3.1006 \(\int \frac{(A+B x) (a+b x+c x^2)^3}{x^{7/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.115129, antiderivative size = 178, normalized size of antiderivative = 1., 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} \[ -\frac{2 a^2 (a B+3 A b)}{3 x^{3/2}}-\frac{2 a^3 A}{5 x^{5/2}}+\frac{2}{3} x^{3/2} \left (3 a A c^2+6 a b B c+3 A b^2 c+b^3 B\right )+\frac{6}{5} c x^{5/2} \left (a B c+A b c+b^2 B\right )+2 \sqrt{x} \left (A \left (6 a b c+b^3\right )+3 a B \left (a c+b^2\right )\right )-\frac{6 a \left (A \left (a c+b^2\right )+a b B\right )}{\sqrt{x}}+\frac{2}{7} c^2 x^{7/2} (A c+3 b B)+\frac{2}{9} B c^3 x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^3)/x^(7/2),x]

[Out]

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

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

Mathematica [A]  time = 0.190773, size = 169, normalized size = 0.95 \[ \frac{2 \left (-315 a^2 x (A (b+3 c x)+3 B x (b-c x))-21 a^3 (3 A+5 B x)+63 a x^2 \left (5 A \left (-3 b^2+6 b c x+c^2 x^2\right )+B x \left (15 b^2+10 b c x+3 c^2 x^2\right )\right )+x^3 \left (9 A \left (35 b^2 c x+35 b^3+21 b c^2 x^2+5 c^3 x^3\right )+B x \left (189 b^2 c x+105 b^3+135 b c^2 x^2+35 c^3 x^3\right )\right )\right )}{315 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^3)/x^(7/2),x]

[Out]

(2*(-21*a^3*(3*A + 5*B*x) - 315*a^2*x*(3*B*x*(b - c*x) + A*(b + 3*c*x)) + 63*a*x^2*(5*A*(-3*b^2 + 6*b*c*x + c^
2*x^2) + B*x*(15*b^2 + 10*b*c*x + 3*c^2*x^2)) + x^3*(9*A*(35*b^3 + 35*b^2*c*x + 21*b*c^2*x^2 + 5*c^3*x^3) + B*
x*(105*b^3 + 189*b^2*c*x + 135*b*c^2*x^2 + 35*c^3*x^3))))/(315*x^(5/2))

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Maple [A]  time = 0.006, size = 192, normalized size = 1.1 \begin{align*} -{\frac{-70\,B{c}^{3}{x}^{7}-90\,A{c}^{3}{x}^{6}-270\,B{x}^{6}b{c}^{2}-378\,A{x}^{5}b{c}^{2}-378\,aB{c}^{2}{x}^{5}-378\,B{x}^{5}{b}^{2}c-630\,aA{c}^{2}{x}^{4}-630\,A{x}^{4}{b}^{2}c-1260\,B{x}^{4}abc-210\,B{x}^{4}{b}^{3}-3780\,A{x}^{3}abc-630\,A{b}^{3}{x}^{3}-1890\,{a}^{2}Bc{x}^{3}-1890\,B{x}^{3}a{b}^{2}+1890\,{a}^{2}Ac{x}^{2}+1890\,A{x}^{2}a{b}^{2}+1890\,B{x}^{2}{a}^{2}b+630\,A{a}^{2}bx+210\,{a}^{3}Bx+126\,A{a}^{3}}{315}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x)

[Out]

-2/315*(-35*B*c^3*x^7-45*A*c^3*x^6-135*B*b*c^2*x^6-189*A*b*c^2*x^5-189*B*a*c^2*x^5-189*B*b^2*c*x^5-315*A*a*c^2
*x^4-315*A*b^2*c*x^4-630*B*a*b*c*x^4-105*B*b^3*x^4-1890*A*a*b*c*x^3-315*A*b^3*x^3-945*B*a^2*c*x^3-945*B*a*b^2*
x^3+945*A*a^2*c*x^2+945*A*a*b^2*x^2+945*B*a^2*b*x^2+315*A*a^2*b*x+105*B*a^3*x+63*A*a^3)/x^(5/2)

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Maxima [A]  time = 1.01628, size = 225, normalized size = 1.26 \begin{align*} \frac{2}{9} \, B c^{3} x^{\frac{9}{2}} + \frac{2}{7} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac{7}{2}} + \frac{6}{5} \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{\frac{5}{2}} + \frac{2}{3} \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{\frac{3}{2}} + 2 \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} \sqrt{x} - \frac{2 \,{\left (3 \, A a^{3} + 45 \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} + 5 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="maxima")

[Out]

2/9*B*c^3*x^(9/2) + 2/7*(3*B*b*c^2 + A*c^3)*x^(7/2) + 6/5*(B*b^2*c + (B*a + A*b)*c^2)*x^(5/2) + 2/3*(B*b^3 + 3
*A*a*c^2 + 3*(2*B*a*b + A*b^2)*c)*x^(3/2) + 2*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*sqrt(x) - 2/15*(3*A*
a^3 + 45*(B*a^2*b + A*a*b^2 + A*a^2*c)*x^2 + 5*(B*a^3 + 3*A*a^2*b)*x)/x^(5/2)

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Fricas [A]  time = 1.05493, size = 383, normalized size = 2.15 \begin{align*} \frac{2 \,{\left (35 \, B c^{3} x^{7} + 45 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \,{\left (B b^{2} c +{\left (B a + A b\right )} c^{2}\right )} x^{5} + 105 \,{\left (B b^{3} + 3 \, A a c^{2} + 3 \,{\left (2 \, B a b + A b^{2}\right )} c\right )} x^{4} - 63 \, A a^{3} + 315 \,{\left (3 \, B a b^{2} + A b^{3} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} c\right )} x^{3} - 945 \,{\left (B a^{2} b + A a b^{2} + A a^{2} c\right )} x^{2} - 105 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} x\right )}}{315 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="fricas")

[Out]

2/315*(35*B*c^3*x^7 + 45*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + (B*a + A*b)*c^2)*x^5 + 105*(B*b^3 + 3*A*a*c^
2 + 3*(2*B*a*b + A*b^2)*c)*x^4 - 63*A*a^3 + 315*(3*B*a*b^2 + A*b^3 + 3*(B*a^2 + 2*A*a*b)*c)*x^3 - 945*(B*a^2*b
+ A*a*b^2 + A*a^2*c)*x^2 - 105*(B*a^3 + 3*A*a^2*b)*x)/x^(5/2)

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Sympy [A]  time = 13.9742, size = 275, normalized size = 1.54 \begin{align*} - \frac{2 A a^{3}}{5 x^{\frac{5}{2}}} - \frac{2 A a^{2} b}{x^{\frac{3}{2}}} - \frac{6 A a^{2} c}{\sqrt{x}} - \frac{6 A a b^{2}}{\sqrt{x}} + 12 A a b c \sqrt{x} + 2 A a c^{2} x^{\frac{3}{2}} + 2 A b^{3} \sqrt{x} + 2 A b^{2} c x^{\frac{3}{2}} + \frac{6 A b c^{2} x^{\frac{5}{2}}}{5} + \frac{2 A c^{3} x^{\frac{7}{2}}}{7} - \frac{2 B a^{3}}{3 x^{\frac{3}{2}}} - \frac{6 B a^{2} b}{\sqrt{x}} + 6 B a^{2} c \sqrt{x} + 6 B a b^{2} \sqrt{x} + 4 B a b c x^{\frac{3}{2}} + \frac{6 B a c^{2} x^{\frac{5}{2}}}{5} + \frac{2 B b^{3} x^{\frac{3}{2}}}{3} + \frac{6 B b^{2} c x^{\frac{5}{2}}}{5} + \frac{6 B b c^{2} x^{\frac{7}{2}}}{7} + \frac{2 B c^{3} x^{\frac{9}{2}}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**3/x**(7/2),x)

[Out]

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

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Giac [A]  time = 1.27805, size = 259, normalized size = 1.46 \begin{align*} \frac{2}{9} \, B c^{3} x^{\frac{9}{2}} + \frac{6}{7} \, B b c^{2} x^{\frac{7}{2}} + \frac{2}{7} \, A c^{3} x^{\frac{7}{2}} + \frac{6}{5} \, B b^{2} c x^{\frac{5}{2}} + \frac{6}{5} \, B a c^{2} x^{\frac{5}{2}} + \frac{6}{5} \, A b c^{2} x^{\frac{5}{2}} + \frac{2}{3} \, B b^{3} x^{\frac{3}{2}} + 4 \, B a b c x^{\frac{3}{2}} + 2 \, A b^{2} c x^{\frac{3}{2}} + 2 \, A a c^{2} x^{\frac{3}{2}} + 6 \, B a b^{2} \sqrt{x} + 2 \, A b^{3} \sqrt{x} + 6 \, B a^{2} c \sqrt{x} + 12 \, A a b c \sqrt{x} - \frac{2 \,{\left (45 \, B a^{2} b x^{2} + 45 \, A a b^{2} x^{2} + 45 \, A a^{2} c x^{2} + 5 \, B a^{3} x + 15 \, A a^{2} b x + 3 \, A a^{3}\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^3/x^(7/2),x, algorithm="giac")

[Out]

2/9*B*c^3*x^(9/2) + 6/7*B*b*c^2*x^(7/2) + 2/7*A*c^3*x^(7/2) + 6/5*B*b^2*c*x^(5/2) + 6/5*B*a*c^2*x^(5/2) + 6/5*
A*b*c^2*x^(5/2) + 2/3*B*b^3*x^(3/2) + 4*B*a*b*c*x^(3/2) + 2*A*b^2*c*x^(3/2) + 2*A*a*c^2*x^(3/2) + 6*B*a*b^2*sq
rt(x) + 2*A*b^3*sqrt(x) + 6*B*a^2*c*sqrt(x) + 12*A*a*b*c*sqrt(x) - 2/15*(45*B*a^2*b*x^2 + 45*A*a*b^2*x^2 + 45*
A*a^2*c*x^2 + 5*B*a^3*x + 15*A*a^2*b*x + 3*A*a^3)/x^(5/2)