3.505 \(\int \frac{(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 \sqrt{b}}+\frac{\sqrt{x} (a+b x)^{5/2} (a B+6 A b)}{3 a}+\frac{5}{12} \sqrt{x} (a+b x)^{3/2} (a B+6 A b)+\frac{5}{8} a \sqrt{x} \sqrt{a+b x} (a B+6 A b)-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}} \]

[Out]

(5*a*(6*A*b + a*B)*Sqrt[x]*Sqrt[a + b*x])/8 + (5*(6*A*b + a*B)*Sqrt[x]*(a + b*x)^(3/2))/12 + ((6*A*b + a*B)*Sq
rt[x]*(a + b*x)^(5/2))/(3*a) - (2*A*(a + b*x)^(7/2))/(a*Sqrt[x]) + (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[
x])/Sqrt[a + b*x]])/(8*Sqrt[b])

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Rubi [A]  time = 0.0626256, antiderivative size = 144, 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 = {78, 50, 63, 217, 206} \[ \frac{5 a^2 (a B+6 A b) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 \sqrt{b}}+\frac{\sqrt{x} (a+b x)^{5/2} (a B+6 A b)}{3 a}+\frac{5}{12} \sqrt{x} (a+b x)^{3/2} (a B+6 A b)+\frac{5}{8} a \sqrt{x} \sqrt{a+b x} (a B+6 A b)-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^(5/2)*(A + B*x))/x^(3/2),x]

[Out]

(5*a*(6*A*b + a*B)*Sqrt[x]*Sqrt[a + b*x])/8 + (5*(6*A*b + a*B)*Sqrt[x]*(a + b*x)^(3/2))/12 + ((6*A*b + a*B)*Sq
rt[x]*(a + b*x)^(5/2))/(3*a) - (2*A*(a + b*x)^(7/2))/(a*Sqrt[x]) + (5*a^2*(6*A*b + a*B)*ArcTanh[(Sqrt[b]*Sqrt[
x])/Sqrt[a + b*x]])/(8*Sqrt[b])

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 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 \frac{(a+b x)^{5/2} (A+B x)}{x^{3/2}} \, dx &=-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{\left (2 \left (3 A b+\frac{a B}{2}\right )\right ) \int \frac{(a+b x)^{5/2}}{\sqrt{x}} \, dx}{a}\\ &=\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{1}{6} (5 (6 A b+a B)) \int \frac{(a+b x)^{3/2}}{\sqrt{x}} \, dx\\ &=\frac{5}{12} (6 A b+a B) \sqrt{x} (a+b x)^{3/2}+\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{1}{8} (5 a (6 A b+a B)) \int \frac{\sqrt{a+b x}}{\sqrt{x}} \, dx\\ &=\frac{5}{8} a (6 A b+a B) \sqrt{x} \sqrt{a+b x}+\frac{5}{12} (6 A b+a B) \sqrt{x} (a+b x)^{3/2}+\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{1}{16} \left (5 a^2 (6 A b+a B)\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx\\ &=\frac{5}{8} a (6 A b+a B) \sqrt{x} \sqrt{a+b x}+\frac{5}{12} (6 A b+a B) \sqrt{x} (a+b x)^{3/2}+\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{1}{8} \left (5 a^2 (6 A b+a B)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{5}{8} a (6 A b+a B) \sqrt{x} \sqrt{a+b x}+\frac{5}{12} (6 A b+a B) \sqrt{x} (a+b x)^{3/2}+\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{1}{8} \left (5 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 )\\ &=\frac{5}{8} a (6 A b+a B) \sqrt{x} \sqrt{a+b x}+\frac{5}{12} (6 A b+a B) \sqrt{x} (a+b x)^{3/2}+\frac{(6 A b+a B) \sqrt{x} (a+b x)^{5/2}}{3 a}-\frac{2 A (a+b x)^{7/2}}{a \sqrt{x}}+\frac{5 a^2 (6 A b+a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{8 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.232181, size = 111, normalized size = 0.77 \[ \frac{1}{24} \sqrt{a+b x} \left (\frac{a^2 (33 B x-48 A)+2 a b x (27 A+13 B x)+4 b^2 x^2 (3 A+2 B x)}{\sqrt{x}}+\frac{15 a^{3/2} (a B+6 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{\frac{b x}{a}+1}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^(5/2)*(A + B*x))/x^(3/2),x]

[Out]

(Sqrt[a + b*x]*((4*b^2*x^2*(3*A + 2*B*x) + 2*a*b*x*(27*A + 13*B*x) + a^2*(-48*A + 33*B*x))/Sqrt[x] + (15*a^(3/
2)*(6*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x)/a])))/24

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Maple [A]  time = 0.013, size = 202, normalized size = 1.4 \begin{align*}{\frac{1}{48}\sqrt{bx+a} \left ( 16\,B\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{3}+24\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}{x}^{2}+52\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}{x}^{2}a+90\,Ab\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{2}+108\,Axa{b}^{3/2}\sqrt{x \left ( bx+a \right ) }+15\,B\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ) x{a}^{3}+66\,Bx{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b}-96\,A{a}^{2}\sqrt{x \left ( bx+a \right ) }\sqrt{b} \right ){\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x)

[Out]

1/48*(b*x+a)^(1/2)*(16*B*(x*(b*x+a))^(1/2)*b^(5/2)*x^3+24*A*(x*(b*x+a))^(1/2)*b^(5/2)*x^2+52*B*(x*(b*x+a))^(1/
2)*b^(3/2)*x^2*a+90*A*b*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^2+108*A*x*a*b^(3/2)*(x*(b*x+
a))^(1/2)+15*B*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*x*a^3+66*B*x*a^2*(x*(b*x+a))^(1/2)*b^(1/2
)-96*A*a^2*(x*(b*x+a))^(1/2)*b^(1/2))/x^(1/2)/(x*(b*x+a))^(1/2)/b^(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)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.74425, size = 570, normalized size = 3.96 \begin{align*} \left [\frac{15 \,{\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt{b} x \log \left (2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b} \sqrt{x} + a\right ) + 2 \,{\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \,{\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \,{\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{48 \, b x}, -\frac{15 \,{\left (B a^{3} + 6 \, A a^{2} b\right )} \sqrt{-b} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-b}}{b \sqrt{x}}\right ) -{\left (8 \, B b^{3} x^{3} - 48 \, A a^{2} b + 2 \,{\left (13 \, B a b^{2} + 6 \, A b^{3}\right )} x^{2} + 3 \,{\left (11 \, B a^{2} b + 18 \, A a b^{2}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{24 \, b x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="fricas")

[Out]

[1/48*(15*(B*a^3 + 6*A*a^2*b)*sqrt(b)*x*log(2*b*x + 2*sqrt(b*x + a)*sqrt(b)*sqrt(x) + a) + 2*(8*B*b^3*x^3 - 48
*A*a^2*b + 2*(13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x), -1/24*(
15*(B*a^3 + 6*A*a^2*b)*sqrt(-b)*x*arctan(sqrt(b*x + a)*sqrt(-b)/(b*sqrt(x))) - (8*B*b^3*x^3 - 48*A*a^2*b + 2*(
13*B*a*b^2 + 6*A*b^3)*x^2 + 3*(11*B*a^2*b + 18*A*a*b^2)*x)*sqrt(b*x + a)*sqrt(x))/(b*x)]

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Sympy [A]  time = 140.612, size = 233, normalized size = 1.62 \begin{align*} A \left (- \frac{2 a^{\frac{5}{2}}}{\sqrt{x} \sqrt{1 + \frac{b x}{a}}} + \frac{a^{\frac{3}{2}} b \sqrt{x}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{11 \sqrt{a} b^{2} x^{\frac{3}{2}}}{4 \sqrt{1 + \frac{b x}{a}}} + \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4} + \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 + \frac{b x}{a}}}\right ) + B \left (\frac{11 a^{\frac{5}{2}} \sqrt{x} \sqrt{1 + \frac{b x}{a}}}{8} + \frac{13 a^{\frac{3}{2}} b x^{\frac{3}{2}} \sqrt{1 + \frac{b x}{a}}}{12} + \frac{\sqrt{a} b^{2} x^{\frac{5}{2}} \sqrt{1 + \frac{b x}{a}}}{3} + \frac{5 a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{8 \sqrt{b}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(3/2),x)

[Out]

A*(-2*a**(5/2)/(sqrt(x)*sqrt(1 + b*x/a)) + a**(3/2)*b*sqrt(x)/(4*sqrt(1 + b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(
4*sqrt(1 + b*x/a)) + 15*a**2*sqrt(b)*asinh(sqrt(b)*sqrt(x)/sqrt(a))/4 + b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 + b*x/
a))) + B*(11*a**(5/2)*sqrt(x)*sqrt(1 + b*x/a)/8 + 13*a**(3/2)*b*x**(3/2)*sqrt(1 + b*x/a)/12 + sqrt(a)*b**2*x**
(5/2)*sqrt(1 + b*x/a)/3 + 5*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b)))

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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)^(5/2)*(B*x+A)/x^(3/2),x, algorithm="giac")

[Out]

Timed out