b*x + c*x^2)^(5/2))/(1287*c^4*Sqrt[x]) + (16*b*(2*b*B - 3*A
*c)*Sqrt[x]*(b*x + c*x^2)^(5/2))/(429*c^3) - (2*(2*b*B - 3*A*c)*x^(3/2)*(b*x + c*x^2)^(5/2))/(39*c^2) + (2*B*x
^(5/2)*(b*x + c*x^2)^(5/2))/(15*c)

Rule 648

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(p + 1)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rule 656

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[(Simplify[m + p]*(2*c*d - b*e))/(c*(m + 2*p + 1)), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g
, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rubi steps

\begin {align*} \int x^{5/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}+\frac {\left (2 \left (\frac {5}{2} (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right )\right ) \int x^{5/2} \left (b x+c x^2\right )^{3/2} \, dx}{15 c}\\ &=-\frac {2 (2 b B-3 A c) x^{3/2} \left (b x+c x^2\right )^{5/2}}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}+\frac {(8 b (2 b B-3 A c)) \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx}{39 c^2}\\ &=\frac {16 b (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{429 c^3}-\frac {2 (2 b B-3 A c) x^{3/2} \left (b x+c x^2\right )^{5/2}}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}-\frac {\left (16 b^2 (2 b B-3 A c)\right ) \int \sqrt {x} \left (b x+c x^2\right )^{3/2} \, dx}{143 c^3}\\ &=-\frac {32 b^2 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{1287 c^4 \sqrt {x}}+\frac {16 b (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{429 c^3}-\frac {2 (2 b B-3 A c) x^{3/2} \left (b x+c x^2\right )^{5/2}}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}+\frac {\left (64 b^3 (2 b B-3 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx}{1287 c^4}\\ &=\frac {128 b^3 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{9009 c^5 x^{3/2}}-\frac {32 b^2 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{1287 c^4 \sqrt {x}}+\frac {16 b (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{429 c^3}-\frac {2 (2 b B-3 A c) x^{3/2} \left (b x+c x^2\right )^{5/2}}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}-\frac {\left (128 b^4 (2 b B-3 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{9009 c^5}\\ &=-\frac {256 b^4 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{45045 c^6 x^{5/2}}+\frac {128 b^3 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{9009 c^5 x^{3/2}}-\frac {32 b^2 (2 b B-3 A c) \left (b x+c x^2\right )^{5/2}}{1287 c^4 \sqrt {x}}+\frac {16 b (2 b B-3 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{429 c^3}-\frac {2 (2 b B-3 A c) x^{3/2} \left (b x+c x^2\right )^{5/2}}{39 c^2}+\frac {2 B x^{5/2} \left (b x+c x^2\right )^{5/2}}{15 c}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 110, normalized size = 0.53 \begin {gather*} \frac {2 (x (b+c x))^{5/2} \left (128 b^4 c (3 A+5 B x)-160 b^3 c^2 x (6 A+7 B x)+1680 b^2 c^3 x^2 (A+B x)-210 b c^4 x^3 (12 A+11 B x)+231 c^5 x^4 (15 A+13 B x)-256 b^5 B\right )}{45045 c^6 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(2*(x*(b + c*x))^(5/2)*(-256*b^5*B + 1680*b^2*c^3*x^2*(A + B*x) + 128*b^4*c*(3*A + 5*B*x) - 160*b^3*c^2*x*(6*A
+ 7*B*x) - 210*b*c^4*x^3*(12*A + 11*B*x) + 231*c^5*x^4*(15*A + 13*B*x)))/(45045*c^6*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.72, size = 131, normalized size = 0.63 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{5/2} \left (384 A b^4 c-960 A b^3 c^2 x+1680 A b^2 c^3 x^2-2520 A b c^4 x^3+3465 A c^5 x^4-256 b^5 B+640 b^4 B c x-1120 b^3 B c^2 x^2+1680 b^2 B c^3 x^3-2310 b B c^4 x^4+3003 B c^5 x^5\right )}{45045 c^6 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(A + B*x)*(b*x + c*x^2)^(3/2),x]

[Out]

(2*(b*x + c*x^2)^(5/2)*(-256*b^5*B + 384*A*b^4*c + 640*b^4*B*c*x - 960*A*b^3*c^2*x - 1120*b^3*B*c^2*x^2 + 1680
*A*b^2*c^3*x^2 + 1680*b^2*B*c^3*x^3 - 2520*A*b*c^4*x^3 - 2310*b*B*c^4*x^4 + 3465*A*c^5*x^4 + 3003*B*c^5*x^5))/
(45045*c^6*x^(5/2))

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fricas [A]  time = 0.41, size = 174, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (3003 \, B c^{7} x^{7} - 256 \, B b^{7} + 384 \, A b^{6} c + 231 \, {\left (16 \, B b c^{6} + 15 \, A c^{7}\right )} x^{6} + 63 \, {\left (B b^{2} c^{5} + 70 \, A b c^{6}\right )} x^{5} - 35 \, {\left (2 \, B b^{3} c^{4} - 3 \, A b^{2} c^{5}\right )} x^{4} + 40 \, {\left (2 \, B b^{4} c^{3} - 3 \, A b^{3} c^{4}\right )} x^{3} - 48 \, {\left (2 \, B b^{5} c^{2} - 3 \, A b^{4} c^{3}\right )} x^{2} + 64 \, {\left (2 \, B b^{6} c - 3 \, A b^{5} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{45045 \, c^{6} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

2/45045*(3003*B*c^7*x^7 - 256*B*b^7 + 384*A*b^6*c + 231*(16*B*b*c^6 + 15*A*c^7)*x^6 + 63*(B*b^2*c^5 + 70*A*b*c
^6)*x^5 - 35*(2*B*b^3*c^4 - 3*A*b^2*c^5)*x^4 + 40*(2*B*b^4*c^3 - 3*A*b^3*c^4)*x^3 - 48*(2*B*b^5*c^2 - 3*A*b^4*
c^3)*x^2 + 64*(2*B*b^6*c - 3*A*b^5*c^2)*x)*sqrt(c*x^2 + b*x)/(c^6*sqrt(x))

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giac [B]  time = 0.25, size = 343, normalized size = 1.66 \begin {gather*} -\frac {2}{45045} \, B c {\left (\frac {1024 \, b^{\frac {15}{2}}}{c^{7}} - \frac {3003 \, {\left (c x + b\right )}^{\frac {15}{2}} - 20790 \, {\left (c x + b\right )}^{\frac {13}{2}} b + 61425 \, {\left (c x + b\right )}^{\frac {11}{2}} b^{2} - 100100 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{3} + 96525 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{4} - 54054 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{5} + 15015 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{6}}{c^{7}}\right )} + \frac {2}{9009} \, B b {\left (\frac {256 \, b^{\frac {13}{2}}}{c^{6}} + \frac {693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}}{c^{6}}\right )} + \frac {2}{9009} \, A c {\left (\frac {256 \, b^{\frac {13}{2}}}{c^{6}} + \frac {693 \, {\left (c x + b\right )}^{\frac {13}{2}} - 4095 \, {\left (c x + b\right )}^{\frac {11}{2}} b + 10010 \, {\left (c x + b\right )}^{\frac {9}{2}} b^{2} - 12870 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{4} - 3003 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{5}}{c^{6}}\right )} - \frac {2}{3465} \, A b {\left (\frac {128 \, b^{\frac {11}{2}}}{c^{5}} - \frac {315 \, {\left (c x + b\right )}^{\frac {11}{2}} - 1540 \, {\left (c x + b\right )}^{\frac {9}{2}} b + 2970 \, {\left (c x + b\right )}^{\frac {7}{2}} b^{2} - 2772 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{4}}{c^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

-2/45045*B*c*(1024*b^(15/2)/c^7 - (3003*(c*x + b)^(15/2) - 20790*(c*x + b)^(13/2)*b + 61425*(c*x + b)^(11/2)*b
^2 - 100100*(c*x + b)^(9/2)*b^3 + 96525*(c*x + b)^(7/2)*b^4 - 54054*(c*x + b)^(5/2)*b^5 + 15015*(c*x + b)^(3/2
)*b^6)/c^7) + 2/9009*B*b*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)
^(9/2)*b^2 - 12870*(c*x + b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) + 2/9009*A*
c*(256*b^(13/2)/c^6 + (693*(c*x + b)^(13/2) - 4095*(c*x + b)^(11/2)*b + 10010*(c*x + b)^(9/2)*b^2 - 12870*(c*x
+ b)^(7/2)*b^3 + 9009*(c*x + b)^(5/2)*b^4 - 3003*(c*x + b)^(3/2)*b^5)/c^6) - 2/3465*A*b*(128*b^(11/2)/c^5 - (
315*(c*x + b)^(11/2) - 1540*(c*x + b)^(9/2)*b + 2970*(c*x + b)^(7/2)*b^2 - 2772*(c*x + b)^(5/2)*b^3 + 1155*(c*
x + b)^(3/2)*b^4)/c^5)

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maple [A]  time = 0.06, size = 131, normalized size = 0.63 \begin {gather*} \frac {2 \left (c x +b \right ) \left (3003 B \,x^{5} c^{5}+3465 A \,c^{5} x^{4}-2310 B b \,c^{4} x^{4}-2520 A b \,c^{4} x^{3}+1680 B \,b^{2} c^{3} x^{3}+1680 A \,b^{2} c^{3} x^{2}-1120 B \,b^{3} c^{2} x^{2}-960 A \,b^{3} c^{2} x +640 B \,b^{4} c x +384 A \,b^{4} c -256 B \,b^{5}\right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{45045 c^{6} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(3/2),x)

[Out]

2/45045*(c*x+b)*(3003*B*c^5*x^5+3465*A*c^5*x^4-2310*B*b*c^4*x^4-2520*A*b*c^4*x^3+1680*B*b^2*c^3*x^3+1680*A*b^2
*c^3*x^2-1120*B*b^3*c^2*x^2-960*A*b^3*c^2*x+640*B*b^4*c*x+384*A*b^4*c-256*B*b^5)*(c*x^2+b*x)^(3/2)/c^6/x^(3/2)

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maxima [A]  time = 0.61, size = 318, normalized size = 1.54 \begin {gather*} \frac {2 \, {\left (5 \, {\left (693 \, c^{6} x^{6} + 63 \, b c^{5} x^{5} - 70 \, b^{2} c^{4} x^{4} + 80 \, b^{3} c^{3} x^{3} - 96 \, b^{4} c^{2} x^{2} + 128 \, b^{5} c x - 256 \, b^{6}\right )} x^{5} + 13 \, {\left (315 \, b c^{5} x^{6} + 35 \, b^{2} c^{4} x^{5} - 40 \, b^{3} c^{3} x^{4} + 48 \, b^{4} c^{2} x^{3} - 64 \, b^{5} c x^{2} + 128 \, b^{6} x\right )} x^{4}\right )} \sqrt {c x + b} A}{45045 \, c^{5} x^{5}} + \frac {2 \, {\left ({\left (3003 \, c^{7} x^{7} + 231 \, b c^{6} x^{6} - 252 \, b^{2} c^{5} x^{5} + 280 \, b^{3} c^{4} x^{4} - 320 \, b^{4} c^{3} x^{3} + 384 \, b^{5} c^{2} x^{2} - 512 \, b^{6} c x + 1024 \, b^{7}\right )} x^{6} + 5 \, {\left (693 \, b c^{6} x^{7} + 63 \, b^{2} c^{5} x^{6} - 70 \, b^{3} c^{4} x^{5} + 80 \, b^{4} c^{3} x^{4} - 96 \, b^{5} c^{2} x^{3} + 128 \, b^{6} c x^{2} - 256 \, b^{7} x\right )} x^{5}\right )} \sqrt {c x + b} B}{45045 \, c^{6} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)*(c*x^2+b*x)^(3/2),x, algorithm="maxima")

[Out]

2/45045*(5*(693*c^6*x^6 + 63*b*c^5*x^5 - 70*b^2*c^4*x^4 + 80*b^3*c^3*x^3 - 96*b^4*c^2*x^2 + 128*b^5*c*x - 256*
b^6)*x^5 + 13*(315*b*c^5*x^6 + 35*b^2*c^4*x^5 - 40*b^3*c^3*x^4 + 48*b^4*c^2*x^3 - 64*b^5*c*x^2 + 128*b^6*x)*x^
4)*sqrt(c*x + b)*A/(c^5*x^5) + 2/45045*((3003*c^7*x^7 + 231*b*c^6*x^6 - 252*b^2*c^5*x^5 + 280*b^3*c^4*x^4 - 32
0*b^4*c^3*x^3 + 384*b^5*c^2*x^2 - 512*b^6*c*x + 1024*b^7)*x^6 + 5*(693*b*c^6*x^7 + 63*b^2*c^5*x^6 - 70*b^3*c^4
*x^5 + 80*b^4*c^3*x^4 - 96*b^5*c^2*x^3 + 128*b^6*c*x^2 - 256*b^7*x)*x^5)*sqrt(c*x + b)*B/(c^6*x^6)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{5/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x + c*x^2)^(3/2)*(A + B*x),x)

[Out]

int(x^(5/2)*(b*x + c*x^2)^(3/2)*(A + B*x), 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**(5/2)*(B*x+A)*(c*x**2+b*x)**(3/2),x)

[Out]

Timed out

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