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

Optimal. Leaf size=159 \[ -\frac{a^2 \sqrt{x} \sqrt{a+b x} (8 A b-5 a B)}{64 b^3}+\frac{a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{7/2}}+\frac{a x^{3/2} \sqrt{a+b x} (8 A b-5 a B)}{96 b^2}+\frac{x^{5/2} \sqrt{a+b x} (8 A b-5 a B)}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b} \]

[Out]

-(a^2*(8*A*b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^3) + (a*(8*A*b - 5*a*B)*x^(3/2)*Sqrt[a + b*x])/(96*b^2) + (
(8*A*b - 5*a*B)*x^(5/2)*Sqrt[a + b*x])/(24*b) + (B*x^(5/2)*(a + b*x)^(3/2))/(4*b) + (a^3*(8*A*b - 5*a*B)*ArcTa
nh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(7/2))

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Rubi [A]  time = 0.0724063, 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} \[ -\frac{a^2 \sqrt{x} \sqrt{a+b x} (8 A b-5 a B)}{64 b^3}+\frac{a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{7/2}}+\frac{a x^{3/2} \sqrt{a+b x} (8 A b-5 a B)}{96 b^2}+\frac{x^{5/2} \sqrt{a+b x} (8 A b-5 a B)}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(8*A*b - 5*a*B)*Sqrt[x]*Sqrt[a + b*x])/(64*b^3) + (a*(8*A*b - 5*a*B)*x^(3/2)*Sqrt[a + b*x])/(96*b^2) + (
(8*A*b - 5*a*B)*x^(5/2)*Sqrt[a + b*x])/(24*b) + (B*x^(5/2)*(a + b*x)^(3/2))/(4*b) + (a^3*(8*A*b - 5*a*B)*ArcTa
nh[(Sqrt[b]*Sqrt[x])/Sqrt[a + b*x]])/(64*b^(7/2))

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 x^{3/2} \sqrt{a+b x} (A+B x) \, dx &=\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{\left (4 A b-\frac{5 a B}{2}\right ) \int x^{3/2} \sqrt{a+b x} \, dx}{4 b}\\ &=\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{(a (8 A b-5 a B)) \int \frac{x^{3/2}}{\sqrt{a+b x}} \, dx}{48 b}\\ &=\frac{a (8 A b-5 a B) x^{3/2} \sqrt{a+b x}}{96 b^2}+\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}-\frac{\left (a^2 (8 A b-5 a B)\right ) \int \frac{\sqrt{x}}{\sqrt{a+b x}} \, dx}{64 b^2}\\ &=-\frac{a^2 (8 A b-5 a B) \sqrt{x} \sqrt{a+b x}}{64 b^3}+\frac{a (8 A b-5 a B) x^{3/2} \sqrt{a+b x}}{96 b^2}+\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{\left (a^3 (8 A b-5 a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx}{128 b^3}\\ &=-\frac{a^2 (8 A b-5 a B) \sqrt{x} \sqrt{a+b x}}{64 b^3}+\frac{a (8 A b-5 a B) x^{3/2} \sqrt{a+b x}}{96 b^2}+\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{\left (a^3 (8 A b-5 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )}{64 b^3}\\ &=-\frac{a^2 (8 A b-5 a B) \sqrt{x} \sqrt{a+b x}}{64 b^3}+\frac{a (8 A b-5 a B) x^{3/2} \sqrt{a+b x}}{96 b^2}+\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{\left (a^3 (8 A b-5 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^3}\\ &=-\frac{a^2 (8 A b-5 a B) \sqrt{x} \sqrt{a+b x}}{64 b^3}+\frac{a (8 A b-5 a B) x^{3/2} \sqrt{a+b x}}{96 b^2}+\frac{(8 A b-5 a B) x^{5/2} \sqrt{a+b x}}{24 b}+\frac{B x^{5/2} (a+b x)^{3/2}}{4 b}+\frac{a^3 (8 A b-5 a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.246723, size = 126, normalized size = 0.79 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (-2 a^2 b (12 A+5 B x)+15 a^3 B+8 a b^2 x (2 A+B x)+16 b^3 x^2 (4 A+3 B x)\right )-\frac{3 a^{5/2} (5 a B-8 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{192 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^3*B + 8*a*b^2*x*(2*A + B*x) + 16*b^3*x^2*(4*A + 3*B*x) - 2*a^2*b*(12*A +
5*B*x)) - (3*a^(5/2)*(-8*A*b + 5*a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(192*b^(7/2))

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Maple [A]  time = 0.011, size = 218, normalized size = 1.4 \begin{align*}{\frac{1}{384}\sqrt{x}\sqrt{bx+a} \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 ) }+16\,B{x}^{2}a{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+32\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}xa-20\,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}-15\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{4}+30\,B\sqrt{x \left ( bx+a \right ) }\sqrt{b}{a}^{3} \right ){b}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*x^(1/2)*(b*x+a)^(1/2)/b^(7/2)*(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)+16
*B*x^2*a*b^(5/2)*(x*(b*x+a))^(1/2)+32*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a-20*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^2-15*B*ln(1/2*(
2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^4+30*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(x^(3/2)*(B*x+A)*(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.16649, size = 599, normalized size = 3.77 \begin{align*} \left [-\frac{3 \,{\left (5 \, 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} + 15 \, B a^{3} b - 24 \, A a^{2} b^{2} + 8 \,{\left (B a b^{3} + 8 \, A b^{4}\right )} x^{2} - 2 \,{\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{384 \, b^{4}}, \frac{3 \,{\left (5 \, 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} + 15 \, B a^{3} b - 24 \, A a^{2} b^{2} + 8 \,{\left (B a b^{3} + 8 \, A b^{4}\right )} x^{2} - 2 \,{\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{192 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(3*(5*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 +
15*B*a^3*b - 24*A*a^2*b^2 + 8*(B*a*b^3 + 8*A*b^4)*x^2 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/
b^4, 1/192*(3*(5*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 + 15*B
*a^3*b - 24*A*a^2*b^2 + 8*(B*a*b^3 + 8*A*b^4)*x^2 - 2*(5*B*a^2*b^2 - 8*A*a*b^3)*x)*sqrt(b*x + a)*sqrt(x))/b^4]

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Sympy [C]  time = 49.2079, size = 1535, normalized size = 9.65 \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)**(1/2),x)

[Out]

-2*A*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)), Ab
s(a + b*x)/Abs(a) > 1), (-I*a**(3/2)*sqrt(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(sqrt(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*A*Piecewise((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)*sqrt(-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 + 2*B*a**2*Piecewise((a**(3/2)*sqrt(a + b*x)/(8*sq
rt(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)*sqrt(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(sqrt(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**3 - 4*B*a*Piecewise(
(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)*sqrt(-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**3 + 2*B*Piecewise((5*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt(b*x/a)) - 5*a**(5/2)*(a + b*x)**(3
/2)/(384*sqrt(b)*sqrt(b*x/a)) - a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt(b*x/a)) - 7*sqrt(a)*(a + b*x)**(7/
2)/(48*sqrt(b)*sqrt(b*x/a)) - 5*a**4*acosh(sqrt(a + b*x)/sqrt(a))/(128*sqrt(b)) + (a + b*x)**(9/2)/(8*sqrt(a)*
sqrt(b)*sqrt(b*x/a)), Abs(a + b*x)/Abs(a) > 1), (-5*I*a**(7/2)*sqrt(a + b*x)/(128*sqrt(b)*sqrt(-b*x/a)) + 5*I*
a**(5/2)*(a + b*x)**(3/2)/(384*sqrt(b)*sqrt(-b*x/a)) + I*a**(3/2)*(a + b*x)**(5/2)/(192*sqrt(b)*sqrt(-b*x/a))
+ 7*I*sqrt(a)*(a + b*x)**(7/2)/(48*sqrt(b)*sqrt(-b*x/a)) + 5*I*a**4*asin(sqrt(a + b*x)/sqrt(a))/(128*sqrt(b))
- I*(a + b*x)**(9/2)/(8*sqrt(a)*sqrt(b)*sqrt(-b*x/a)), True))/b**3

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

[Out]

Timed out