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

Optimal. Leaf size=109 \[ \frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}}-\frac {\sqrt {x} (5 b B-3 A c)}{c^3}+\frac {x^{3/2} (5 b B-3 A c)}{3 b c^2}-\frac {x^{5/2} (b B-A c)}{b c (b+c x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {781, 78, 50, 63, 205} \begin {gather*} \frac {x^{3/2} (5 b B-3 A c)}{3 b c^2}-\frac {\sqrt {x} (5 b B-3 A c)}{c^3}+\frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}}-\frac {x^{5/2} (b B-A c)}{b c (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 88, normalized size = 0.81 \begin {gather*} \frac {\sqrt {x} \left (b c (9 A-10 B x)+2 c^2 x (3 A+B x)-15 b^2 B\right )}{3 c^3 (b+c x)}+\frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-15*b^2*B + b*c*(9*A - 10*B*x) + 2*c^2*x*(3*A + B*x)))/(3*c^3*(b + c*x)) + (Sqrt[b]*(5*b*B - 3*A*c)*
ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(7/2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.13, size = 95, normalized size = 0.87 \begin {gather*} \frac {\left (5 b^{3/2} B-3 A \sqrt {b} c\right ) \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{7/2}}+\frac {\sqrt {x} \left (9 A b c+6 A c^2 x-15 b^2 B-10 b B c x+2 B c^2 x^2\right )}{3 c^3 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x]*(-15*b^2*B + 9*A*b*c - 10*b*B*c*x + 6*A*c^2*x + 2*B*c^2*x^2))/(3*c^3*(b + c*x)) + ((5*b^(3/2)*B - 3*A
*Sqrt[b]*c)*ArcTan[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/c^(7/2)

________________________________________________________________________________________

fricas [A]  time = 0.41, size = 231, normalized size = 2.12 \begin {gather*} \left [-\frac {3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x\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 (2 \, B c^{2} x^{2} - 15 \, B b^{2} + 9 \, A b c - 2 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt {x}}{6 \, {\left (c^{4} x + b c^{3}\right )}}, \frac {3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c \sqrt {x} \sqrt {\frac {b}{c}}}{b}\right ) + {\left (2 \, B c^{2} x^{2} - 15 \, B b^{2} + 9 \, A b c - 2 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x\right )} \sqrt {x}}{3 \, {\left (c^{4} x + b c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(3*(5*B*b^2 - 3*A*b*c + (5*B*b*c - 3*A*c^2)*x)*sqrt(-b/c)*log((c*x - 2*c*sqrt(x)*sqrt(-b/c) - b)/(c*x +
b)) - 2*(2*B*c^2*x^2 - 15*B*b^2 + 9*A*b*c - 2*(5*B*b*c - 3*A*c^2)*x)*sqrt(x))/(c^4*x + b*c^3), 1/3*(3*(5*B*b^2
- 3*A*b*c + (5*B*b*c - 3*A*c^2)*x)*sqrt(b/c)*arctan(c*sqrt(x)*sqrt(b/c)/b) + (2*B*c^2*x^2 - 15*B*b^2 + 9*A*b*
c - 2*(5*B*b*c - 3*A*c^2)*x)*sqrt(x))/(c^4*x + b*c^3)]

________________________________________________________________________________________

giac [A]  time = 0.16, size = 95, normalized size = 0.87 \begin {gather*} \frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} - \frac {B b^{2} \sqrt {x} - A b c \sqrt {x}}{{\left (c x + b\right )} c^{3}} + \frac {2 \, {\left (B c^{4} x^{\frac {3}{2}} - 6 \, B b c^{3} \sqrt {x} + 3 \, A c^{4} \sqrt {x}\right )}}{3 \, c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*B*b^2 - 3*A*b*c)*arctan(c*sqrt(x)/sqrt(b*c))/(sqrt(b*c)*c^3) - (B*b^2*sqrt(x) - A*b*c*sqrt(x))/((c*x + b)*c
^3) + 2/3*(B*c^4*x^(3/2) - 6*B*b*c^3*sqrt(x) + 3*A*c^4*sqrt(x))/c^6

________________________________________________________________________________________

maple [A]  time = 0.07, size = 113, normalized size = 1.04 \begin {gather*} -\frac {3 A b \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{2}}+\frac {5 B \,b^{2} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{3}}+\frac {A b \sqrt {x}}{\left (c x +b \right ) c^{2}}-\frac {B \,b^{2} \sqrt {x}}{\left (c x +b \right ) c^{3}}+\frac {2 B \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {2 A \sqrt {x}}{c^{2}}-\frac {4 B b \sqrt {x}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.20, size = 88, normalized size = 0.81 \begin {gather*} -\frac {{\left (B b^{2} - A b c\right )} \sqrt {x}}{c^{4} x + b c^{3}} + \frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {2 \, {\left (B c x^{\frac {3}{2}} - 3 \, {\left (2 \, B b - A c\right )} \sqrt {x}\right )}}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.09, size = 107, normalized size = 0.98 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{c^2}-\frac {4\,B\,b}{c^3}\right )-\frac {\sqrt {x}\,\left (B\,b^2-A\,b\,c\right )}{x\,c^4+b\,c^3}+\frac {2\,B\,x^{3/2}}{3\,c^2}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,\sqrt {x}\,\left (3\,A\,c-5\,B\,b\right )}{5\,B\,b^2-3\,A\,b\,c}\right )\,\left (3\,A\,c-5\,B\,b\right )}{c^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________