3.10.28 \(\int \frac {(A+B x) (a+b x+c x^2)^3}{\sqrt {x}} \, dx\)

Optimal. Leaf size=180 \[ 2 a^3 A \sqrt {x}+\frac {2}{3} a^2 x^{3/2} (a B+3 A b)+\frac {6}{11} c x^{11/2} \left (a B c+A b c+b^2 B\right )+\frac {6}{5} a x^{5/2} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{9} x^{9/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {2}{7} x^{7/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{13} c^2 x^{13/2} (A c+3 b B)+\frac {2}{15} B c^3 x^{15/2} \]

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Rubi [A]  time = 0.11, antiderivative size = 180, 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}{3} a^2 x^{3/2} (a B+3 A b)+2 a^3 A \sqrt {x}+\frac {2}{9} x^{9/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {6}{11} c x^{11/2} \left (a B c+A b c+b^2 B\right )+\frac {2}{7} x^{7/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {6}{5} a x^{5/2} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{13} c^2 x^{13/2} (A c+3 b B)+\frac {2}{15} B c^3 x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.20, size = 176, normalized size = 0.98 \begin {gather*} \frac {2 \sqrt {x} \left (15015 a^3 (3 A+B x)+1287 a^2 x (7 A (5 b+3 c x)+3 B x (7 b+5 c x))+39 a x^2 \left (11 A \left (63 b^2+90 b c x+35 c^2 x^2\right )+5 B x \left (99 b^2+154 b c x+63 c^2 x^2\right )\right )+x^3 \left (15 A \left (429 b^3+1001 b^2 c x+819 b c^2 x^2+231 c^3 x^3\right )+7 B x \left (715 b^3+1755 b^2 c x+1485 b c^2 x^2+429 c^3 x^3\right )\right )\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(15015*a^3*(3*A + B*x) + 1287*a^2*x*(7*A*(5*b + 3*c*x) + 3*B*x*(7*b + 5*c*x)) + 39*a*x^2*(11*A*(63*
b^2 + 90*b*c*x + 35*c^2*x^2) + 5*B*x*(99*b^2 + 154*b*c*x + 63*c^2*x^2)) + x^3*(15*A*(429*b^3 + 1001*b^2*c*x +
819*b*c^2*x^2 + 231*c^3*x^3) + 7*B*x*(715*b^3 + 1755*b^2*c*x + 1485*b*c^2*x^2 + 429*c^3*x^3))))/45045

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IntegrateAlgebraic [A]  time = 0.13, size = 237, normalized size = 1.32 \begin {gather*} \frac {2 \left (45045 a^3 A \sqrt {x}+15015 a^3 B x^{3/2}+45045 a^2 A b x^{3/2}+27027 a^2 A c x^{5/2}+27027 a^2 b B x^{5/2}+19305 a^2 B c x^{7/2}+27027 a A b^2 x^{5/2}+38610 a A b c x^{7/2}+15015 a A c^2 x^{9/2}+19305 a b^2 B x^{7/2}+30030 a b B c x^{9/2}+12285 a B c^2 x^{11/2}+6435 A b^3 x^{7/2}+15015 A b^2 c x^{9/2}+12285 A b c^2 x^{11/2}+3465 A c^3 x^{13/2}+5005 b^3 B x^{9/2}+12285 b^2 B c x^{11/2}+10395 b B c^2 x^{13/2}+3003 B c^3 x^{15/2}\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(45045*a^3*A*Sqrt[x] + 45045*a^2*A*b*x^(3/2) + 15015*a^3*B*x^(3/2) + 27027*a*A*b^2*x^(5/2) + 27027*a^2*b*B*
x^(5/2) + 27027*a^2*A*c*x^(5/2) + 6435*A*b^3*x^(7/2) + 19305*a*b^2*B*x^(7/2) + 38610*a*A*b*c*x^(7/2) + 19305*a
^2*B*c*x^(7/2) + 5005*b^3*B*x^(9/2) + 15015*A*b^2*c*x^(9/2) + 30030*a*b*B*c*x^(9/2) + 15015*a*A*c^2*x^(9/2) +
12285*b^2*B*c*x^(11/2) + 12285*A*b*c^2*x^(11/2) + 12285*a*B*c^2*x^(11/2) + 10395*b*B*c^2*x^(13/2) + 3465*A*c^3
*x^(13/2) + 3003*B*c^3*x^(15/2)))/45045

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fricas [A]  time = 0.44, size = 166, normalized size = 0.92 \begin {gather*} \frac {2}{45045} \, {\left (3003 \, B c^{3} x^{7} + 3465 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 12285 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 5005 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} + 45045 \, A a^{3} + 6435 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 27027 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 15015 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(1/2),x, algorithm="fricas")

[Out]

2/45045*(3003*B*c^3*x^7 + 3465*(3*B*b*c^2 + A*c^3)*x^6 + 12285*(B*b^2*c + (B*a + A*b)*c^2)*x^5 + 5005*(B*b^3 +
3*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 + 45045*A*a^3 + 6435*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 +
27027*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 15015*(B*a^3 + 3*A*a^2*b)*x)*sqrt(x)

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giac [A]  time = 0.17, size = 193, normalized size = 1.07 \begin {gather*} \frac {2}{15} \, B c^{3} x^{\frac {15}{2}} + \frac {6}{13} \, B b c^{2} x^{\frac {13}{2}} + \frac {2}{13} \, A c^{3} x^{\frac {13}{2}} + \frac {6}{11} \, B b^{2} c x^{\frac {11}{2}} + \frac {6}{11} \, B a c^{2} x^{\frac {11}{2}} + \frac {6}{11} \, A b c^{2} x^{\frac {11}{2}} + \frac {2}{9} \, B b^{3} x^{\frac {9}{2}} + \frac {4}{3} \, B a b c x^{\frac {9}{2}} + \frac {2}{3} \, A b^{2} c x^{\frac {9}{2}} + \frac {2}{3} \, A a c^{2} x^{\frac {9}{2}} + \frac {6}{7} \, B a b^{2} x^{\frac {7}{2}} + \frac {2}{7} \, A b^{3} x^{\frac {7}{2}} + \frac {6}{7} \, B a^{2} c x^{\frac {7}{2}} + \frac {12}{7} \, A a b c x^{\frac {7}{2}} + \frac {6}{5} \, B a^{2} b x^{\frac {5}{2}} + \frac {6}{5} \, A a b^{2} x^{\frac {5}{2}} + \frac {6}{5} \, A a^{2} c x^{\frac {5}{2}} + \frac {2}{3} \, B a^{3} x^{\frac {3}{2}} + 2 \, A a^{2} b x^{\frac {3}{2}} + 2 \, A a^{3} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(1/2),x, algorithm="giac")

[Out]

2/15*B*c^3*x^(15/2) + 6/13*B*b*c^2*x^(13/2) + 2/13*A*c^3*x^(13/2) + 6/11*B*b^2*c*x^(11/2) + 6/11*B*a*c^2*x^(11
/2) + 6/11*A*b*c^2*x^(11/2) + 2/9*B*b^3*x^(9/2) + 4/3*B*a*b*c*x^(9/2) + 2/3*A*b^2*c*x^(9/2) + 2/3*A*a*c^2*x^(9
/2) + 6/7*B*a*b^2*x^(7/2) + 2/7*A*b^3*x^(7/2) + 6/7*B*a^2*c*x^(7/2) + 12/7*A*a*b*c*x^(7/2) + 6/5*B*a^2*b*x^(5/
2) + 6/5*A*a*b^2*x^(5/2) + 6/5*A*a^2*c*x^(5/2) + 2/3*B*a^3*x^(3/2) + 2*A*a^2*b*x^(3/2) + 2*A*a^3*sqrt(x)

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maple [A]  time = 0.05, size = 192, normalized size = 1.07 \begin {gather*} \frac {2 \left (3003 B \,c^{3} x^{7}+3465 A \,c^{3} x^{6}+10395 x^{6} B b \,c^{2}+12285 x^{5} A b \,c^{2}+12285 B a \,c^{2} x^{5}+12285 x^{5} B \,b^{2} c +15015 A a \,c^{2} x^{4}+15015 x^{4} A \,b^{2} c +30030 x^{4} a b B c +5005 x^{4} b^{3} B +38610 x^{3} A a b c +6435 A \,b^{3} x^{3}+19305 B \,a^{2} c \,x^{3}+19305 x^{3} B a \,b^{2}+27027 A \,a^{2} c \,x^{2}+27027 x^{2} A a \,b^{2}+27027 B \,a^{2} b \,x^{2}+45045 x A \,a^{2} b +15015 B \,a^{3} x +45045 A \,a^{3}\right ) \sqrt {x}}{45045} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^3/x^(1/2),x)

[Out]

2/45045*x^(1/2)*(3003*B*c^3*x^7+3465*A*c^3*x^6+10395*B*b*c^2*x^6+12285*A*b*c^2*x^5+12285*B*a*c^2*x^5+12285*B*b
^2*c*x^5+15015*A*a*c^2*x^4+15015*A*b^2*c*x^4+30030*B*a*b*c*x^4+5005*B*b^3*x^4+38610*A*a*b*c*x^3+6435*A*b^3*x^3
+19305*B*a^2*c*x^3+19305*B*a*b^2*x^3+27027*A*a^2*c*x^2+27027*A*a*b^2*x^2+27027*B*a^2*b*x^2+45045*A*a^2*b*x+150
15*B*a^3*x+45045*A*a^3)

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maxima [A]  time = 0.56, size = 166, normalized size = 0.92 \begin {gather*} \frac {2}{15} \, B c^{3} x^{\frac {15}{2}} + \frac {2}{13} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {13}{2}} + \frac {6}{11} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {9}{2}} + 2 \, A a^{3} \sqrt {x} + \frac {2}{7} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{\frac {7}{2}} + \frac {6}{5} \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 169, normalized size = 0.94 \begin {gather*} x^{7/2}\,\left (\frac {6\,B\,c\,a^2}{7}+\frac {6\,B\,a\,b^2}{7}+\frac {12\,A\,c\,a\,b}{7}+\frac {2\,A\,b^3}{7}\right )+x^{9/2}\,\left (\frac {2\,B\,b^3}{9}+\frac {2\,A\,b^2\,c}{3}+\frac {4\,B\,a\,b\,c}{3}+\frac {2\,A\,a\,c^2}{3}\right )+x^{3/2}\,\left (\frac {2\,B\,a^3}{3}+2\,A\,b\,a^2\right )+x^{13/2}\,\left (\frac {2\,A\,c^3}{13}+\frac {6\,B\,b\,c^2}{13}\right )+x^{5/2}\,\left (\frac {6\,B\,a^2\,b}{5}+\frac {6\,A\,c\,a^2}{5}+\frac {6\,A\,a\,b^2}{5}\right )+x^{11/2}\,\left (\frac {6\,B\,b^2\,c}{11}+\frac {6\,A\,b\,c^2}{11}+\frac {6\,B\,a\,c^2}{11}\right )+2\,A\,a^3\,\sqrt {x}+\frac {2\,B\,c^3\,x^{15/2}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^3)/x^(1/2),x)

[Out]

x^(7/2)*((2*A*b^3)/7 + (6*B*a*b^2)/7 + (6*B*a^2*c)/7 + (12*A*a*b*c)/7) + x^(9/2)*((2*B*b^3)/9 + (2*A*a*c^2)/3
+ (2*A*b^2*c)/3 + (4*B*a*b*c)/3) + x^(3/2)*((2*B*a^3)/3 + 2*A*a^2*b) + x^(13/2)*((2*A*c^3)/13 + (6*B*b*c^2)/13
) + x^(5/2)*((6*A*a*b^2)/5 + (6*A*a^2*c)/5 + (6*B*a^2*b)/5) + x^(11/2)*((6*A*b*c^2)/11 + (6*B*a*c^2)/11 + (6*B
*b^2*c)/11) + 2*A*a^3*x^(1/2) + (2*B*c^3*x^(15/2))/15

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sympy [A]  time = 5.90, size = 291, normalized size = 1.62 \begin {gather*} 2 A a^{3} \sqrt {x} + 2 A a^{2} b x^{\frac {3}{2}} + \frac {6 A a^{2} c x^{\frac {5}{2}}}{5} + \frac {6 A a b^{2} x^{\frac {5}{2}}}{5} + \frac {12 A a b c x^{\frac {7}{2}}}{7} + \frac {2 A a c^{2} x^{\frac {9}{2}}}{3} + \frac {2 A b^{3} x^{\frac {7}{2}}}{7} + \frac {2 A b^{2} c x^{\frac {9}{2}}}{3} + \frac {6 A b c^{2} x^{\frac {11}{2}}}{11} + \frac {2 A c^{3} x^{\frac {13}{2}}}{13} + \frac {2 B a^{3} x^{\frac {3}{2}}}{3} + \frac {6 B a^{2} b x^{\frac {5}{2}}}{5} + \frac {6 B a^{2} c x^{\frac {7}{2}}}{7} + \frac {6 B a b^{2} x^{\frac {7}{2}}}{7} + \frac {4 B a b c x^{\frac {9}{2}}}{3} + \frac {6 B a c^{2} x^{\frac {11}{2}}}{11} + \frac {2 B b^{3} x^{\frac {9}{2}}}{9} + \frac {6 B b^{2} c x^{\frac {11}{2}}}{11} + \frac {6 B b c^{2} x^{\frac {13}{2}}}{13} + \frac {2 B c^{3} x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**3/x**(1/2),x)

[Out]

2*A*a**3*sqrt(x) + 2*A*a**2*b*x**(3/2) + 6*A*a**2*c*x**(5/2)/5 + 6*A*a*b**2*x**(5/2)/5 + 12*A*a*b*c*x**(7/2)/7
+ 2*A*a*c**2*x**(9/2)/3 + 2*A*b**3*x**(7/2)/7 + 2*A*b**2*c*x**(9/2)/3 + 6*A*b*c**2*x**(11/2)/11 + 2*A*c**3*x*
*(13/2)/13 + 2*B*a**3*x**(3/2)/3 + 6*B*a**2*b*x**(5/2)/5 + 6*B*a**2*c*x**(7/2)/7 + 6*B*a*b**2*x**(7/2)/7 + 4*B
*a*b*c*x**(9/2)/3 + 6*B*a*c**2*x**(11/2)/11 + 2*B*b**3*x**(9/2)/9 + 6*B*b**2*c*x**(11/2)/11 + 6*B*b*c**2*x**(1
3/2)/13 + 2*B*c**3*x**(15/2)/15

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