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

Optimal. Leaf size=100 \[ \frac{(A c+3 b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{3/2} c^{5/2}}-\frac{\sqrt{x} (A c+3 b B)}{4 b c^2 (b+c x)}-\frac{x^{3/2} (b B-A c)}{2 b c (b+c x)^2} \]

[Out]

-((b*B - A*c)*x^(3/2))/(2*b*c*(b + c*x)^2) - ((3*b*B + A*c)*Sqrt[x])/(4*b*c^2*(b + c*x)) + ((3*b*B + A*c)*ArcT
an[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(3/2)*c^(5/2))

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Rubi [A]  time = 0.0491069, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {781, 78, 47, 63, 205} \[ \frac{(A c+3 b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{3/2} c^{5/2}}-\frac{\sqrt{x} (A c+3 b B)}{4 b c^2 (b+c x)}-\frac{x^{3/2} (b B-A c)}{2 b c (b+c x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B - A*c)*x^(3/2))/(2*b*c*(b + c*x)^2) - ((3*b*B + A*c)*Sqrt[x])/(4*b*c^2*(b + c*x)) + ((3*b*B + A*c)*ArcT
an[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(4*b^(3/2)*c^(5/2))

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 78

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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)}{\left (b x+c x^2\right )^3} \, dx &=\int \frac{\sqrt{x} (A+B x)}{(b+c x)^3} \, dx\\ &=-\frac{(b B-A c) x^{3/2}}{2 b c (b+c x)^2}+\frac{(3 b B+A c) \int \frac{\sqrt{x}}{(b+c x)^2} \, dx}{4 b c}\\ &=-\frac{(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac{(3 b B+A c) \sqrt{x}}{4 b c^2 (b+c x)}+\frac{(3 b B+A c) \int \frac{1}{\sqrt{x} (b+c x)} \, dx}{8 b c^2}\\ &=-\frac{(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac{(3 b B+A c) \sqrt{x}}{4 b c^2 (b+c x)}+\frac{(3 b B+A c) \operatorname{Subst}\left (\int \frac{1}{b+c x^2} \, dx,x,\sqrt{x}\right )}{4 b c^2}\\ &=-\frac{(b B-A c) x^{3/2}}{2 b c (b+c x)^2}-\frac{(3 b B+A c) \sqrt{x}}{4 b c^2 (b+c x)}+\frac{(3 b B+A c) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{3/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0649169, size = 85, normalized size = 0.85 \[ \frac{\sqrt{x} \left (-b c (A+5 B x)+A c^2 x-3 b^2 B\right )}{4 b c^2 (b+c x)^2}+\frac{(A c+3 b B) \tan ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 b^{3/2} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 94, normalized size = 0.9 \begin{align*} 2\,{\frac{1}{ \left ( cx+b \right ) ^{2}} \left ( 1/8\,{\frac{ \left ( Ac-5\,bB \right ){x}^{3/2}}{bc}}-1/8\,{\frac{ \left ( Ac+3\,bB \right ) \sqrt{x}}{{c}^{2}}} \right ) }+{\frac{A}{4\,bc}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{3\,B}{4\,{c}^{2}}\arctan \left ({c\sqrt{x}{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \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)^3,x)

[Out]

2*(1/8*(A*c-5*B*b)/b/c*x^(3/2)-1/8*(A*c+3*B*b)/c^2*x^(1/2))/(c*x+b)^2+1/4/c/b/(b*c)^(1/2)*arctan(x^(1/2)*c/(b*
c)^(1/2))*A+3/4/c^2/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.7455, size = 633, normalized size = 6.33 \begin{align*} \left [-\frac{{\left (3 \, B b^{3} + A b^{2} c +{\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} c + A b c^{2}\right )} x\right )} \sqrt{-b c} \log \left (\frac{c x - b - 2 \, \sqrt{-b c} \sqrt{x}}{c x + b}\right ) + 2 \,{\left (3 \, B b^{3} c + A b^{2} c^{2} +{\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt{x}}{8 \,{\left (b^{2} c^{5} x^{2} + 2 \, b^{3} c^{4} x + b^{4} c^{3}\right )}}, -\frac{{\left (3 \, B b^{3} + A b^{2} c +{\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 2 \,{\left (3 \, B b^{2} c + A b c^{2}\right )} x\right )} \sqrt{b c} \arctan \left (\frac{\sqrt{b c}}{c \sqrt{x}}\right ) +{\left (3 \, B b^{3} c + A b^{2} c^{2} +{\left (5 \, B b^{2} c^{2} - A b c^{3}\right )} x\right )} \sqrt{x}}{4 \,{\left (b^{2} c^{5} x^{2} + 2 \, b^{3} c^{4} x + b^{4} c^{3}\right )}}\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)^3,x, algorithm="fricas")

[Out]

[-1/8*((3*B*b^3 + A*b^2*c + (3*B*b*c^2 + A*c^3)*x^2 + 2*(3*B*b^2*c + A*b*c^2)*x)*sqrt(-b*c)*log((c*x - b - 2*s
qrt(-b*c)*sqrt(x))/(c*x + b)) + 2*(3*B*b^3*c + A*b^2*c^2 + (5*B*b^2*c^2 - A*b*c^3)*x)*sqrt(x))/(b^2*c^5*x^2 +
2*b^3*c^4*x + b^4*c^3), -1/4*((3*B*b^3 + A*b^2*c + (3*B*b*c^2 + A*c^3)*x^2 + 2*(3*B*b^2*c + A*b*c^2)*x)*sqrt(b
*c)*arctan(sqrt(b*c)/(c*sqrt(x))) + (3*B*b^3*c + A*b^2*c^2 + (5*B*b^2*c^2 - A*b*c^3)*x)*sqrt(x))/(b^2*c^5*x^2
+ 2*b^3*c^4*x + b^4*c^3)]

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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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.11938, size = 111, normalized size = 1.11 \begin{align*} \frac{{\left (3 \, B b + A c\right )} \arctan \left (\frac{c \sqrt{x}}{\sqrt{b c}}\right )}{4 \, \sqrt{b c} b c^{2}} - \frac{5 \, B b c x^{\frac{3}{2}} - A c^{2} x^{\frac{3}{2}} + 3 \, B b^{2} \sqrt{x} + A b c \sqrt{x}}{4 \,{\left (c x + b\right )}^{2} b c^{2}} \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)^3,x, algorithm="giac")

[Out]

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