3.481 \(\int \sqrt{x} \sqrt{a+b x} (A+B x) \, dx\)

Optimal. Leaf size=126 \[ -\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{5/2}}+\frac{a \sqrt{x} \sqrt{a+b x} (2 A b-a B)}{8 b^2}+\frac{x^{3/2} \sqrt{a+b x} (2 A b-a B)}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b} \]

[Out]

(a*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^2) + ((2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(4*b) + (B*x^(3/2)*(a
+ b*x)^(3/2))/(3*b) - (a^2*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(5/2))

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Rubi [A]  time = 0.0560503, 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} \[ -\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{5/2}}+\frac{a \sqrt{x} \sqrt{a+b x} (2 A b-a B)}{8 b^2}+\frac{x^{3/2} \sqrt{a+b x} (2 A b-a B)}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[a + b*x]*(A + B*x),x]

[Out]

(a*(2*A*b - a*B)*Sqrt[x]*Sqrt[a + b*x])/(8*b^2) + ((2*A*b - a*B)*x^(3/2)*Sqrt[a + b*x])/(4*b) + (B*x^(3/2)*(a
+ b*x)^(3/2))/(3*b) - (a^2*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(8*b^(5/2))

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 \sqrt{x} \sqrt{a+b x} (A+B x) \, dx &=\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}+\frac{\left (3 A b-\frac{3 a B}{2}\right ) \int \sqrt{x} \sqrt{a+b x} \, dx}{3 b}\\ &=\frac{(2 A b-a B) x^{3/2} \sqrt{a+b x}}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}+\frac{(a (2 A b-a B)) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{8 b}\\ &=\frac{a (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b^2}+\frac{(2 A b-a B) x^{3/2} \sqrt{a+b x}}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}-\frac{\left (a^2 (2 A b-a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{16 b^2}\\ &=\frac{a (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b^2}+\frac{(2 A b-a B) x^{3/2} \sqrt{a+b x}}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}-\frac{\left (a^2 (2 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{8 b^2}\\ &=\frac{a (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b^2}+\frac{(2 A b-a B) x^{3/2} \sqrt{a+b x}}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}-\frac{\left (a^2 (2 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^2}\\ &=\frac{a (2 A b-a B) \sqrt{x} \sqrt{a+b x}}{8 b^2}+\frac{(2 A b-a B) x^{3/2} \sqrt{a+b x}}{4 b}+\frac{B x^{3/2} (a+b x)^{3/2}}{3 b}-\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.198585, size = 106, normalized size = 0.84 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (-3 a^2 B+2 a b (3 A+B x)+4 b^2 x (3 A+2 B x)\right )+\frac{3 a^{3/2} (a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{24 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[a + b*x]*(A + B*x),x]

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(-3*a^2*B + 2*a*b*(3*A + B*x) + 4*b^2*x*(3*A + 2*B*x)) + (3*a^(3/2)*(-2*A*b +
a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(24*b^(5/2))

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Maple [A]  time = 0.009, size = 176, normalized size = 1.4 \begin{align*} -{\frac{1}{48}\sqrt{x}\sqrt{bx+a} \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-4\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}xa+6\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{2}b-12\,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{5}{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)*x^(1/2)*(b*x+a)^(1/2),x)

[Out]

-1/48*x^(1/2)*(b*x+a)^(1/2)/b^(5/2)*(-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-4*B*
(x*(b*x+a))^(1/2)*b^(3/2)*x*a+6*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^2*b-12*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)*x^(1/2)*(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.20866, size = 487, normalized size = 3.87 \begin{align*} \left [-\frac{3 \,{\left (B a^{3} - 2 \, 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 + 6 \, A a b^{2} + 2 \,{\left (B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{48 \, b^{3}}, -\frac{3 \,{\left (B a^{3} - 2 \, 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 + 6 \, A a b^{2} + 2 \,{\left (B a b^{2} + 6 \, A b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 32.6595, size = 677, normalized size = 5.37 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a**(3/2)*sqrt(x)/(4*b*sqrt(1 + b*x/a)) + 3*A*sqrt(a)*x**(3/2)/(4*sqrt(1 + b*x/a)) - A*a**2*asinh(sqrt(b)*sqr
t(x)/sqrt(a))/(4*b**(3/2)) + A*b*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/a)) - 2*B*a*Piecewise((a**(3/2)*sqrt(a + b*x
)/(8*sqrt(b)*sqrt(b*x/a)) - 3*sqrt(a)*(a + b*x)**(3/2)/(8*sqrt(b)*sqrt(b*x/a)) - a**2*acosh(sqrt(a + b*x)/sqrt
(a))/(8*sqrt(b)) + (a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-I*a**(3/2)*sq
rt(a + b*x)/(8*sqrt(b)*sqrt(-b*x/a)) + 3*I*sqrt(a)*(a + b*x)**(3/2)/(8*sqrt(b)*sqrt(-b*x/a)) + I*a**2*asin(sqr
t(a + b*x)/sqrt(a))/(8*sqrt(b)) - I*(a + b*x)**(5/2)/(4*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**2 + 2*B*Piece
wise((a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(b*x/a)) - 5
*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*sqrt(b*x/a)) - a**3*acosh(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) + (a + b*x
)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-I*a**(5/2)*sqrt(a + b*x)/(16*sqrt(b)*sqr
t(-b*x/a)) + I*a**(3/2)*(a + b*x)**(3/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*sqrt(a)*(a + b*x)**(5/2)/(24*sqrt(b)*
sqrt(-b*x/a)) + I*a**3*asin(sqrt(a + b*x)/sqrt(a))/(16*sqrt(b)) - I*(a + b*x)**(7/2)/(6*sqrt(a)*sqrt(b)*sqrt(-
b*x/a)), True))/b**2

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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)*x^(1/2)*(b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out