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

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

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Rubi [A]  time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 )^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*}

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Mathematica [A]  time = 0.06, size = 85, normalized size = 0.85 \begin {gather*} \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} \left (-b c (A+5 B x)+A c^2 x-3 b^2 B\right )}{4 b c^2 (b+c x)^2} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.17, size = 86, normalized size = 0.86 \begin {gather*} \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} \left (A b c-A c^2 x+3 b^2 B+5 b B c x\right )}{4 b c^2 (b+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(Sqrt[x]*(3*b^2*B + A*b*c + 5*b*B*c*x - A*c^2*x))/(b*c^2*(b + c*x)^2) + ((3*b*B + A*c)*ArcTan[(Sqrt[c]*Sq
rt[x])/Sqrt[b]])/(4*b^(3/2)*c^(5/2))

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fricas [A]  time = 0.42, size = 291, normalized size = 2.91 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 82, normalized size = 0.82 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.06, size = 94, normalized size = 0.94 \begin {gather*} \frac {A \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, b c}+\frac {3 B \arctan \left (\frac {c \sqrt {x}}{\sqrt {b c}}\right )}{4 \sqrt {b c}\, c^{2}}+\frac {\frac {\left (A c -5 b B \right ) x^{\frac {3}{2}}}{4 b c}-\frac {\left (A c +3 b B \right ) \sqrt {x}}{4 c^{2}}}{\left (c x +b \right )^{2}} \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)^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(1/(b*c)^(1/2)
*c*x^(1/2))*A+3/4/c^2/(b*c)^(1/2)*arctan(1/(b*c)^(1/2)*c*x^(1/2))*B

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maxima [A]  time = 1.29, size = 94, normalized size = 0.94 \begin {gather*} -\frac {{\left (5 \, B b c - A c^{2}\right )} x^{\frac {3}{2}} + {\left (3 \, B b^{2} + A b c\right )} \sqrt {x}}{4 \, {\left (b c^{4} x^{2} + 2 \, b^{2} c^{3} x + b^{3} c^{2}\right )}} + \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}} \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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.10, size = 84, normalized size = 0.84 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {x}}{\sqrt {b}}\right )\,\left (A\,c+3\,B\,b\right )}{4\,b^{3/2}\,c^{5/2}}-\frac {\frac {\sqrt {x}\,\left (A\,c+3\,B\,b\right )}{4\,c^2}-\frac {x^{3/2}\,\left (A\,c-5\,B\,b\right )}{4\,b\,c}}{b^2+2\,b\,c\,x+c^2\,x^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)^3,x)

[Out]

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

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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)**3,x)

[Out]

Timed out

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