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

Optimal. Leaf size=134 \[ -\frac{x^{3/2} \sqrt{a+b x} (4 A b-5 a B)}{2 a b^2}+\frac{3 \sqrt{x} \sqrt{a+b x} (4 A b-5 a B)}{4 b^3}-\frac{3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{7/2}}+\frac{2 x^{5/2} (A b-a B)}{a b \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.0519871, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {78, 50, 63, 217, 206} \[ -\frac{x^{3/2} \sqrt{a+b x} (4 A b-5 a B)}{2 a b^2}+\frac{3 \sqrt{x} \sqrt{a+b x} (4 A b-5 a B)}{4 b^3}-\frac{3 a (4 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{4 b^{7/2}}+\frac{2 x^{5/2} (A b-a B)}{a b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*x^(5/2))/(a*b*Sqrt[a + b*x]) + (3*(4*A*b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/(4*b^3) - ((4*A*b - 5*
a*B)*x^(3/2)*Sqrt[a + b*x])/(2*a*b^2) - (3*a*(4*A*b - 5*a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(4*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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0709659, size = 106, normalized size = 0.79 \[ \frac{\sqrt{b} \sqrt{x} \left (-15 a^2 B+a b (12 A-5 B x)+2 b^2 x (2 A+B x)\right )+3 a^{3/2} \sqrt{\frac{b x}{a}+1} (5 a B-4 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{7/2} \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*Sqrt[x]*(-15*a^2*B + a*b*(12*A - 5*B*x) + 2*b^2*x*(2*A + B*x)) + 3*a^(3/2)*(-4*A*b + 5*a*B)*Sqrt[1 +
(b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(7/2)*Sqrt[a + b*x])

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Maple [B]  time = 0.013, size = 244, normalized size = 1.8 \begin{align*} -{\frac{1}{8} \left ( -4\,B{x}^{2}{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+12\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) xa{b}^{2}-8\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}x-15\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{2}b+10\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}xa+12\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}b-24\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}a-15\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}+30\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{2} \right ) \sqrt{x}{b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{bx+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(-4*B*x^2*b^(5/2)*(x*(b*x+a))^(1/2)+12*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a*b^2-8*
A*(x*(b*x+a))^(1/2)*b^(5/2)*x-15*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^2*b+10*B*(x*(b*x+
a))^(1/2)*b^(3/2)*x*a+12*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2*b-24*A*(x*(b*x+a))^(1/2)*
b^(3/2)*a-15*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3+30*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^2)/b
^(7/2)*x^(1/2)/(x*(b*x+a))^(1/2)/(b*x+a)^(1/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69028, size = 603, normalized size = 4.5 \begin{align*} \left [-\frac{3 \,{\left (5 \, B a^{3} - 4 \, A a^{2} b +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{8 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{3 \,{\left (5 \, B a^{3} - 4 \, A a^{2} b +{\left (5 \, B a^{2} b - 4 \, A a b^{2}\right )} x\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (2 \, B b^{3} x^{2} - 15 \, B a^{2} b + 12 \, A a b^{2} -{\left (5 \, B a b^{2} - 4 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{4 \,{\left (b^{5} x + a b^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x)
+ a) - 2*(2*B*b^3*x^2 - 15*B*a^2*b + 12*A*a*b^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*
b^4), -1/4*(3*(5*B*a^3 - 4*A*a^2*b + (5*B*a^2*b - 4*A*a*b^2)*x)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt
(x))) - (2*B*b^3*x^2 - 15*B*a^2*b + 12*A*a*b^2 - (5*B*a*b^2 - 4*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/(b^5*x + a*b^
4)]

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Sympy [A]  time = 154.232, size = 182, normalized size = 1.36 \begin{align*} A \left (\frac{3 \sqrt{a} \sqrt{x}}{b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{b^{\frac{5}{2}}} + \frac{x^{\frac{3}{2}}}{\sqrt{a} b \sqrt{1 + \frac{b x}{a}}}\right ) + B \left (- \frac{15 a^{\frac{3}{2}} \sqrt{x}}{4 b^{3} \sqrt{1 + \frac{b x}{a}}} - \frac{5 \sqrt{a} x^{\frac{3}{2}}}{4 b^{2} \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 b^{\frac{7}{2}}} + \frac{x^{\frac{5}{2}}}{2 \sqrt{a} b \sqrt{1 + \frac{b x}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*(3*sqrt(a)*sqrt(x)/(b**2*sqrt(1 + b*x/a)) - 3*a*asinh(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) + x**(3/2)/(sqrt(a)*
b*sqrt(1 + b*x/a))) + B*(-15*a**(3/2)*sqrt(x)/(4*b**3*sqrt(1 + b*x/a)) - 5*sqrt(a)*x**(3/2)/(4*b**2*sqrt(1 + b
*x/a)) + 15*a**2*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(4*b**(7/2)) + x**(5/2)/(2*sqrt(a)*b*sqrt(1 + b*x/a)))

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Giac [A]  time = 91.2904, size = 244, normalized size = 1.82 \begin{align*} \frac{1}{4} \, \sqrt{{\left (b x + a\right )} b - a b} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} B{\left | b \right |}}{b^{5}} - \frac{9 \, B a b^{9}{\left | b \right |} - 4 \, A b^{10}{\left | b \right |}}{b^{14}}\right )} - \frac{3 \,{\left (5 \, B a^{2} \sqrt{b}{\left | b \right |} - 4 \, A a b^{\frac{3}{2}}{\left | b \right |}\right )} \log \left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2}\right )}{8 \, b^{5}} - \frac{4 \,{\left (B a^{3} \sqrt{b}{\left | b \right |} - A a^{2} b^{\frac{3}{2}}{\left | b \right |}\right )}}{{\left ({\left (\sqrt{b x + a} \sqrt{b} - \sqrt{{\left (b x + a\right )} b - a b}\right )}^{2} + a b\right )} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt((b*x + a)*b - a*b)*sqrt(b*x + a)*(2*(b*x + a)*B*abs(b)/b^5 - (9*B*a*b^9*abs(b) - 4*A*b^10*abs(b))/b^1
4) - 3/8*(5*B*a^2*sqrt(b)*abs(b) - 4*A*a*b^(3/2)*abs(b))*log((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))
^2)/b^5 - 4*(B*a^3*sqrt(b)*abs(b) - A*a^2*b^(3/2)*abs(b))/(((sqrt(b*x + a)*sqrt(b) - sqrt((b*x + a)*b - a*b))^
2 + a*b)*b^4)