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

Optimal. Leaf size=170 \[ \frac {32 b^3 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^{5/2}}-\frac {16 b^2 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^{3/2}}+\frac {4 b \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{429 c^3 \sqrt {x}}-\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c} \]

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Rubi [A]  time = 0.16, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, 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 {32 b^3 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{15015 c^5 x^{5/2}}-\frac {16 b^2 \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{3003 c^4 x^{3/2}}+\frac {4 b \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{429 c^3 \sqrt {x}}-\frac {2 \sqrt {x} \left (b x+c x^2\right )^{5/2} (8 b B-13 A c)}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(32*b^3*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(15015*c^5*x^(5/2)) - (16*b^2*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/
2))/(3003*c^4*x^(3/2)) + (4*b*(8*b*B - 13*A*c)*(b*x + c*x^2)^(5/2))/(429*c^3*Sqrt[x]) - (2*(8*b*B - 13*A*c)*Sq
rt[x]*(b*x + c*x^2)^(5/2))/(143*c^2) + (2*B*x^(3/2)*(b*x + c*x^2)^(5/2))/(13*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^{3/2} (A+B x) \left (b x+c x^2\right )^{3/2} \, dx &=\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac {\left (2 \left (\frac {3}{2} (-b B+A c)+\frac {5}{2} (-b B+2 A c)\right )\right ) \int x^{3/2} \left (b x+c x^2\right )^{3/2} \, dx}{13 c}\\ &=-\frac {2 (8 b B-13 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac {(6 b (8 b B-13 A c)) \int \sqrt {x} \left (b x+c x^2\right )^{3/2} \, dx}{143 c^2}\\ &=\frac {4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt {x}}-\frac {2 (8 b B-13 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}-\frac {\left (8 b^2 (8 b B-13 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{\sqrt {x}} \, dx}{429 c^3}\\ &=-\frac {16 b^2 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac {4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt {x}}-\frac {2 (8 b B-13 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}+\frac {\left (16 b^3 (8 b B-13 A c)\right ) \int \frac {\left (b x+c x^2\right )^{3/2}}{x^{3/2}} \, dx}{3003 c^4}\\ &=\frac {32 b^3 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{15015 c^5 x^{5/2}}-\frac {16 b^2 (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{3003 c^4 x^{3/2}}+\frac {4 b (8 b B-13 A c) \left (b x+c x^2\right )^{5/2}}{429 c^3 \sqrt {x}}-\frac {2 (8 b B-13 A c) \sqrt {x} \left (b x+c x^2\right )^{5/2}}{143 c^2}+\frac {2 B x^{3/2} \left (b x+c x^2\right )^{5/2}}{13 c}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 94, normalized size = 0.55 \begin {gather*} \frac {2 (x (b+c x))^{5/2} \left (-16 b^3 c (13 A+20 B x)+40 b^2 c^2 x (13 A+14 B x)-70 b c^3 x^2 (13 A+12 B x)+105 c^4 x^3 (13 A+11 B x)+128 b^4 B\right )}{15015 c^5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(128*b^4*B + 105*c^4*x^3*(13*A + 11*B*x) - 70*b*c^3*x^2*(13*A + 12*B*x) + 40*b^2*c^2*x*
(13*A + 14*B*x) - 16*b^3*c*(13*A + 20*B*x)))/(15015*c^5*x^(5/2))

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IntegrateAlgebraic [A]  time = 0.65, size = 107, normalized size = 0.63 \begin {gather*} \frac {2 \left (b x+c x^2\right )^{5/2} \left (-208 A b^3 c+520 A b^2 c^2 x-910 A b c^3 x^2+1365 A c^4 x^3+128 b^4 B-320 b^3 B c x+560 b^2 B c^2 x^2-840 b B c^3 x^3+1155 B c^4 x^4\right )}{15015 c^5 x^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b*x + c*x^2)^(5/2)*(128*b^4*B - 208*A*b^3*c - 320*b^3*B*c*x + 520*A*b^2*c^2*x + 560*b^2*B*c^2*x^2 - 910*A*
b*c^3*x^2 - 840*b*B*c^3*x^3 + 1365*A*c^4*x^3 + 1155*B*c^4*x^4))/(15015*c^5*x^(5/2))

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fricas [A]  time = 0.41, size = 150, normalized size = 0.88 \begin {gather*} \frac {2 \, {\left (1155 \, B c^{6} x^{6} + 128 \, B b^{6} - 208 \, A b^{5} c + 105 \, {\left (14 \, B b c^{5} + 13 \, A c^{6}\right )} x^{5} + 35 \, {\left (B b^{2} c^{4} + 52 \, A b c^{5}\right )} x^{4} - 5 \, {\left (8 \, B b^{3} c^{3} - 13 \, A b^{2} c^{4}\right )} x^{3} + 6 \, {\left (8 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{2} - 8 \, {\left (8 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{15015 \, c^{5} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(1155*B*c^6*x^6 + 128*B*b^6 - 208*A*b^5*c + 105*(14*B*b*c^5 + 13*A*c^6)*x^5 + 35*(B*b^2*c^4 + 52*A*b*c
^5)*x^4 - 5*(8*B*b^3*c^3 - 13*A*b^2*c^4)*x^3 + 6*(8*B*b^4*c^2 - 13*A*b^3*c^3)*x^2 - 8*(8*B*b^5*c - 13*A*b^4*c^
2)*x)*sqrt(c*x^2 + b*x)/(c^5*sqrt(x))

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giac [B]  time = 0.29, size = 295, normalized size = 1.74 \begin {gather*} \frac {2}{9009} \, B 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} \, B 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 )} - \frac {2}{3465} \, A c {\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 )} + \frac {2}{315} \, A b {\left (\frac {16 \, b^{\frac {9}{2}}}{c^{4}} + \frac {35 \, {\left (c x + b\right )}^{\frac {9}{2}} - 135 \, {\left (c x + b\right )}^{\frac {7}{2}} b + 189 \, {\left (c x + b\right )}^{\frac {5}{2}} b^{2} - 105 \, {\left (c x + b\right )}^{\frac {3}{2}} b^{3}}{c^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9009*B*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 - 1
2870*(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*B*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) - 2/3465*A*c*(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) + 2/315*A*b*(16*b^(9/
2)/c^4 + (35*(c*x + b)^(9/2) - 135*(c*x + b)^(7/2)*b + 189*(c*x + b)^(5/2)*b^2 - 105*(c*x + b)^(3/2)*b^3)/c^4)

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maple [A]  time = 0.05, size = 107, normalized size = 0.63 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (-1155 B \,x^{4} c^{4}-1365 A \,c^{4} x^{3}+840 B b \,c^{3} x^{3}+910 A b \,c^{3} x^{2}-560 B \,b^{2} c^{2} x^{2}-520 A \,b^{2} c^{2} x +320 B \,b^{3} c x +208 A \,b^{3} c -128 b^{4} B \right ) \left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{15015 c^{5} x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15015*(c*x+b)*(-1155*B*c^4*x^4-1365*A*c^4*x^3+840*B*b*c^3*x^3+910*A*b*c^3*x^2-560*B*b^2*c^2*x^2-520*A*b^2*c
^2*x+320*B*b^3*c*x+208*A*b^3*c-128*B*b^4)*(c*x^2+b*x)^(3/2)/c^5/x^(3/2)

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maxima [A]  time = 0.59, size = 274, normalized size = 1.61 \begin {gather*} \frac {2 \, {\left ({\left (315 \, c^{5} x^{5} + 35 \, b c^{4} x^{4} - 40 \, b^{2} c^{3} x^{3} + 48 \, b^{3} c^{2} x^{2} - 64 \, b^{4} c x + 128 \, b^{5}\right )} x^{4} + 11 \, {\left (35 \, b c^{4} x^{5} + 5 \, b^{2} c^{3} x^{4} - 6 \, b^{3} c^{2} x^{3} + 8 \, b^{4} c x^{2} - 16 \, b^{5} x\right )} x^{3}\right )} \sqrt {c x + b} A}{3465 \, c^{4} x^{4}} + \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} B}{45045 \, c^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*((315*c^5*x^5 + 35*b*c^4*x^4 - 40*b^2*c^3*x^3 + 48*b^3*c^2*x^2 - 64*b^4*c*x + 128*b^5)*x^4 + 11*(35*b*c
^4*x^5 + 5*b^2*c^3*x^4 - 6*b^3*c^2*x^3 + 8*b^4*c*x^2 - 16*b^5*x)*x^3)*sqrt(c*x + b)*A/(c^4*x^4) + 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 + 1
3*(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)*B/(c^5*x^5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{3/2}\,{\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^(3/2)*(b*x + c*x^2)^(3/2)*(A + B*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(3/2)*(x*(b + c*x))**(3/2)*(A + B*x), x)

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