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

Optimal. Leaf size=115 \[ \frac {\sqrt {x} (a+b x)^{3/2} (a B+4 A b)}{2 a}+\frac {3}{4} \sqrt {x} \sqrt {a+b x} (a B+4 A b)+\frac {3 a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 \sqrt {b}}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}} \]

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Rubi [A]  time = 0.05, antiderivative size = 115, 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 = {78, 50, 63, 217, 206} \begin {gather*} \frac {\sqrt {x} (a+b x)^{3/2} (a B+4 A b)}{2 a}+\frac {3}{4} \sqrt {x} \sqrt {a+b x} (a B+4 A b)+\frac {3 a (a B+4 A b) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a+b x}}\right )}{4 \sqrt {b}}-\frac {2 A (a+b x)^{5/2}}{a \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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

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Mathematica [A]  time = 0.23, size = 91, normalized size = 0.79 \begin {gather*} \frac {1}{4} \sqrt {a+b x} \left (\frac {a (5 B x-8 A)+2 b x (2 A+B x)}{\sqrt {x}}+\frac {3 \sqrt {a} (a B+4 A b) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b x}{a}+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*((2*b*x*(2*A + B*x) + a*(-8*A + 5*B*x))/Sqrt[x] + (3*Sqrt[a]*(4*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqr
t[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x)/a])))/4

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IntegrateAlgebraic [A]  time = 0.25, size = 84, normalized size = 0.73 \begin {gather*} \frac {\sqrt {a+b x} \left (-8 a A+5 a B x+4 A b x+2 b B x^2\right )}{4 \sqrt {x}}-\frac {3 \left (a^2 B+4 a A b\right ) \log \left (\sqrt {a+b x}-\sqrt {b} \sqrt {x}\right )}{4 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.43, size = 179, normalized size = 1.56 \begin {gather*} \left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {b} x \log \left (2 \, b x + 2 \, \sqrt {b x + a} \sqrt {b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, B b^{2} x^{2} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{8 \, b x}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b} x \arctan \left (\frac {\sqrt {b x + a} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (2 \, B b^{2} x^{2} - 8 \, A a b + {\left (5 \, B a b + 4 \, A b^{2}\right )} x\right )} \sqrt {b x + a} \sqrt {x}}{4 \, b x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(B*a^2 + 4*A*a*b)*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(2*B*b^2*x^2 - 8*A*a*
b + (5*B*a*b + 4*A*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x), -1/4*(3*(B*a^2 + 4*A*a*b)*sqrt(-b)*x*arctan(sqrt(b*x
+ a)*sqrt(-b)/(b*sqrt(x))) - (2*B*b^2*x^2 - 8*A*a*b + (5*B*a*b + 4*A*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x)]

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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^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 158, normalized size = 1.37 \begin {gather*} \frac {\sqrt {b x +a}\, \left (12 A a b x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+3 B \,a^{2} x \ln \left (\frac {2 b x +a +2 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}}{2 \sqrt {b}}\right )+4 \sqrt {\left (b x +a \right ) x}\, B \,b^{\frac {3}{2}} x^{2}+8 \sqrt {\left (b x +a \right ) x}\, A \,b^{\frac {3}{2}} x +10 \sqrt {\left (b x +a \right ) x}\, B a \sqrt {b}\, x -16 \sqrt {\left (b x +a \right ) x}\, A a \sqrt {b}\right )}{8 \sqrt {\left (b x +a \right ) x}\, \sqrt {b}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(b*x+a)^(1/2)*(4*B*b^(3/2)*((b*x+a)*x)^(1/2)*x^2+12*A*b*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/
2))*x*a+8*((b*x+a)*x)^(1/2)*A*b^(3/2)*x+3*B*ln(1/2*(2*b*x+a+2*((b*x+a)*x)^(1/2)*b^(1/2))/b^(1/2))*x*a^2+10*((b
*x+a)*x)^(1/2)*B*a*b^(1/2)*x-16*((b*x+a)*x)^(1/2)*A*a*b^(1/2))/x^(1/2)/((b*x+a)*x)^(1/2)/b^(1/2)

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maxima [A]  time = 0.91, size = 129, normalized size = 1.12 \begin {gather*} \frac {3 \, B a^{2} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{8 \, \sqrt {b}} + \frac {3}{2} \, A a \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \frac {3}{4} \, \sqrt {b x^{2} + a x} B a + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} B}{2 \, x} - \frac {3 \, \sqrt {b x^{2} + a x} A a}{x} + \frac {{\left (b x^{2} + a x\right )}^{\frac {3}{2}} A}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{x^{3/2}} \,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^(3/2),x)

[Out]

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

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sympy [A]  time = 39.65, size = 172, normalized size = 1.50 \begin {gather*} A \left (- \frac {2 a^{\frac {3}{2}}}{\sqrt {x} \sqrt {1 + \frac {b x}{a}}} - \frac {\sqrt {a} b \sqrt {x}}{\sqrt {1 + \frac {b x}{a}}} + 3 a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )} + \frac {b^{2} x^{\frac {3}{2}}}{\sqrt {a} \sqrt {1 + \frac {b x}{a}}}\right ) + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)*(B*x+A)/x**(3/2),x)

[Out]

A*(-2*a**(3/2)/(sqrt(x)*sqrt(1 + b*x/a)) - sqrt(a)*b*sqrt(x)/sqrt(1 + b*x/a) + 3*a*sqrt(b)*asinh(sqrt(b)*sqrt(
x)/sqrt(a)) + b**2*x**(3/2)/(sqrt(a)*sqrt(1 + b*x/a))) + B*(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)))

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