3.364 \(\int \frac{x^{5/2} (A+B x)}{(a+b x)^3} \, dx\)

Optimal. Leaf size=147 \[ \frac{x^{5/2} (3 A b-7 a B)}{4 a b^2 (a+b x)}-\frac{5 x^{3/2} (3 A b-7 a B)}{12 a b^3}+\frac{5 \sqrt{x} (3 A b-7 a B)}{4 b^4}-\frac{5 \sqrt{a} (3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}+\frac{x^{7/2} (A b-a B)}{2 a b (a+b x)^2} \]

[Out]

(5*(3*A*b - 7*a*B)*Sqrt[x])/(4*b^4) - (5*(3*A*b - 7*a*B)*x^(3/2))/(12*a*b^3) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a
+ b*x)^2) + ((3*A*b - 7*a*B)*x^(5/2))/(4*a*b^2*(a + b*x)) - (5*Sqrt[a]*(3*A*b - 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x
])/Sqrt[a]])/(4*b^(9/2))

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Rubi [A]  time = 0.0623048, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 205} \[ \frac{x^{5/2} (3 A b-7 a B)}{4 a b^2 (a+b x)}-\frac{5 x^{3/2} (3 A b-7 a B)}{12 a b^3}+\frac{5 \sqrt{x} (3 A b-7 a B)}{4 b^4}-\frac{5 \sqrt{a} (3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}+\frac{x^{7/2} (A b-a B)}{2 a b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(a + b*x)^3,x]

[Out]

(5*(3*A*b - 7*a*B)*Sqrt[x])/(4*b^4) - (5*(3*A*b - 7*a*B)*x^(3/2))/(12*a*b^3) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a
+ b*x)^2) + ((3*A*b - 7*a*B)*x^(5/2))/(4*a*b^2*(a + b*x)) - (5*Sqrt[a]*(3*A*b - 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x
])/Sqrt[a]])/(4*b^(9/2))

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 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^{5/2} (A+B x)}{(a+b x)^3} \, dx &=\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}-\frac{\left (\frac{3 A b}{2}-\frac{7 a B}{2}\right ) \int \frac{x^{5/2}}{(a+b x)^2} \, dx}{2 a b}\\ &=\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}-\frac{(5 (3 A b-7 a B)) \int \frac{x^{3/2}}{a+b x} \, dx}{8 a b^2}\\ &=-\frac{5 (3 A b-7 a B) x^{3/2}}{12 a b^3}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}+\frac{(5 (3 A b-7 a B)) \int \frac{\sqrt{x}}{a+b x} \, dx}{8 b^3}\\ &=\frac{5 (3 A b-7 a B) \sqrt{x}}{4 b^4}-\frac{5 (3 A b-7 a B) x^{3/2}}{12 a b^3}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}-\frac{(5 a (3 A b-7 a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{8 b^4}\\ &=\frac{5 (3 A b-7 a B) \sqrt{x}}{4 b^4}-\frac{5 (3 A b-7 a B) x^{3/2}}{12 a b^3}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}-\frac{(5 a (3 A b-7 a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{4 b^4}\\ &=\frac{5 (3 A b-7 a B) \sqrt{x}}{4 b^4}-\frac{5 (3 A b-7 a B) x^{3/2}}{12 a b^3}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x)^2}+\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 (a+b x)}-\frac{5 \sqrt{a} (3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0243585, size = 61, normalized size = 0.41 \[ \frac{x^{7/2} \left (\frac{7 a^2 (A b-a B)}{(a+b x)^2}+(7 a B-3 A b) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )\right )}{14 a^3 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(a + b*x)^3,x]

[Out]

(x^(7/2)*((7*a^2*(A*b - a*B))/(a + b*x)^2 + (-3*A*b + 7*a*B)*Hypergeometric2F1[2, 7/2, 9/2, -((b*x)/a)]))/(14*
a^3*b)

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Maple [A]  time = 0.015, size = 152, normalized size = 1. \begin{align*}{\frac{2\,B}{3\,{b}^{3}}{x}^{{\frac{3}{2}}}}+2\,{\frac{A\sqrt{x}}{{b}^{3}}}-6\,{\frac{Ba\sqrt{x}}{{b}^{4}}}+{\frac{9\,Aa}{4\,{b}^{2} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}-{\frac{13\,B{a}^{2}}{4\,{b}^{3} \left ( bx+a \right ) ^{2}}{x}^{{\frac{3}{2}}}}+{\frac{7\,A{a}^{2}}{4\,{b}^{3} \left ( bx+a \right ) ^{2}}\sqrt{x}}-{\frac{11\,B{a}^{3}}{4\,{b}^{4} \left ( bx+a \right ) ^{2}}\sqrt{x}}-{\frac{15\,Aa}{4\,{b}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{35\,B{a}^{2}}{4\,{b}^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x+A)/(b*x+a)^3,x)

[Out]

2/3/b^3*B*x^(3/2)+2/b^3*A*x^(1/2)-6/b^4*B*a*x^(1/2)+9/4*a/b^2/(b*x+a)^2*x^(3/2)*A-13/4*a^2/b^3/(b*x+a)^2*x^(3/
2)*B+7/4*a^2/b^3/(b*x+a)^2*A*x^(1/2)-11/4*a^3/b^4/(b*x+a)^2*B*x^(1/2)-15/4*a/b^3/(a*b)^(1/2)*arctan(b*x^(1/2)/
(a*b)^(1/2))*A+35/4*a^2/b^4/(a*b)^(1/2)*arctan(b*x^(1/2)/(a*b)^(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^(5/2)*(B*x+A)/(b*x+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.67727, size = 775, normalized size = 5.27 \begin{align*} \left [-\frac{15 \,{\left (7 \, B a^{3} - 3 \, A a^{2} b +{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) - 2 \,{\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \,{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{24 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{15 \,{\left (7 \, B a^{3} - 3 \, A a^{2} b +{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \,{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{12 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^3,x, algorithm="fricas")

[Out]

[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x
- 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*
x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(15*(7*B*a^3 - 3*A*a^2*b +
(7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^
3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2
*a*b^5*x + a^2*b^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**(5/2)*(B*x+A)/(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17535, size = 161, normalized size = 1.1 \begin{align*} \frac{5 \,{\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{4}} - \frac{13 \, B a^{2} b x^{\frac{3}{2}} - 9 \, A a b^{2} x^{\frac{3}{2}} + 11 \, B a^{3} \sqrt{x} - 7 \, A a^{2} b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{4}} + \frac{2 \,{\left (B b^{6} x^{\frac{3}{2}} - 9 \, B a b^{5} \sqrt{x} + 3 \, A b^{6} \sqrt{x}\right )}}{3 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b*x+a)^3,x, algorithm="giac")

[Out]

5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/4*(13*B*a^2*b*x^(3/2) - 9*A*a*b^2*x^(3
/2) + 11*B*a^3*sqrt(x) - 7*A*a^2*b*sqrt(x))/((b*x + a)^2*b^4) + 2/3*(B*b^6*x^(3/2) - 9*B*a*b^5*sqrt(x) + 3*A*b
^6*sqrt(x))/b^9