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

Optimal. Leaf size=108 \[ -\frac{x^{3/2} (3 A b-5 a B)}{3 a b^2}+\frac{\sqrt{x} (3 A b-5 a B)}{b^3}-\frac{\sqrt{a} (3 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}}+\frac{x^{5/2} (A b-a B)}{a b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0463346, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 50, 63, 205} \[ -\frac{x^{3/2} (3 A b-5 a B)}{3 a b^2}+\frac{\sqrt{x} (3 A b-5 a B)}{b^3}-\frac{\sqrt{a} (3 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}}+\frac{x^{5/2} (A b-a B)}{a b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0592108, size = 88, normalized size = 0.81 \[ \frac{\sqrt{x} \left (-15 a^2 B+a b (9 A-10 B x)+2 b^2 x (3 A+B x)\right )}{3 b^3 (a+b x)}+\frac{\sqrt{a} (5 a B-3 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 113, normalized size = 1.1 \begin{align*}{\frac{2\,B}{3\,{b}^{2}}{x}^{{\frac{3}{2}}}}+2\,{\frac{A\sqrt{x}}{{b}^{2}}}-4\,{\frac{Ba\sqrt{x}}{{b}^{3}}}+{\frac{Aa}{{b}^{2} \left ( bx+a \right ) }\sqrt{x}}-{\frac{B{a}^{2}}{{b}^{3} \left ( bx+a \right ) }\sqrt{x}}-3\,{\frac{Aa}{{b}^{2}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) }+5\,{\frac{B{a}^{2}}{{b}^{3}\sqrt{ab}}\arctan \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.49922, size = 524, normalized size = 4.85 \begin{align*} \left [-\frac{3 \,{\left (5 \, B a^{2} - 3 \, A a b +{\left (5 \, B a b - 3 \, 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 (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \,{\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt{x}}{6 \,{\left (b^{4} x + a b^{3}\right )}}, \frac{3 \,{\left (5 \, B a^{2} - 3 \, A a b +{\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (2 \, B b^{2} x^{2} - 15 \, B a^{2} + 9 \, A a b - 2 \,{\left (5 \, B a b - 3 \, A b^{2}\right )} x\right )} \sqrt{x}}{3 \,{\left (b^{4} x + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 26.9693, size = 932, normalized size = 8.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(B*x+A)/(b*x+a)**2,x)

[Out]

Piecewise((zoo*(2*A*sqrt(x) + 2*B*x**(3/2)/3), Eq(a, 0) & Eq(b, 0)), ((2*A*sqrt(x) + 2*B*x**(3/2)/3)/b**2, Eq(
a, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(7/2)/7)/a**2, Eq(b, 0)), (18*I*A*a**(3/2)*b**2*sqrt(x)*sqrt(1/b)/(6*I*a**(3
/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 12*I*A*sqrt(a)*b**3*x**(3/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4
*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 9*A*a**2*b*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4
*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 9*A*a**2*b*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*
sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 9*A*a*b**2*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**
4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) + 9*A*a*b**2*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b*
*4*sqrt(1/b) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 30*I*B*a**(5/2)*b*sqrt(x)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b
) + 6*I*sqrt(a)*b**5*x*sqrt(1/b)) - 20*I*B*a**(3/2)*b**2*x**(3/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I
*sqrt(a)*b**5*x*sqrt(1/b)) + 4*I*B*sqrt(a)*b**3*x**(5/2)*sqrt(1/b)/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*
b**5*x*sqrt(1/b)) + 15*B*a**3*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b
**5*x*sqrt(1/b)) - 15*B*a**3*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*b**
5*x*sqrt(1/b)) + 15*B*a**2*b*x*log(-I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a)*
b**5*x*sqrt(1/b)) - 15*B*a**2*b*x*log(I*sqrt(a)*sqrt(1/b) + sqrt(x))/(6*I*a**(3/2)*b**4*sqrt(1/b) + 6*I*sqrt(a
)*b**5*x*sqrt(1/b)), True))

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Giac [A]  time = 1.2631, size = 128, normalized size = 1.19 \begin{align*} \frac{{\left (5 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} b^{3}} - \frac{B a^{2} \sqrt{x} - A a b \sqrt{x}}{{\left (b x + a\right )} b^{3}} + \frac{2 \,{\left (B b^{4} x^{\frac{3}{2}} - 6 \, B a b^{3} \sqrt{x} + 3 \, A b^{4} \sqrt{x}\right )}}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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