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

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

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Rubi [A]  time = 0.07, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {80, 50, 63, 217, 206} \begin {gather*} \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} \end {gather*}

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 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 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 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])

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]

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*}

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Mathematica [A]  time = 0.27, size = 126, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a+b x} \left (\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}}+\sqrt {b} \sqrt {x} \left (-9 a^3 B+6 a^2 b (4 A+B x)+8 a b^2 x (14 A+9 B x)+16 b^3 x^2 (4 A+3 B x)\right )\right )}{192 b^{5/2}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.23, size = 145, normalized size = 0.91 \begin {gather*} \frac {\left (8 a^3 A b-3 a^4 B\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{64 b^{5/2}}+\frac {\sqrt {a+b x} \left (-9 a^3 B \sqrt {x}+24 a^2 A b \sqrt {x}+6 a^2 b B x^{3/2}+112 a A b^2 x^{3/2}+72 a b^2 B x^{5/2}+64 A b^3 x^{5/2}+48 b^3 B x^{7/2}\right )}{192 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(24*a^2*A*b*Sqrt[x] - 9*a^3*B*Sqrt[x] + 112*a*A*b^2*x^(3/2) + 6*a^2*b*B*x^(3/2) + 64*A*b^3*x^(5
/2) + 72*a*b^2*B*x^(5/2) + 48*b^3*B*x^(7/2)))/(192*b^2) + ((8*a^3*A*b - 3*a^4*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt
[a + b*x]])/(64*b^(5/2))

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fricas [A]  time = 1.50, size = 249, normalized size = 1.57 \begin {gather*} \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 {gather*}

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

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maple [A]  time = 0.01, size = 218, normalized size = 1.37 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (-96 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {7}{2}} x^{3}-128 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {7}{2}} x^{2}-144 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {5}{2}} x^{2}+24 A \,a^{3} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-9 B \,a^{4} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-224 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {5}{2}} x -12 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} b^{\frac {3}{2}} x -48 \sqrt {\left (b x +a \right ) x}\, A \,a^{2} b^{\frac {3}{2}}+18 \sqrt {\left (b x +a \right ) x}\, B \,a^{3} \sqrt {b}\right ) \sqrt {x}}{384 \sqrt {\left (b x +a \right ) x}\, b^{\frac {5}{2}}} \end {gather*}

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+a)*x)^(1/2)*B*b^(7/2)*x^3-128*((b*x+a)*x)^(1/2)*A*b^(7/2)*x^2-
144*((b*x+a)*x)^(1/2)*B*a*b^(5/2)*x^2-224*((b*x+a)*x)^(1/2)*A*a*b^(5/2)*x-12*((b*x+a)*x)^(1/2)*B*a^2*b^(3/2)*x
+24*A*a^3*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))-48*((b*x+a)*x)^(1/2)*A*a^2*b^(3/2)-9*B*a^4*l
n(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))+18*((b*x+a)*x)^(1/2)*B*a^3*b^(1/2))/((b*x+a)*x)^(1/2)

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maxima [B]  time = 0.84, size = 287, normalized size = 1.81 \begin {gather*} \frac {1}{4} \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B x + \frac {1}{2} \, \sqrt {b x^{2} + a x} A a x + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2} x}{32 \, b} - \frac {5 \, B a^{4} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{128 \, b^{\frac {5}{2}}} - \frac {A a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {3}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{3}}{64 \, b^{2}} - \frac {5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} B a}{24 \, b} + \frac {\sqrt {b x^{2} + a x} A a^{2}}{4 \, b} - \frac {\sqrt {b x^{2} + a x} {\left (B a + A b\right )} a x}{4 \, b} + \frac {{\left (B a + A b\right )} a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {5}{2}}} - \frac {\sqrt {b x^{2} + a x} {\left (B a + A b\right )} a^{2}}{8 \, b^{2}} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} {\left (B a + A b\right )}}{3 \, b} \end {gather*}

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]

1/4*(b*x^2 + a*x)^(3/2)*B*x + 1/2*sqrt(b*x^2 + a*x)*A*a*x + 5/32*sqrt(b*x^2 + a*x)*B*a^2*x/b - 5/128*B*a^4*log
(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/2) - 1/8*A*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(
3/2) + 5/64*sqrt(b*x^2 + a*x)*B*a^3/b^2 - 5/24*(b*x^2 + a*x)^(3/2)*B*a/b + 1/4*sqrt(b*x^2 + a*x)*A*a^2/b - 1/4
*sqrt(b*x^2 + a*x)*(B*a + A*b)*a*x/b + 1/16*(B*a + A*b)*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(5/
2) - 1/8*sqrt(b*x^2 + a*x)*(B*a + A*b)*a^2/b^2 + 1/3*(b*x^2 + a*x)^(3/2)*(B*a + A*b)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {x}\,\left (A+B\,x\right )\,{\left (a+b\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 26.20, size = 298, normalized size = 1.87 \begin {gather*} \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 {gather*}

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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