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

Optimal. Leaf size=113 \[ -\frac{2 b^2 \sqrt{x} (b B-A c)}{c^4}+\frac{2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}}-\frac{2 x^{5/2} (b B-A c)}{5 c^2}+\frac{2 b x^{3/2} (b B-A c)}{3 c^3}+\frac{2 B x^{7/2}}{7 c} \]

[Out]

(-2*b^2*(b*B - A*c)*Sqrt[x])/c^4 + (2*b*(b*B - A*c)*x^(3/2))/(3*c^3) - (2*(b*B - A*c)*x^(5/2))/(5*c^2) + (2*B*
x^(7/2))/(7*c) + (2*b^(5/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

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Rubi [A]  time = 0.0712113, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {781, 80, 50, 63, 205} \[ -\frac{2 b^2 \sqrt{x} (b B-A c)}{c^4}+\frac{2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}}-\frac{2 x^{5/2} (b B-A c)}{5 c^2}+\frac{2 b x^{3/2} (b B-A c)}{3 c^3}+\frac{2 B x^{7/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(b*B - A*c)*Sqrt[x])/c^4 + (2*b*(b*B - A*c)*x^(3/2))/(3*c^3) - (2*(b*B - A*c)*x^(5/2))/(5*c^2) + (2*B*
x^(7/2))/(7*c) + (2*b^(5/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

Rule 781

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e^p, Int[(e
*x)^(m + p)*(f + g*x)*(b + c*x)^p, x], x] /; FreeQ[{b, c, e, f, g, m}, x] && IntegerQ[p]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^{7/2} (A+B x)}{b x+c x^2} \, dx &=\int \frac{x^{5/2} (A+B x)}{b+c x} \, dx\\ &=\frac{2 B x^{7/2}}{7 c}+\frac{\left (2 \left (-\frac{7 b B}{2}+\frac{7 A c}{2}\right )\right ) \int \frac{x^{5/2}}{b+c x} \, dx}{7 c}\\ &=-\frac{2 (b B-A c) x^{5/2}}{5 c^2}+\frac{2 B x^{7/2}}{7 c}+\frac{(b (b B-A c)) \int \frac{x^{3/2}}{b+c x} \, dx}{c^2}\\ &=\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{5/2}}{5 c^2}+\frac{2 B x^{7/2}}{7 c}-\frac{\left (b^2 (b B-A c)\right ) \int \frac{\sqrt{x}}{b+c x} \, dx}{c^3}\\ &=-\frac{2 b^2 (b B-A c) \sqrt{x}}{c^4}+\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{5/2}}{5 c^2}+\frac{2 B x^{7/2}}{7 c}+\frac{\left (b^3 (b B-A c)\right ) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{c^4}\\ &=-\frac{2 b^2 (b B-A c) \sqrt{x}}{c^4}+\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{5/2}}{5 c^2}+\frac{2 B x^{7/2}}{7 c}+\frac{\left (2 b^3 (b B-A c)\right ) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{c^4}\\ &=-\frac{2 b^2 (b B-A c) \sqrt{x}}{c^4}+\frac{2 b (b B-A c) x^{3/2}}{3 c^3}-\frac{2 (b B-A c) x^{5/2}}{5 c^2}+\frac{2 B x^{7/2}}{7 c}+\frac{2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0659556, size = 101, normalized size = 0.89 \[ \frac{2 \sqrt{x} \left (35 b^2 c (3 A+B x)-7 b c^2 x (5 A+3 B x)+3 c^3 x^2 (7 A+5 B x)-105 b^3 B\right )}{105 c^4}+\frac{2 b^{5/2} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{c^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(-105*b^3*B + 35*b^2*c*(3*A + B*x) - 7*b*c^2*x*(5*A + 3*B*x) + 3*c^3*x^2*(7*A + 5*B*x)))/(105*c^4)
+ (2*b^(5/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(9/2)

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Maple [A]  time = 0.014, size = 126, normalized size = 1.1 \begin{align*}{\frac{2\,B}{7\,c}{x}^{{\frac{7}{2}}}}+{\frac{2\,A}{5\,c}{x}^{{\frac{5}{2}}}}-{\frac{2\,bB}{5\,{c}^{2}}{x}^{{\frac{5}{2}}}}-{\frac{2\,Ab}{3\,{c}^{2}}{x}^{{\frac{3}{2}}}}+{\frac{2\,{b}^{2}B}{3\,{c}^{3}}{x}^{{\frac{3}{2}}}}+2\,{\frac{A{b}^{2}\sqrt{x}}{{c}^{3}}}-2\,{\frac{{b}^{3}B\sqrt{x}}{{c}^{4}}}-2\,{\frac{{b}^{3}A}{{c}^{3}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) }+2\,{\frac{{b}^{4}B}{{c}^{4}\sqrt{bc}}\arctan \left ({\frac{\sqrt{x}c}{\sqrt{bc}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*B*x^(7/2)/c+2/5/c*A*x^(5/2)-2/5/c^2*B*x^(5/2)*b-2/3/c^2*A*x^(3/2)*b+2/3/c^3*B*x^(3/2)*b^2+2/c^3*A*b^2*x^(1
/2)-2/c^4*b^3*B*x^(1/2)-2*b^3/c^3/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*c)^(1/2))*A+2*b^4/c^4/(b*c)^(1/2)*arctan(x^(
1/2)*c/(b*c)^(1/2))*B

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.75962, size = 524, normalized size = 4.64 \begin{align*} \left [-\frac{105 \,{\left (B b^{3} - A b^{2} c\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x - 2 \, c \sqrt{x} \sqrt{-\frac{b}{c}} - b}{c x + b}\right ) - 2 \,{\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \,{\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \,{\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt{x}}{105 \, c^{4}}, \frac{2 \,{\left (105 \,{\left (B b^{3} - A b^{2} c\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c \sqrt{x} \sqrt{\frac{b}{c}}}{b}\right ) +{\left (15 \, B c^{3} x^{3} - 105 \, B b^{3} + 105 \, A b^{2} c - 21 \,{\left (B b c^{2} - A c^{3}\right )} x^{2} + 35 \,{\left (B b^{2} c - A b c^{2}\right )} x\right )} \sqrt{x}\right )}}{105 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/105*(105*(B*b^3 - A*b^2*c)*sqrt(-b/c)*log((c*x - 2*c*sqrt(x)*sqrt(-b/c) - b)/(c*x + b)) - 2*(15*B*c^3*x^3
- 105*B*b^3 + 105*A*b^2*c - 21*(B*b*c^2 - A*c^3)*x^2 + 35*(B*b^2*c - A*b*c^2)*x)*sqrt(x))/c^4, 2/105*(105*(B*b
^3 - A*b^2*c)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) + (15*B*c^3*x^3 - 105*B*b^3 + 105*A*b^2*c - 21*(B*b*c^2
- A*c^3)*x^2 + 35*(B*b^2*c - A*b*c^2)*x)*sqrt(x))/c^4]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1163, size = 155, normalized size = 1.37 \begin{align*} \frac{2 \,{\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{\sqrt{b c} c^{4}} + \frac{2 \,{\left (15 \, B c^{6} x^{\frac{7}{2}} - 21 \, B b c^{5} x^{\frac{5}{2}} + 21 \, A c^{6} x^{\frac{5}{2}} + 35 \, B b^{2} c^{4} x^{\frac{3}{2}} - 35 \, A b c^{5} x^{\frac{3}{2}} - 105 \, B b^{3} c^{3} \sqrt{x} + 105 \, A b^{2} c^{4} \sqrt{x}\right )}}{105 \, c^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(B*b^4 - A*b^3*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^4) + 2/105*(15*B*c^6*x^(7/2) - 21*B*b*c^5*x^(5/2)
+ 21*A*c^6*x^(5/2) + 35*B*b^2*c^4*x^(3/2) - 35*A*b*c^5*x^(3/2) - 105*B*b^3*c^3*sqrt(x) + 105*A*b^2*c^4*sqrt(x
))/c^7