3.1005 \(\int \frac {(A+B x) (a+b x+c x^2)^3}{x^{5/2}} \, dx\)

Optimal. Leaf size=178 \[ -\frac {2 a^3 A}{3 x^{3/2}}-\frac {2 a^2 (a B+3 A b)}{\sqrt {x}}+\frac {6}{7} c x^{7/2} \left (a B c+A b c+b^2 B\right )+6 a \sqrt {x} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{5} x^{5/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {2}{3} x^{3/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+\frac {2}{9} c^2 x^{9/2} (A c+3 b B)+\frac {2}{11} B c^3 x^{11/2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 178, 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} \[ -\frac {2 a^2 (a B+3 A b)}{\sqrt {x}}-\frac {2 a^3 A}{3 x^{3/2}}+\frac {2}{5} x^{5/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac {6}{7} c x^{7/2} \left (a B c+A b c+b^2 B\right )+\frac {2}{3} x^{3/2} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )+6 a \sqrt {x} \left (A \left (a c+b^2\right )+a b B\right )+\frac {2}{9} c^2 x^{9/2} (A c+3 b B)+\frac {2}{11} B c^3 x^{11/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 170, normalized size = 0.96 \[ \frac {2 \left (-1155 a^3 (A+3 B x)+3465 a^2 x (B x (3 b+c x)-3 A (b-c x))+99 a x^2 \left (7 A \left (15 b^2+10 b c x+3 c^2 x^2\right )+B x \left (35 b^2+42 b c x+15 c^2 x^2\right )\right )+x^3 \left (11 A \left (105 b^3+189 b^2 c x+135 b c^2 x^2+35 c^3 x^3\right )+3 B x \left (231 b^3+495 b^2 c x+385 b c^2 x^2+105 c^3 x^3\right )\right )\right )}{3465 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1155*a^3*(A + 3*B*x) + 3465*a^2*x*(-3*A*(b - c*x) + B*x*(3*b + c*x)) + 99*a*x^2*(7*A*(15*b^2 + 10*b*c*x +
 3*c^2*x^2) + B*x*(35*b^2 + 42*b*c*x + 15*c^2*x^2)) + x^3*(11*A*(105*b^3 + 189*b^2*c*x + 135*b*c^2*x^2 + 35*c^
3*x^3) + 3*B*x*(231*b^3 + 495*b^2*c*x + 385*b*c^2*x^2 + 105*c^3*x^3))))/(3465*x^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 166, normalized size = 0.93 \[ \frac {2 \, {\left (315 \, B c^{3} x^{7} + 385 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 1485 \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{5} + 693 \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 1155 \, A a^{3} + 1155 \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} + 10395 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 3465 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{3465 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3465*(315*B*c^3*x^7 + 385*(3*B*b*c^2 + A*c^3)*x^6 + 1485*(B*b^2*c + (B*a + A*b)*c^2)*x^5 + 693*(B*b^3 + 3*A*
a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 1155*A*a^3 + 1155*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 + 10395
*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 - 3465*(B*a^3 + 3*A*a^2*b)*x)/x^(3/2)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 191, normalized size = 1.07 \[ \frac {2}{11} \, B c^{3} x^{\frac {11}{2}} + \frac {2}{3} \, B b c^{2} x^{\frac {9}{2}} + \frac {2}{9} \, A c^{3} x^{\frac {9}{2}} + \frac {6}{7} \, B b^{2} c x^{\frac {7}{2}} + \frac {6}{7} \, B a c^{2} x^{\frac {7}{2}} + \frac {6}{7} \, A b c^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B b^{3} x^{\frac {5}{2}} + \frac {12}{5} \, B a b c x^{\frac {5}{2}} + \frac {6}{5} \, A b^{2} c x^{\frac {5}{2}} + \frac {6}{5} \, A a c^{2} x^{\frac {5}{2}} + 2 \, B a b^{2} x^{\frac {3}{2}} + \frac {2}{3} \, A b^{3} x^{\frac {3}{2}} + 2 \, B a^{2} c x^{\frac {3}{2}} + 4 \, A a b c x^{\frac {3}{2}} + 6 \, B a^{2} b \sqrt {x} + 6 \, A a b^{2} \sqrt {x} + 6 \, A a^{2} c \sqrt {x} - \frac {2 \, {\left (3 \, B a^{3} x + 9 \, A a^{2} b x + A a^{3}\right )}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 192, normalized size = 1.08 \[ -\frac {2 \left (-315 B \,c^{3} x^{7}-385 A \,c^{3} x^{6}-1155 x^{6} B b \,c^{2}-1485 x^{5} A b \,c^{2}-1485 B a \,c^{2} x^{5}-1485 x^{5} B \,b^{2} c -2079 A a \,c^{2} x^{4}-2079 x^{4} A \,b^{2} c -4158 x^{4} a b B c -693 x^{4} b^{3} B -6930 x^{3} A a b c -1155 A \,b^{3} x^{3}-3465 B \,a^{2} c \,x^{3}-3465 x^{3} B a \,b^{2}-10395 A \,a^{2} c \,x^{2}-10395 x^{2} A a \,b^{2}-10395 B \,a^{2} b \,x^{2}+10395 x A \,a^{2} b +3465 B \,a^{3} x +1155 A \,a^{3}\right )}{3465 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(-315*B*c^3*x^7-385*A*c^3*x^6-1155*B*b*c^2*x^6-1485*A*b*c^2*x^5-1485*B*a*c^2*x^5-1485*B*b^2*c*x^5-2079
*A*a*c^2*x^4-2079*A*b^2*c*x^4-4158*B*a*b*c*x^4-693*B*b^3*x^4-6930*A*a*b*c*x^3-1155*A*b^3*x^3-3465*B*a^2*c*x^3-
3465*B*a*b^2*x^3-10395*A*a^2*c*x^2-10395*A*a*b^2*x^2-10395*B*a^2*b*x^2+10395*A*a^2*b*x+3465*B*a^3*x+1155*A*a^3
)/x^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.57, size = 166, normalized size = 0.93 \[ \frac {2}{11} \, B c^{3} x^{\frac {11}{2}} + \frac {2}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac {9}{2}} + \frac {6}{7} \, {\left (B b^{2} c + {\left (B a + A b\right )} c^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B b^{3} + 3 \, A a c^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (3 \, B a b^{2} + A b^{3} + 3 \, {\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{\frac {3}{2}} + 6 \, {\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} \sqrt {x} - \frac {2 \, {\left (A a^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 170, normalized size = 0.96 \[ x^{3/2}\,\left (2\,B\,c\,a^2+2\,B\,a\,b^2+4\,A\,c\,a\,b+\frac {2\,A\,b^3}{3}\right )+x^{5/2}\,\left (\frac {2\,B\,b^3}{5}+\frac {6\,A\,b^2\,c}{5}+\frac {12\,B\,a\,b\,c}{5}+\frac {6\,A\,a\,c^2}{5}\right )-\frac {x\,\left (2\,B\,a^3+6\,A\,b\,a^2\right )+\frac {2\,A\,a^3}{3}}{x^{3/2}}+x^{9/2}\,\left (\frac {2\,A\,c^3}{9}+\frac {2\,B\,b\,c^2}{3}\right )+\sqrt {x}\,\left (6\,B\,a^2\,b+6\,A\,c\,a^2+6\,A\,a\,b^2\right )+x^{7/2}\,\left (\frac {6\,B\,b^2\,c}{7}+\frac {6\,A\,b\,c^2}{7}+\frac {6\,B\,a\,c^2}{7}\right )+\frac {2\,B\,c^3\,x^{11/2}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 7.05, size = 280, normalized size = 1.57 \[ - \frac {2 A a^{3}}{3 x^{\frac {3}{2}}} - \frac {6 A a^{2} b}{\sqrt {x}} + 6 A a^{2} c \sqrt {x} + 6 A a b^{2} \sqrt {x} + 4 A a b c x^{\frac {3}{2}} + \frac {6 A a c^{2} x^{\frac {5}{2}}}{5} + \frac {2 A b^{3} x^{\frac {3}{2}}}{3} + \frac {6 A b^{2} c x^{\frac {5}{2}}}{5} + \frac {6 A b c^{2} x^{\frac {7}{2}}}{7} + \frac {2 A c^{3} x^{\frac {9}{2}}}{9} - \frac {2 B a^{3}}{\sqrt {x}} + 6 B a^{2} b \sqrt {x} + 2 B a^{2} c x^{\frac {3}{2}} + 2 B a b^{2} x^{\frac {3}{2}} + \frac {12 B a b c x^{\frac {5}{2}}}{5} + \frac {6 B a c^{2} x^{\frac {7}{2}}}{7} + \frac {2 B b^{3} x^{\frac {5}{2}}}{5} + \frac {6 B b^{2} c x^{\frac {7}{2}}}{7} + \frac {2 B b c^{2} x^{\frac {9}{2}}}{3} + \frac {2 B c^{3} x^{\frac {11}{2}}}{11} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**3/x**(5/2),x)

[Out]

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

________________________________________________________________________________________