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

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

[Out]

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

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Rubi [A]  time = 0.065605, 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} \[ \frac{5 a^3 (8 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a+b x}}\right )}{64 b^{3/2}}+\frac{5 a^2 \sqrt{x} \sqrt{a+b x} (8 A b-a B)}{64 b}+\frac{\sqrt{x} (a+b x)^{5/2} (8 A b-a B)}{24 b}+\frac{5 a \sqrt{x} (a+b x)^{3/2} (8 A b-a B)}{96 b}+\frac{B \sqrt{x} (a+b x)^{7/2}}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.232588, size = 126, normalized size = 0.79 \[ \frac{\sqrt{a+b x} \left (\sqrt{b} \sqrt{x} \left (2 a^2 b (132 A+59 B x)+15 a^3 B+8 a b^2 x (26 A+17 B x)+16 b^3 x^2 (4 A+3 B x)\right )-\frac{15 a^{5/2} (a B-8 A b) \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{\sqrt{\frac{b x}{a}+1}}\right )}{192 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(Sqrt[b]*Sqrt[x]*(15*a^3*B + 16*b^3*x^2*(4*A + 3*B*x) + 8*a*b^2*x*(26*A + 17*B*x) + 2*a^2*b*(13
2*A + 59*B*x)) - (15*a^(5/2)*(-8*A*b + a*B)*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/Sqrt[1 + (b*x)/a]))/(192*b^(3/
2))

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Maple [A]  time = 0.01, size = 218, normalized size = 1.4 \begin{align*}{\frac{1}{384}\sqrt{bx+a}\sqrt{x} \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 ) }+272\,B{x}^{2}a{b}^{5/2}\sqrt{x \left ( bx+a \right ) }+416\,A\sqrt{x \left ( bx+a \right ) }{b}^{5/2}xa+236\,B\sqrt{x \left ( bx+a \right ) }{b}^{3/2}x{a}^{2}+120\,A\ln \left ( 1/2\,{\frac{2\,\sqrt{x \left ( bx+a \right ) }\sqrt{b}+2\,bx+a}{\sqrt{b}}} \right ){a}^{3}b+528\,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{3}{2}}}{\frac{1}{\sqrt{x \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(5/2)*(B*x+A)/x^(1/2),x)

[Out]

1/384*(b*x+a)^(1/2)*x^(1/2)/b^(3/2)*(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)+27
2*B*x^2*a*b^(5/2)*(x*(b*x+a))^(1/2)+416*A*(x*(b*x+a))^(1/2)*b^(5/2)*x*a+236*B*(x*(b*x+a))^(1/2)*b^(3/2)*x*a^2+
120*A*ln(1/2*(2*(x*(b*x+a))^(1/2)*b^(1/2)+2*b*x+a)/b^(1/2))*a^3*b+528*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((b*x+a)^(5/2)*(B*x+A)/x^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.61389, size = 616, normalized size = 3.87 \begin{align*} \left [-\frac{15 \,{\left (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 + 264 \, A a^{2} b^{2} + 8 \,{\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \,{\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{384 \, b^{2}}, \frac{15 \,{\left (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 + 264 \, A a^{2} b^{2} + 8 \,{\left (17 \, B a b^{3} + 8 \, A b^{4}\right )} x^{2} + 2 \,{\left (59 \, B a^{2} b^{2} + 104 \, A a b^{3}\right )} x\right )} \sqrt{b x + a} \sqrt{x}}{192 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(5/2)*(B*x+A)/x^(1/2),x, algorithm="fricas")

[Out]

[-1/384*(15*(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 + 1
5*B*a^3*b + 264*A*a^2*b^2 + 8*(17*B*a*b^3 + 8*A*b^4)*x^2 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x)*sqrt(b*x + a)*sqr
t(x))/b^2, 1/192*(15*(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 + 264*A*a^2*b^2 + 8*(17*B*a*b^3 + 8*A*b^4)*x^2 + 2*(59*B*a^2*b^2 + 104*A*a*b^3)*x)*sqrt(b*x + a)*s
qrt(x))/b^2]

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Sympy [A]  time = 130.176, size = 262, normalized size = 1.65 \begin{align*} A \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 ) + B \left (\frac{5 a^{\frac{7}{2}} \sqrt{x}}{64 b \sqrt{1 + \frac{b x}{a}}} + \frac{133 a^{\frac{5}{2}} x^{\frac{3}{2}}}{192 \sqrt{1 + \frac{b x}{a}}} + \frac{127 a^{\frac{3}{2}} b x^{\frac{5}{2}}}{96 \sqrt{1 + \frac{b x}{a}}} + \frac{23 \sqrt{a} b^{2} x^{\frac{7}{2}}}{24 \sqrt{1 + \frac{b x}{a}}} - \frac{5 a^{4} \operatorname{asinh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{64 b^{\frac{3}{2}}} + \frac{b^{3} x^{\frac{9}{2}}}{4 \sqrt{a} \sqrt{1 + \frac{b x}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(5/2)*(B*x+A)/x**(1/2),x)

[Out]

A*(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)*s
qrt(1 + b*x/a)/3 + 5*a**3*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(8*sqrt(b))) + B*(5*a**(7/2)*sqrt(x)/(64*b*sqrt(1 + b
*x/a)) + 133*a**(5/2)*x**(3/2)/(192*sqrt(1 + b*x/a)) + 127*a**(3/2)*b*x**(5/2)/(96*sqrt(1 + b*x/a)) + 23*sqrt(
a)*b**2*x**(7/2)/(24*sqrt(1 + b*x/a)) - 5*a**4*asinh(sqrt(b)*sqrt(x)/sqrt(a))/(64*b**(3/2)) + b**3*x**(9/2)/(4
*sqrt(a)*sqrt(1 + b*x/a)))

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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^(1/2),x, algorithm="giac")

[Out]

Timed out