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

Optimal. Leaf size=159 \[ \frac{a^2 \sqrt{x} \sqrt{a+b x} (8 A b-3 a B)}{64 b^2}-\frac{a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{5/2}}+\frac{a x^{3/2} \sqrt{a+b x} (8 A b-3 a B)}{32 b}+\frac{x^{3/2} (a+b x)^{3/2} (8 A b-3 a B)}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b} \]

[Out]

(a^2*(8*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^2) + (a*(8*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(32*b) + ((8*
A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(24*b) + (B*x^(3/2)*(a + b*x)^(5/2))/(4*b) - (a^3*(8*A*b - 3*a*B)*ArcTan
h[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(5/2))

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Rubi [A]  time = 0.0672959, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{a^2 \sqrt{x} \sqrt{a+b x} (8 A b-3 a B)}{64 b^2}-\frac{a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{5/2}}+\frac{a x^{3/2} \sqrt{a+b x} (8 A b-3 a B)}{32 b}+\frac{x^{3/2} (a+b x)^{3/2} (8 A b-3 a B)}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*A*b - 3*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^2) + (a*(8*A*b - 3*a*B)*x^(3/2)*Sqrt[a + b*x])/(32*b) + ((8*
A*b - 3*a*B)*x^(3/2)*(a + b*x)^(3/2))/(24*b) + (B*x^(3/2)*(a + b*x)^(5/2))/(4*b) - (a^3*(8*A*b - 3*a*B)*ArcTan
h[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*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} (a+b x)^{3/2} (A+B x) \, dx &=\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}+\frac{\left (4 A b-\frac{3 a B}{2}\right ) \int \sqrt{x} (a+b x)^{3/2} \, dx}{4 b}\\ &=\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}+\frac{(a (8 A b-3 a B)) \int \sqrt{x} \sqrt{a+b x} \, dx}{16 b}\\ &=\frac{a (8 A b-3 a B) x^{3/2} \sqrt{a+b x}}{32 b}+\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}+\frac{\left (a^2 (8 A b-3 a B)\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{64 b}\\ &=\frac{a^2 (8 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{64 b^2}+\frac{a (8 A b-3 a B) x^{3/2} \sqrt{a+b x}}{32 b}+\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac{\left (a^3 (8 A b-3 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{128 b^2}\\ &=\frac{a^2 (8 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{64 b^2}+\frac{a (8 A b-3 a B) x^{3/2} \sqrt{a+b x}}{32 b}+\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac{\left (a^3 (8 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{64 b^2}\\ &=\frac{a^2 (8 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{64 b^2}+\frac{a (8 A b-3 a B) x^{3/2} \sqrt{a+b x}}{32 b}+\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac{\left (a^3 (8 A b-3 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^2}\\ &=\frac{a^2 (8 A b-3 a B) \sqrt{x} \sqrt{a+b x}}{64 b^2}+\frac{a (8 A b-3 a B) x^{3/2} \sqrt{a+b x}}{32 b}+\frac{(8 A b-3 a B) x^{3/2} (a+b x)^{3/2}}{24 b}+\frac{B x^{3/2} (a+b x)^{5/2}}{4 b}-\frac{a^3 (8 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.218456, size = 126, normalized size = 0.79 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (6 a^2 b (4 A+B x)-9 a^3 B+8 a b^2 x (14 A+9 B x)+16 b^3 x^2 (4 A+3 B x)\right )+\frac{3 a^{5/2} (3 a B-8 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{192 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(-9*a^3*B + 6*a^2*b*(4*A + B*x) + 16*b^3*x^2*(4*A + 3*B*x) + 8*a*b^2*x*(14*A +
9*B*x)) + (3*a^(5/2)*(-8*A*b + 3*a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(192*b^(5/2))

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Maple [A]  time = 0.01, size = 218, normalized size = 1.4 \begin{align*} -{\frac{1}{384}\sqrt{bx+a}\sqrt{x} \left ( -96\,B{x}^{3}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-128\,A{x}^{2}{b}^{7/2}\sqrt{x \left ( bx+a \right ) }-144\,B{x}^{2}a{b}^{5/2}\sqrt{x \left ( bx+a \right ) }-224\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}xa-12\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{2}+24\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b-48\,A\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{a}^{2}-9\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}+18\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{3} \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)^(3/2)*(B*x+A)*x^(1/2),x)

[Out]

-1/384*(b*x+a)^(1/2)*x^(1/2)/b^(5/2)*(-96*B*x^3*b^(7/2)*(x*(b*x+a))^(1/2)-128*A*x^2*b^(7/2)*(x*(b*x+a))^(1/2)-
144*B*x^2*a*b^(5/2)*(x*(b*x+a))^(1/2)-224*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a-12*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^2
+24*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b-48*A*(x*(b*x+a))^(1/2)*b^(3/2)*a^2-9*B*ln(1/
2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4+18*B*(x*(b*x+a))^(1/2)*b^(1/2)*a^3)/(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 = 2.76704, size = 606, normalized size = 3.81 \begin{align*} \left [-\frac{3 \,{\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt{b} \log \left (2 \, b x - 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) - 2 \,{\left (48 \, B b^{4} x^{3} - 9 \, B a^{3} b + 24 \, A a^{2} b^{2} + 8 \,{\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \,{\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{384 \, b^{3}}, -\frac{3 \,{\left (3 \, B a^{4} - 8 \, A a^{3} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (48 \, B b^{4} x^{3} - 9 \, B a^{3} b + 24 \, A a^{2} b^{2} + 8 \,{\left (9 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \,{\left (3 \, B a^{2} b^{2} + 56 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{192 \, b^{3}}\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]

[-1/384*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(b)*log(2*b*x - 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) - 2*(48*B*b^4*x^3 -
9*B*a^3*b + 24*A*a^2*b^2 + 8*(9*B*a*b^3 + 8*A*b^4)*x^2 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x)
)/b^3, -1/192*(3*(3*B*a^4 - 8*A*a^3*b)*sqrt(-b)*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (48*B*b^4*x^3 - 9
*B*a^3*b + 24*A*a^2*b^2 + 8*(9*B*a*b^3 + 8*A*b^4)*x^2 + 2*(3*B*a^2*b^2 + 56*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))
/b^3]

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Sympy [B]  time = 34.418, size = 298, normalized size = 1.87 \begin{align*} \frac{A a^{\frac{5}{2}} \sqrt{x}}{8 b \sqrt{1 + \frac{b x}{a}}} + \frac{17 A a^{\frac{3}{2}} x^{\frac{3}{2}}}{24 \sqrt{1 + \frac{b x}{a}}} + \frac{11 A \sqrt{a} b x^{\frac{5}{2}}}{12 \sqrt{1 + \frac{b x}{a}}} - \frac{A a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 b^{\frac{3}{2}}} + \frac{A b^{2} x^{\frac{7}{2}}}{3 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} - \frac{3 B a^{\frac{7}{2}} \sqrt{x}}{64 b^{2} \sqrt{1 + \frac{b x}{a}}} - \frac{B a^{\frac{5}{2}} x^{\frac{3}{2}}}{64 b \sqrt{1 + \frac{b x}{a}}} + \frac{13 B a^{\frac{3}{2}} x^{\frac{5}{2}}}{32 \sqrt{1 + \frac{b x}{a}}} + \frac{5 B \sqrt{a} b x^{\frac{7}{2}}}{8 \sqrt{1 + \frac{b x}{a}}} + \frac{3 B a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{64 b^{\frac{5}{2}}} + \frac{B b^{2} x^{\frac{9}{2}}}{4 \sqrt{a} \sqrt{1 + \frac{b x}{a}}} \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*a**(5/2)*sqrt(x)/(8*b*sqrt(1 + b*x/a)) + 17*A*a**(3/2)*x**(3/2)/(24*sqrt(1 + b*x/a)) + 11*A*sqrt(a)*b*x**(5/
2)/(12*sqrt(1 + b*x/a)) - A*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(3/2)) + A*b**2*x**(7/2)/(3*sqrt(a)*sqrt
(1 + b*x/a)) - 3*B*a**(7/2)*sqrt(x)/(64*b**2*sqrt(1 + b*x/a)) - B*a**(5/2)*x**(3/2)/(64*b*sqrt(1 + b*x/a)) + 1
3*B*a**(3/2)*x**(5/2)/(32*sqrt(1 + b*x/a)) + 5*B*sqrt(a)*b*x**(7/2)/(8*sqrt(1 + b*x/a)) + 3*B*a**4*asinh(sqrt(
b)*sqrt(x)/sqrt(a))/(64*b**(5/2)) + B*b**2*x**(9/2)/(4*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