3.103 \(\int \frac{(A+B x) (b x+c x^2)^{5/2}}{x^7} \, dx\)

Optimal. Leaf size=119 \[ -\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}-\frac{2 B c^2 \sqrt{b x+c x^2}}{x}+2 B c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.13935, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {792, 662, 620, 206} \[ -\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}-\frac{2 B c^2 \sqrt{b x+c x^2}}{x}+2 B c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^(5/2))/x^7,x]

[Out]

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

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 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (b x+c x^2\right )^{5/2}}{x^7} \, dx &=-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+B \int \frac{\left (b x+c x^2\right )^{5/2}}{x^6} \, dx\\ &=-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+(B c) \int \frac{\left (b x+c x^2\right )^{3/2}}{x^4} \, dx\\ &=-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+\left (B c^2\right ) \int \frac{\sqrt{b x+c x^2}}{x^2} \, dx\\ &=-\frac{2 B c^2 \sqrt{b x+c x^2}}{x}-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+\left (B c^3\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=-\frac{2 B c^2 \sqrt{b x+c x^2}}{x}-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+\left (2 B c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )\\ &=-\frac{2 B c^2 \sqrt{b x+c x^2}}{x}-\frac{2 B c \left (b x+c x^2\right )^{3/2}}{3 x^3}-\frac{2 B \left (b x+c x^2\right )^{5/2}}{5 x^5}-\frac{2 A \left (b x+c x^2\right )^{7/2}}{7 b x^7}+2 B c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0643442, size = 83, normalized size = 0.7 \[ \frac{2 (x (b+c x))^{5/2} \left ((b+c x)^3 (b B-A c)-\frac{b^4 B \, _2F_1\left (-\frac{7}{2},-\frac{7}{2};-\frac{5}{2};-\frac{c x}{b}\right )}{\sqrt{\frac{c x}{b}+1}}\right )}{7 b c x^6 (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^(5/2))/x^7,x]

[Out]

(2*(x*(b + c*x))^(5/2)*((b*B - A*c)*(b + c*x)^3 - (b^4*B*Hypergeometric2F1[-7/2, -7/2, -5/2, -((c*x)/b)])/Sqrt
[1 + (c*x)/b]))/(7*b*c*x^6*(b + c*x)^2)

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Maple [B]  time = 0.011, size = 263, normalized size = 2.2 \begin{align*} -{\frac{2\,A}{7\,b{x}^{7}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{2\,B}{5\,b{x}^{6}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{4\,Bc}{15\,{b}^{2}{x}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{16\,B{c}^{2}}{15\,{b}^{3}{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{32\,B{c}^{3}}{5\,{b}^{4}{x}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}-{\frac{256\,B{c}^{4}}{15\,{b}^{5}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{7}{2}}}}+{\frac{256\,B{c}^{5}}{15\,{b}^{5}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{32\,B{c}^{5}x}{3\,{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{16\,B{c}^{4}}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-4\,{\frac{B{c}^{4}\sqrt{c{x}^{2}+bx}x}{{b}^{2}}}-2\,{\frac{B{c}^{3}\sqrt{c{x}^{2}+bx}}{b}}+B{c}^{{\frac{5}{2}}}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^(5/2)/x^7,x)

[Out]

-2/7*A*(c*x^2+b*x)^(7/2)/b/x^7-2/5*B/b/x^6*(c*x^2+b*x)^(7/2)-4/15*B/b^2*c/x^5*(c*x^2+b*x)^(7/2)-16/15*B/b^3*c^
2/x^4*(c*x^2+b*x)^(7/2)+32/5*B/b^4*c^3/x^3*(c*x^2+b*x)^(7/2)-256/15*B/b^5*c^4/x^2*(c*x^2+b*x)^(7/2)+256/15*B/b
^5*c^5*(c*x^2+b*x)^(5/2)+32/3*B/b^4*c^5*(c*x^2+b*x)^(3/2)*x+16/3*B/b^3*c^4*(c*x^2+b*x)^(3/2)-4*B/b^2*c^4*(c*x^
2+b*x)^(1/2)*x-2*B/b*c^3*(c*x^2+b*x)^(1/2)+B*c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))

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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)^(5/2)/x^7,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85464, size = 564, normalized size = 4.74 \begin{align*} \left [\frac{105 \, B b c^{\frac{5}{2}} x^{4} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (15 \, A b^{3} +{\left (161 \, B b c^{2} + 15 \, A c^{3}\right )} x^{3} +{\left (77 \, B b^{2} c + 45 \, A b c^{2}\right )} x^{2} + 3 \,{\left (7 \, B b^{3} + 15 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x}}{105 \, b x^{4}}, -\frac{2 \,{\left (105 \, B b \sqrt{-c} c^{2} x^{4} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (15 \, A b^{3} +{\left (161 \, B b c^{2} + 15 \, A c^{3}\right )} x^{3} +{\left (77 \, B b^{2} c + 45 \, A b c^{2}\right )} x^{2} + 3 \,{\left (7 \, B b^{3} + 15 \, A b^{2} c\right )} x\right )} \sqrt{c x^{2} + b x}\right )}}{105 \, b x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^7,x, algorithm="fricas")

[Out]

[1/105*(105*B*b*c^(5/2)*x^4*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(15*A*b^3 + (161*B*b*c^2 + 15*A*c
^3)*x^3 + (77*B*b^2*c + 45*A*b*c^2)*x^2 + 3*(7*B*b^3 + 15*A*b^2*c)*x)*sqrt(c*x^2 + b*x))/(b*x^4), -2/105*(105*
B*b*sqrt(-c)*c^2*x^4*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (15*A*b^3 + (161*B*b*c^2 + 15*A*c^3)*x^3 + (77
*B*b^2*c + 45*A*b*c^2)*x^2 + 3*(7*B*b^3 + 15*A*b^2*c)*x)*sqrt(c*x^2 + b*x))/(b*x^4)]

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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{5}{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)**(5/2)/x**7,x)

[Out]

Integral((x*(b + c*x))**(5/2)*(A + B*x)/x**7, x)

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Giac [B]  time = 1.18853, size = 527, normalized size = 4.43 \begin{align*} -B c^{\frac{5}{2}} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right ) + \frac{2 \,{\left (315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} B b c^{\frac{5}{2}} + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{6} A c^{\frac{7}{2}} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} B b^{2} c^{2} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{5} A b c^{3} + 245 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} B b^{3} c^{\frac{3}{2}} + 525 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{4} A b^{2} c^{\frac{5}{2}} + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} B b^{4} c + 525 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3} A b^{3} c^{2} + 21 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} B b^{5} \sqrt{c} + 315 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} A b^{4} c^{\frac{3}{2}} + 105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} A b^{5} c + 15 \, A b^{6} \sqrt{c}\right )}}{105 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{7} \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^(5/2)/x^7,x, algorithm="giac")

[Out]

-B*c^(5/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/105*(315*(sqrt(c)*x - sqrt(c*x^2 + b*x
))^6*B*b*c^(5/2) + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^6*A*c^(7/2) + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*B*b
^2*c^2 + 315*(sqrt(c)*x - sqrt(c*x^2 + b*x))^5*A*b*c^3 + 245*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*B*b^3*c^(3/2) +
525*(sqrt(c)*x - sqrt(c*x^2 + b*x))^4*A*b^2*c^(5/2) + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*b^4*c + 525*(sq
rt(c)*x - sqrt(c*x^2 + b*x))^3*A*b^3*c^2 + 21*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b^5*sqrt(c) + 315*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^2*A*b^4*c^(3/2) + 105*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b^5*c + 15*A*b^6*sqrt(c))/((sqrt
(c)*x - sqrt(c*x^2 + b*x))^7*sqrt(c))