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

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

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, 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 (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} \end {gather*}

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 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 \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*}

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Mathematica [A]  time = 0.29, size = 107, normalized size = 0.85 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.20, size = 116, normalized size = 0.92 \begin {gather*} \frac {\sqrt {a+b x} \left (3 a^2 B \sqrt {x}+30 a A b \sqrt {x}+14 a b B x^{3/2}+12 A b^2 x^{3/2}+8 b^2 B x^{5/2}\right )}{24 b}+\frac {\left (a^3 B-6 a^2 A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{8 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(30*a*A*b*Sqrt[x] + 3*a^2*B*Sqrt[x] + 12*A*b^2*x^(3/2) + 14*a*b*B*x^(3/2) + 8*b^2*B*x^(5/2)))/(
24*b) + ((-6*a^2*A*b + a^3*B)*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]])/(8*b^(3/2))

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fricas [A]  time = 2.03, size = 198, normalized size = 1.57 \begin {gather*} \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 {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]

[-1/48*(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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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 = 176, normalized size = 1.40 \begin {gather*} \frac {\sqrt {b x +a}\, \left (16 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {5}{2}} x^{2}+18 A \,a^{2} b \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )-3 B \,a^{3} \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+24 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {5}{2}} x +28 \sqrt {\left (b x +a \right ) x}\, B a \,b^{\frac {3}{2}} x +60 \sqrt {\left (b x +a \right ) x}\, A a \,b^{\frac {3}{2}}+6 \sqrt {\left (b x +a \right ) x}\, B \,a^{2} \sqrt {b}\right ) \sqrt {x}}{48 \sqrt {\left (b x +a \right ) x}\, b^{\frac {3}{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]

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

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maxima [B]  time = 0.93, size = 282, normalized size = 2.24 \begin {gather*} \frac {1}{3} \, \sqrt {b x^{2} + a x} B b x^{2} - \frac {5}{12} \, \sqrt {b x^{2} + a x} B a x - \frac {5 \, B a^{3} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{16 \, b^{\frac {3}{2}}} + \frac {A a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{\sqrt {b}} + \frac {5 \, \sqrt {b x^{2} + a x} B a^{2}}{8 \, b} + \frac {{\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} x}{2 \, b} + \frac {3 \, {\left (2 \, B a b + A b^{2}\right )} a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, b^{\frac {5}{2}}} - \frac {{\left (B a^{2} + 2 \, A a b\right )} a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {3}{2}}} - \frac {3 \, {\left (2 \, B a b + A b^{2}\right )} \sqrt {b x^{2} + a x} a}{4 \, b^{2}} + \frac {{\left (B a^{2} + 2 \, A a b\right )} \sqrt {b x^{2} + a x}}{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]

1/3*sqrt(b*x^2 + a*x)*B*b*x^2 - 5/12*sqrt(b*x^2 + a*x)*B*a*x - 5/16*B*a^3*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*
sqrt(b))/b^(3/2) + A*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/sqrt(b) + 5/8*sqrt(b*x^2 + a*x)*B*a^2/b
+ 1/2*(2*B*a*b + A*b^2)*sqrt(b*x^2 + a*x)*x/b + 3/8*(2*B*a*b + A*b^2)*a^2*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*
sqrt(b))/b^(5/2) - 1/2*(B*a^2 + 2*A*a*b)*a*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b))/b^(3/2) - 3/4*(2*B*a*b
+ A*b^2)*sqrt(b*x^2 + a*x)*a/b^2 + (B*a^2 + 2*A*a*b)*sqrt(b*x^2 + a*x)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 46.99, size = 204, normalized size = 1.62 \begin {gather*} 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 {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*(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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