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

Optimal. Leaf size=126 \[ \frac{a^2 (6 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{3/2}}+\frac{\sqrt{x} (a+b x)^{3/2} (6 A b-a B)}{12 b}+\frac{a \sqrt{x} \sqrt{a+b x} (6 A b-a B)}{8 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b} \]

[Out]

(a*(6*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b) + ((6*A*b - a*B)*Sqrt[x]*(a + b*x)^(3/2))/(12*b) + (B*Sqrt[x]*(a
+ b*x)^(5/2))/(3*b) + (a^2*(6*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(3/2))

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Rubi [A]  time = 0.0538404, antiderivative size = 126, 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 = {80, 50, 63, 217, 206} \[ \frac{a^2 (6 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{3/2}}+\frac{\sqrt{x} (a+b x)^{3/2} (6 A b-a B)}{12 b}+\frac{a \sqrt{x} \sqrt{a+b x} (6 A b-a B)}{8 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b) + ((6*A*b - a*B)*Sqrt[x]*(a + b*x)^(3/2))/(12*b) + (B*Sqrt[x]*(a
+ b*x)^(5/2))/(3*b) + (a^2*(6*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(3/2))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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{(a+b x)^{3/2} (A+B x)}{\sqrt{x}} \, dx &=\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{\left (3 A b-\frac{a B}{2}\right ) \int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx}{3 b}\\ &=\frac{(6 A b-a B) \sqrt{x} (a+b x)^{3/2}}{12 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{(a (6 A b-a B)) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx}{8 b}\\ &=\frac{a (6 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b}+\frac{(6 A b-a B) \sqrt{x} (a+b x)^{3/2}}{12 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{\left (a^2 (6 A b-a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{16 b}\\ &=\frac{a (6 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b}+\frac{(6 A b-a B) \sqrt{x} (a+b x)^{3/2}}{12 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{\left (a^2 (6 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{8 b}\\ &=\frac{a (6 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b}+\frac{(6 A b-a B) \sqrt{x} (a+b x)^{3/2}}{12 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{\left (a^2 (6 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{8 b}\\ &=\frac{a (6 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b}+\frac{(6 A b-a B) \sqrt{x} (a+b x)^{3/2}}{12 b}+\frac{B \sqrt{x} (a+b x)^{5/2}}{3 b}+\frac{a^2 (6 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.230532, size = 107, normalized size = 0.85 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (3 a^2 B+2 a b (15 A+7 B x)+4 b^2 x (3 A+2 B x)\right )-\frac{3 a^{3/2} (a B-6 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{24 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(3*a^2*B + 4*b^2*x*(3*A + 2*B*x) + 2*a*b*(15*A + 7*B*x)) - (3*a^(3/2)*(-6*A*b
+ a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(24*b^(3/2))

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Maple [A]  time = 0.01, size = 176, normalized size = 1.4 \begin{align*}{\frac{1}{48}\sqrt{bx+a}\sqrt{x} \left ( 16\,B{x}^{2}{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+24\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}x+28\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}xa+18\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}b+60\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}a-3\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}+6\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{2} \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)*(B*x+A)/x^(1/2),x)

[Out]

1/48*(b*x+a)^(1/2)*x^(1/2)/b^(3/2)*(16*B*x^2*b^(5/2)*(x*(b*x+a))^(1/2)+24*A*(x*(b*x+a))^(1/2)*b^(5/2)*x+28*B*(
x*(b*x+a))^(1/2)*b^(3/2)*x*a+18*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2*b+60*A*(x*(b*x+a))
^(1/2)*b^(3/2)*a-3*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3+6*B*(x*(b*x+a))^(1/2)*b^(1/2)*a
^2)/(x*(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((b*x+a)^(3/2)*(B*x+A)/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.62616, size = 494, normalized size = 3.92 \begin{align*} \left [-\frac{3 \,{\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt{b} \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \,{\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{48 \, b^{2}}, \frac{3 \,{\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) +{\left (8 \, B b^{3} x^{2} + 3 \, B a^{2} b + 30 \, A a b^{2} + 2 \,{\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)*(B*x+A)/x^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(B*a^3 - 6*A*a^2*b)*sqrt(b)*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(8*B*b^3*x^2 + 3*B*
a^2*b + 30*A*a*b^2 + 2*(7*B*a*b^2 + 6*A*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^2, 1/24*(3*(B*a^3 - 6*A*a^2*b)*sqrt(-
b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) + (8*B*b^3*x^2 + 3*B*a^2*b + 30*A*a*b^2 + 2*(7*B*a*b^2 + 6*A*b^3
)*x)*sqrt(b*x + a)*sqrt(x))/b^2]

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Sympy [A]  time = 67.7662, size = 204, normalized size = 1.62 \begin{align*} A \left (\frac{5 a^{\frac{3}{2}} \sqrt{x} \sqrt{1 + \frac{b x}{a}}}{4} + \frac{\sqrt{a} b x^{\frac{3}{2}} \sqrt{1 + \frac{b x}{a}}}{2} + \frac{3 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4 \sqrt{b}}\right ) + B \left (\frac{a^{\frac{5}{2}} \sqrt{x}}{8 b \sqrt{1 + \frac{b x}{a}}} + \frac{17 a^{\frac{3}{2}} x^{\frac{3}{2}}}{24 \sqrt{1 + \frac{b x}{a}}} + \frac{11 \sqrt{a} b x^{\frac{5}{2}}}{12 \sqrt{1 + \frac{b x}{a}}} - \frac{a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 b^{\frac{3}{2}}} + \frac{b^{2} x^{\frac{7}{2}}}{3 \sqrt{a} \sqrt{1 + \frac{b x}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**(1/2),x)

[Out]

A*(5*a**(3/2)*sqrt(x)*sqrt(1 + b*x/a)/4 + sqrt(a)*b*x**(3/2)*sqrt(1 + b*x/a)/2 + 3*a**2*asinh(sqrt(b)*sqrt(x)/
sqrt(a))/(4*sqrt(b))) + B*(a**(5/2)*sqrt(x)/(8*b*sqrt(1 + b*x/a)) + 17*a**(3/2)*x**(3/2)/(24*sqrt(1 + b*x/a))
+ 11*sqrt(a)*b*x**(5/2)/(12*sqrt(1 + b*x/a)) - a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(3/2)) + b**2*x**(7/2
)/(3*sqrt(a)*sqrt(1 + b*x/a)))

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Giac [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((b*x+a)^(3/2)*(B*x+A)/x^(1/2),x, algorithm="giac")

[Out]

Timed out