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

Optimal. Leaf size=90 \[ \frac{4 c \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^5}-\frac{2 \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^6}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(9*b*x^7) - (2*(9*b*B - 4*A*c)*(b*x + c*x^2)^(5/2))/(63*b^2*x^6) + (4*c*(9*b*B - 4*
A*c)*(b*x + c*x^2)^(5/2))/(315*b^3*x^5)

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Rubi [A]  time = 0.0864704, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {792, 658, 650} \[ \frac{4 c \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{315 b^3 x^5}-\frac{2 \left (b x+c x^2\right )^{5/2} (9 b B-4 A c)}{63 b^2 x^6}-\frac{2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*(b*x + c*x^2)^(5/2))/(9*b*x^7) - (2*(9*b*B - 4*A*c)*(b*x + c*x^2)^(5/2))/(63*b^2*x^6) + (4*c*(9*b*B - 4*
A*c)*(b*x + c*x^2)^(5/2))/(315*b^3*x^5)

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(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] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), 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 + 2*p + 2, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (b x+c x^2\right )^{3/2}}{x^7} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}+\frac{\left (2 \left (-7 (-b B+A c)+\frac{5}{2} (-b B+2 A c)\right )\right ) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^6} \, dx}{9 b}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac{2 (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}-\frac{(2 c (9 b B-4 A c)) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^5} \, dx}{63 b^2}\\ &=-\frac{2 A \left (b x+c x^2\right )^{5/2}}{9 b x^7}-\frac{2 (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{63 b^2 x^6}+\frac{4 c (9 b B-4 A c) \left (b x+c x^2\right )^{5/2}}{315 b^3 x^5}\\ \end{align*}

Mathematica [A]  time = 0.0279867, size = 56, normalized size = 0.62 \[ \frac{2 (x (b+c x))^{5/2} \left (A \left (-35 b^2+20 b c x-8 c^2 x^2\right )+9 b B x (2 c x-5 b)\right )}{315 b^3 x^7} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(x*(b + c*x))^(5/2)*(9*b*B*x*(-5*b + 2*c*x) + A*(-35*b^2 + 20*b*c*x - 8*c^2*x^2)))/(315*b^3*x^7)

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Maple [A]  time = 0.004, size = 62, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 8\,A{c}^{2}{x}^{2}-18\,B{x}^{2}bc-20\,Abcx+45\,{b}^{2}Bx+35\,A{b}^{2} \right ) }{315\,{x}^{6}{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(c*x+b)*(8*A*c^2*x^2-18*B*b*c*x^2-20*A*b*c*x+45*B*b^2*x+35*A*b^2)*(c*x^2+b*x)^(3/2)/x^6/b^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88289, size = 231, normalized size = 2.57 \begin{align*} -\frac{2 \,{\left (35 \, A b^{4} - 2 \,{\left (9 \, B b c^{3} - 4 \, A c^{4}\right )} x^{4} +{\left (9 \, B b^{2} c^{2} - 4 \, A b c^{3}\right )} x^{3} + 3 \,{\left (24 \, B b^{3} c + A b^{2} c^{2}\right )} x^{2} + 5 \,{\left (9 \, B b^{4} + 10 \, A b^{3} c\right )} x\right )} \sqrt{c x^{2} + b x}}{315 \, b^{3} x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(35*A*b^4 - 2*(9*B*b*c^3 - 4*A*c^4)*x^4 + (9*B*b^2*c^2 - 4*A*b*c^3)*x^3 + 3*(24*B*b^3*c + A*b^2*c^2)*x^
2 + 5*(9*B*b^4 + 10*A*b^3*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x \left (b + c x\right )\right )^{\frac{3}{2}} \left (A + B x\right )}{x^{7}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.17588, size = 501, normalized size = 5.57 \begin{align*} \frac{2 \,{\left (315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} B c^{\frac{5}{2}} + 945 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B b c^{2} + 420 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} A c^{3} + 1260 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b^{2} c^{\frac{3}{2}} + 1575 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A b c^{\frac{5}{2}} + 882 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{3} c + 2583 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b^{2} c^{2} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{4} \sqrt{c} + 2310 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{3} c^{\frac{3}{2}} + 45 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{5} + 1170 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{4} c + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{5} \sqrt{c} + 35 \, A b^{6}\right )}}{315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315*(315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^7*B*c^(5/2) + 945*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*B*b*c^2 + 420*(
sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*c^3 + 1260*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b^2*c^(3/2) + 1575*(sqrt(c)*
x - sqrt(c*x^2 + b*x))^5*A*b*c^(5/2) + 882*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^3*c + 2583*(sqrt(c)*x - sqrt(
c*x^2 + b*x))^4*A*b^2*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*sqrt(c) + 2310*(sqrt(c)*x - sqrt(c*x^2
+ b*x))^3*A*b^3*c^(3/2) + 45*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^5 + 1170*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2
*A*b^4*c + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^5*sqrt(c) + 35*A*b^6)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^9