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

Optimal. Leaf size=182 \[ \frac {2}{5} a^3 A x^{5/2}+\frac {2}{7} a^2 x^{7/2} (a B+3 A b)+\frac {2}{5} c x^{15/2} \left (a B c+A b c+b^2 B\right )+\frac {2}{3} a x^{9/2} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{13} x^{13/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {2}{11} x^{11/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{17} c^2 x^{17/2} (A c+3 b B)+\frac {2}{19} B c^3 x^{19/2} \]

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Rubi [A]  time = 0.11, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {765} \begin {gather*} \frac {2}{7} a^2 x^{7/2} (a B+3 A b)+\frac {2}{5} a^3 A x^{5/2}+\frac {2}{13} x^{13/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {2}{5} c x^{15/2} \left (a B c+A b c+b^2 B\right )+\frac {2}{11} x^{11/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{3} a x^{9/2} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{17} c^2 x^{17/2} (A c+3 b B)+\frac {2}{19} B c^3 x^{19/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*A*x^(5/2))/5 + (2*a^2*(3*A*b + a*B)*x^(7/2))/7 + (2*a*(a*b*B + A*(b^2 + a*c))*x^(9/2))/3 + (2*(3*a*B*(b
^2 + a*c) + A*(b^3 + 6*a*b*c))*x^(11/2))/11 + (2*(b^3*B + 3*A*b^2*c + 6*a*b*B*c + 3*a*A*c^2)*x^(13/2))/13 + (2
*c*(b^2*B + A*b*c + a*B*c)*x^(15/2))/5 + (2*c^2*(3*b*B + A*c)*x^(17/2))/17 + (2*B*c^3*x^(19/2))/19

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int x^{3/2} (A+B x) \left (a+b x+c x^2\right )^3 \, dx &=\int \left (a^3 A x^{3/2}+a^2 (3 A b+a B) x^{5/2}+3 a \left (a b B+A \left (b^2+a c\right )\right ) x^{7/2}+\left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^{9/2}+\left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{11/2}+3 c \left (b^2 B+A b c+a B c\right ) x^{13/2}+c^2 (3 b B+A c) x^{15/2}+B c^3 x^{17/2}\right ) \, dx\\ &=\frac {2}{5} a^3 A x^{5/2}+\frac {2}{7} a^2 (3 A b+a B) x^{7/2}+\frac {2}{3} a \left (a b B+A \left (b^2+a c\right )\right ) x^{9/2}+\frac {2}{11} \left (3 a B \left (b^2+a c\right )+A \left (b^3+6 a b c\right )\right ) x^{11/2}+\frac {2}{13} \left (b^3 B+3 A b^2 c+6 a b B c+3 a A c^2\right ) x^{13/2}+\frac {2}{5} c \left (b^2 B+A b c+a B c\right ) x^{15/2}+\frac {2}{17} c^2 (3 b B+A c) x^{17/2}+\frac {2}{19} B c^3 x^{19/2}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 178, normalized size = 0.98 \begin {gather*} \frac {2 x^{5/2} \left (138567 a^3 (7 A+5 B x)+20995 a^2 x (11 A (9 b+7 c x)+7 B x (11 b+9 c x))+2261 a x^2 \left (5 A \left (143 b^2+234 b c x+99 c^2 x^2\right )+3 B x \left (195 b^2+330 b c x+143 c^2 x^2\right )\right )+21 x^3 \left (19 A \left (1105 b^3+2805 b^2 c x+2431 b c^2 x^2+715 c^3 x^3\right )+11 B x \left (1615 b^3+4199 b^2 c x+3705 b c^2 x^2+1105 c^3 x^3\right )\right )\right )}{4849845} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(138567*a^3*(7*A + 5*B*x) + 20995*a^2*x*(11*A*(9*b + 7*c*x) + 7*B*x*(11*b + 9*c*x)) + 2261*a*x^2*(5
*A*(143*b^2 + 234*b*c*x + 99*c^2*x^2) + 3*B*x*(195*b^2 + 330*b*c*x + 143*c^2*x^2)) + 21*x^3*(19*A*(1105*b^3 +
2805*b^2*c*x + 2431*b*c^2*x^2 + 715*c^3*x^3) + 11*B*x*(1615*b^3 + 4199*b^2*c*x + 3705*b*c^2*x^2 + 1105*c^3*x^3
))))/4849845

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IntegrateAlgebraic [A]  time = 0.10, size = 237, normalized size = 1.30 \begin {gather*} \frac {2 \left (969969 a^3 A x^{5/2}+692835 a^3 B x^{7/2}+2078505 a^2 A b x^{7/2}+1616615 a^2 A c x^{9/2}+1616615 a^2 b B x^{9/2}+1322685 a^2 B c x^{11/2}+1616615 a A b^2 x^{9/2}+2645370 a A b c x^{11/2}+1119195 a A c^2 x^{13/2}+1322685 a b^2 B x^{11/2}+2238390 a b B c x^{13/2}+969969 a B c^2 x^{15/2}+440895 A b^3 x^{11/2}+1119195 A b^2 c x^{13/2}+969969 A b c^2 x^{15/2}+285285 A c^3 x^{17/2}+373065 b^3 B x^{13/2}+969969 b^2 B c x^{15/2}+855855 b B c^2 x^{17/2}+255255 B c^3 x^{19/2}\right )}{4849845} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(969969*a^3*A*x^(5/2) + 2078505*a^2*A*b*x^(7/2) + 692835*a^3*B*x^(7/2) + 1616615*a*A*b^2*x^(9/2) + 1616615*
a^2*b*B*x^(9/2) + 1616615*a^2*A*c*x^(9/2) + 440895*A*b^3*x^(11/2) + 1322685*a*b^2*B*x^(11/2) + 2645370*a*A*b*c
*x^(11/2) + 1322685*a^2*B*c*x^(11/2) + 373065*b^3*B*x^(13/2) + 1119195*A*b^2*c*x^(13/2) + 2238390*a*b*B*c*x^(1
3/2) + 1119195*a*A*c^2*x^(13/2) + 969969*b^2*B*c*x^(15/2) + 969969*A*b*c^2*x^(15/2) + 969969*a*B*c^2*x^(15/2)
+ 855855*b*B*c^2*x^(17/2) + 285285*A*c^3*x^(17/2) + 255255*B*c^3*x^(19/2)))/4849845

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fricas [A]  time = 0.44, size = 171, normalized size = 0.94 \begin {gather*} \frac {2}{4849845} \, {\left (255255 \, B c^{3} x^{9} + 285285 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{8} + 969969 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{7} + 373065 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{6} + 969969 \, A a^{3} x^{2} + 440895 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{5} + 1616615 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{4} + 692835 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{3}\right )} \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+a)^3,x, algorithm="fricas")

[Out]

2/4849845*(255255*B*c^3*x^9 + 285285*(3*B*b*c^2 + A*c^3)*x^8 + 969969*(B*b^2*c + (B*a + A*b)*c^2)*x^7 + 373065
*(B*b^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^6 + 969969*A*a^3*x^2 + 440895*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2
*A*a*b)*c)*x^5 + 1616615*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^4 + 692835*(B*a^3 + 3*A*a^2*b)*x^3)*sqrt(x)

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giac [A]  time = 0.18, size = 193, normalized size = 1.06 \begin {gather*} \frac {2}{19} \, B c^{3} x^{\frac {19}{2}} + \frac {6}{17} \, B b c^{2} x^{\frac {17}{2}} + \frac {2}{17} \, A c^{3} x^{\frac {17}{2}} + \frac {2}{5} \, B b^{2} c x^{\frac {15}{2}} + \frac {2}{5} \, B a c^{2} x^{\frac {15}{2}} + \frac {2}{5} \, A b c^{2} x^{\frac {15}{2}} + \frac {2}{13} \, B b^{3} x^{\frac {13}{2}} + \frac {12}{13} \, B a b c x^{\frac {13}{2}} + \frac {6}{13} \, A b^{2} c x^{\frac {13}{2}} + \frac {6}{13} \, A a c^{2} x^{\frac {13}{2}} + \frac {6}{11} \, B a b^{2} x^{\frac {11}{2}} + \frac {2}{11} \, A b^{3} x^{\frac {11}{2}} + \frac {6}{11} \, B a^{2} c x^{\frac {11}{2}} + \frac {12}{11} \, A a b c x^{\frac {11}{2}} + \frac {2}{3} \, B a^{2} b x^{\frac {9}{2}} + \frac {2}{3} \, A a b^{2} x^{\frac {9}{2}} + \frac {2}{3} \, A a^{2} c x^{\frac {9}{2}} + \frac {2}{7} \, B a^{3} x^{\frac {7}{2}} + \frac {6}{7} \, A a^{2} b x^{\frac {7}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} \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+a)^3,x, algorithm="giac")

[Out]

2/19*B*c^3*x^(19/2) + 6/17*B*b*c^2*x^(17/2) + 2/17*A*c^3*x^(17/2) + 2/5*B*b^2*c*x^(15/2) + 2/5*B*a*c^2*x^(15/2
) + 2/5*A*b*c^2*x^(15/2) + 2/13*B*b^3*x^(13/2) + 12/13*B*a*b*c*x^(13/2) + 6/13*A*b^2*c*x^(13/2) + 6/13*A*a*c^2
*x^(13/2) + 6/11*B*a*b^2*x^(11/2) + 2/11*A*b^3*x^(11/2) + 6/11*B*a^2*c*x^(11/2) + 12/11*A*a*b*c*x^(11/2) + 2/3
*B*a^2*b*x^(9/2) + 2/3*A*a*b^2*x^(9/2) + 2/3*A*a^2*c*x^(9/2) + 2/7*B*a^3*x^(7/2) + 6/7*A*a^2*b*x^(7/2) + 2/5*A
*a^3*x^(5/2)

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maple [A]  time = 0.05, size = 192, normalized size = 1.05 \begin {gather*} \frac {2 \left (255255 B \,c^{3} x^{7}+285285 A \,c^{3} x^{6}+855855 x^{6} B b \,c^{2}+969969 x^{5} A b \,c^{2}+969969 B a \,c^{2} x^{5}+969969 x^{5} B \,b^{2} c +1119195 A a \,c^{2} x^{4}+1119195 x^{4} A \,b^{2} c +2238390 x^{4} a b B c +373065 x^{4} b^{3} B +2645370 x^{3} A a b c +440895 A \,b^{3} x^{3}+1322685 B \,a^{2} c \,x^{3}+1322685 x^{3} B a \,b^{2}+1616615 A \,a^{2} c \,x^{2}+1616615 x^{2} A a \,b^{2}+1616615 B \,a^{2} b \,x^{2}+2078505 x A \,a^{2} b +692835 B \,a^{3} x +969969 A \,a^{3}\right ) x^{\frac {5}{2}}}{4849845} \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+a)^3,x)

[Out]

2/4849845*x^(5/2)*(255255*B*c^3*x^7+285285*A*c^3*x^6+855855*B*b*c^2*x^6+969969*A*b*c^2*x^5+969969*B*a*c^2*x^5+
969969*B*b^2*c*x^5+1119195*A*a*c^2*x^4+1119195*A*b^2*c*x^4+2238390*B*a*b*c*x^4+373065*B*b^3*x^4+2645370*A*a*b*
c*x^3+440895*A*b^3*x^3+1322685*B*a^2*c*x^3+1322685*B*a*b^2*x^3+1616615*A*a^2*c*x^2+1616615*A*a*b^2*x^2+1616615
*B*a^2*b*x^2+2078505*A*a^2*b*x+692835*B*a^3*x+969969*A*a^3)

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maxima [A]  time = 0.66, size = 166, normalized size = 0.91 \begin {gather*} \frac {2}{19} \, B c^{3} x^{\frac {19}{2}} + \frac {2}{17} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {17}{2}} + \frac {2}{5} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {15}{2}} + \frac {2}{13} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {13}{2}} + \frac {2}{5} \, A a^{3} x^{\frac {5}{2}} + \frac {2}{11} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{\frac {11}{2}} + \frac {2}{3} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {7}{2}} \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+a)^3,x, algorithm="maxima")

[Out]

2/19*B*c^3*x^(19/2) + 2/17*(3*B*b*c^2 + A*c^3)*x^(17/2) + 2/5*(B*b^2*c + (B*a + A*b)*c^2)*x^(15/2) + 2/13*(B*b
^3 + 3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^(13/2) + 2/5*A*a^3*x^(5/2) + 2/11*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*
A*a*b)*c)*x^(11/2) + 2/3*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^(9/2) + 2/7*(B*a^3 + 3*A*a^2*b)*x^(7/2)

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mupad [B]  time = 0.05, size = 169, normalized size = 0.93 \begin {gather*} x^{11/2}\,\left (\frac {6\,B\,c\,a^2}{11}+\frac {6\,B\,a\,b^2}{11}+\frac {12\,A\,c\,a\,b}{11}+\frac {2\,A\,b^3}{11}\right )+x^{13/2}\,\left (\frac {2\,B\,b^3}{13}+\frac {6\,A\,b^2\,c}{13}+\frac {12\,B\,a\,b\,c}{13}+\frac {6\,A\,a\,c^2}{13}\right )+x^{7/2}\,\left (\frac {2\,B\,a^3}{7}+\frac {6\,A\,b\,a^2}{7}\right )+x^{17/2}\,\left (\frac {2\,A\,c^3}{17}+\frac {6\,B\,b\,c^2}{17}\right )+x^{9/2}\,\left (\frac {2\,B\,a^2\,b}{3}+\frac {2\,A\,c\,a^2}{3}+\frac {2\,A\,a\,b^2}{3}\right )+x^{15/2}\,\left (\frac {2\,B\,b^2\,c}{5}+\frac {2\,A\,b\,c^2}{5}+\frac {2\,B\,a\,c^2}{5}\right )+\frac {2\,A\,a^3\,x^{5/2}}{5}+\frac {2\,B\,c^3\,x^{19/2}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(11/2)*((2*A*b^3)/11 + (6*B*a*b^2)/11 + (6*B*a^2*c)/11 + (12*A*a*b*c)/11) + x^(13/2)*((2*B*b^3)/13 + (6*A*a*
c^2)/13 + (6*A*b^2*c)/13 + (12*B*a*b*c)/13) + x^(7/2)*((2*B*a^3)/7 + (6*A*a^2*b)/7) + x^(17/2)*((2*A*c^3)/17 +
(6*B*b*c^2)/17) + x^(9/2)*((2*A*a*b^2)/3 + (2*A*a^2*c)/3 + (2*B*a^2*b)/3) + x^(15/2)*((2*A*b*c^2)/5 + (2*B*a*
c^2)/5 + (2*B*b^2*c)/5) + (2*A*a^3*x^(5/2))/5 + (2*B*c^3*x^(19/2))/19

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sympy [A]  time = 9.59, size = 294, normalized size = 1.62 \begin {gather*} \frac {2 A a^{3} x^{\frac {5}{2}}}{5} + \frac {6 A a^{2} b x^{\frac {7}{2}}}{7} + \frac {2 A a^{2} c x^{\frac {9}{2}}}{3} + \frac {2 A a b^{2} x^{\frac {9}{2}}}{3} + \frac {12 A a b c x^{\frac {11}{2}}}{11} + \frac {6 A a c^{2} x^{\frac {13}{2}}}{13} + \frac {2 A b^{3} x^{\frac {11}{2}}}{11} + \frac {6 A b^{2} c x^{\frac {13}{2}}}{13} + \frac {2 A b c^{2} x^{\frac {15}{2}}}{5} + \frac {2 A c^{3} x^{\frac {17}{2}}}{17} + \frac {2 B a^{3} x^{\frac {7}{2}}}{7} + \frac {2 B a^{2} b x^{\frac {9}{2}}}{3} + \frac {6 B a^{2} c x^{\frac {11}{2}}}{11} + \frac {6 B a b^{2} x^{\frac {11}{2}}}{11} + \frac {12 B a b c x^{\frac {13}{2}}}{13} + \frac {2 B a c^{2} x^{\frac {15}{2}}}{5} + \frac {2 B b^{3} x^{\frac {13}{2}}}{13} + \frac {2 B b^{2} c x^{\frac {15}{2}}}{5} + \frac {6 B b c^{2} x^{\frac {17}{2}}}{17} + \frac {2 B c^{3} x^{\frac {19}{2}}}{19} \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+a)**3,x)

[Out]

2*A*a**3*x**(5/2)/5 + 6*A*a**2*b*x**(7/2)/7 + 2*A*a**2*c*x**(9/2)/3 + 2*A*a*b**2*x**(9/2)/3 + 12*A*a*b*c*x**(1
1/2)/11 + 6*A*a*c**2*x**(13/2)/13 + 2*A*b**3*x**(11/2)/11 + 6*A*b**2*c*x**(13/2)/13 + 2*A*b*c**2*x**(15/2)/5 +
2*A*c**3*x**(17/2)/17 + 2*B*a**3*x**(7/2)/7 + 2*B*a**2*b*x**(9/2)/3 + 6*B*a**2*c*x**(11/2)/11 + 6*B*a*b**2*x*
*(11/2)/11 + 12*B*a*b*c*x**(13/2)/13 + 2*B*a*c**2*x**(15/2)/5 + 2*B*b**3*x**(13/2)/13 + 2*B*b**2*c*x**(15/2)/5
+ 6*B*b*c**2*x**(17/2)/17 + 2*B*c**3*x**(19/2)/19

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