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

Optimal. Leaf size=118 \[ -a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{(a+b x)^{5/2} (2 a B+5 A b)}{5 a}+\frac{1}{3} (a+b x)^{3/2} (2 a B+5 A b)+a \sqrt{a+b x} (2 a B+5 A b)-\frac{A (a+b x)^{7/2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0532098, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {78, 50, 63, 208} \[ -a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )+\frac{(a+b x)^{5/2} (2 a B+5 A b)}{5 a}+\frac{1}{3} (a+b x)^{3/2} (2 a B+5 A b)+a \sqrt{a+b x} (2 a B+5 A b)-\frac{A (a+b x)^{7/2}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0599408, size = 91, normalized size = 0.77 \[ \frac{\sqrt{a+b x} \left (a^2 (46 B x-15 A)+2 a b x (35 A+11 B x)+2 b^2 x^2 (5 A+3 B x)\right )}{15 x}-a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x]*(2*b^2*x^2*(5*A + 3*B*x) + 2*a*b*x*(35*A + 11*B*x) + a^2*(-15*A + 46*B*x)))/(15*x) - a^(3/2)*(5
*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]

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Maple [A]  time = 0.01, size = 104, normalized size = 0.9 \begin{align*}{\frac{2\,B}{5} \left ( bx+a \right ) ^{{\frac{5}{2}}}}+{\frac{2\,Ab}{3} \left ( bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{2\,Ba}{3} \left ( bx+a \right ) ^{{\frac{3}{2}}}}+4\,abA\sqrt{bx+a}+2\,{a}^{2}B\sqrt{bx+a}+2\,{a}^{2} \left ( -1/2\,{\frac{A\sqrt{bx+a}}{x}}-1/2\,{\frac{5\,Ab+2\,Ba}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*B*(b*x+a)^(5/2)+2/3*A*b*(b*x+a)^(3/2)+2/3*B*(b*x+a)^(3/2)*a+4*a*b*A*(b*x+a)^(1/2)+2*a^2*B*(b*x+a)^(1/2)+2*
a^2*(-1/2*A*(b*x+a)^(1/2)/x-1/2*(5*A*b+2*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.35086, size = 495, normalized size = 4.19 \begin{align*} \left [\frac{15 \,{\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt{a} x \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (6 \, B b^{2} x^{3} - 15 \, A a^{2} + 2 \,{\left (11 \, B a b + 5 \, A b^{2}\right )} x^{2} + 2 \,{\left (23 \, B a^{2} + 35 \, A a b\right )} x\right )} \sqrt{b x + a}}{30 \, x}, \frac{15 \,{\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (6 \, B b^{2} x^{3} - 15 \, A a^{2} + 2 \,{\left (11 \, B a b + 5 \, A b^{2}\right )} x^{2} + 2 \,{\left (23 \, B a^{2} + 35 \, A a b\right )} x\right )} \sqrt{b x + a}}{15 \, 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^2,x, algorithm="fricas")

[Out]

[1/30*(15*(2*B*a^2 + 5*A*a*b)*sqrt(a)*x*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(6*B*b^2*x^3 - 15*A*a
^2 + 2*(11*B*a*b + 5*A*b^2)*x^2 + 2*(23*B*a^2 + 35*A*a*b)*x)*sqrt(b*x + a))/x, 1/15*(15*(2*B*a^2 + 5*A*a*b)*sq
rt(-a)*x*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (6*B*b^2*x^3 - 15*A*a^2 + 2*(11*B*a*b + 5*A*b^2)*x^2 + 2*(23*B*a^2
+ 35*A*a*b)*x)*sqrt(b*x + a))/x]

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Sympy [A]  time = 24.2532, size = 267, normalized size = 2.26 \begin{align*} - \frac{A a^{3} b \sqrt{\frac{1}{a^{3}}} \log{\left (- a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{A a^{3} b \sqrt{\frac{1}{a^{3}}} \log{\left (a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{6 A a^{2} b \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - \frac{A a^{2} \sqrt{a + b x}}{x} + 4 A a b \sqrt{a + b x} + A b^{2} \left (\begin{cases} \sqrt{a} x & \text{for}\: b = 0 \\\frac{2 \left (a + b x\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) + \frac{2 B a^{3} \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + 2 B a^{2} \sqrt{a + b x} + 2 B a b \left (\begin{cases} \sqrt{a} x & \text{for}\: b = 0 \\\frac{2 \left (a + b x\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\right ) - \frac{2 B a \left (a + b x\right )^{\frac{3}{2}}}{3} + \frac{2 B \left (a + b x\right )^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*a**3*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + A*a**3*b*sqrt(a**(-3))*log(a**2*sqrt(a**(
-3)) + sqrt(a + b*x))/2 + 6*A*a**2*b*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - A*a**2*sqrt(a + b*x)/x + 4*A*a*b*
sqrt(a + b*x) + A*b**2*Piecewise((sqrt(a)*x, Eq(b, 0)), (2*(a + b*x)**(3/2)/(3*b), True)) + 2*B*a**3*atan(sqrt
(a + b*x)/sqrt(-a))/sqrt(-a) + 2*B*a**2*sqrt(a + b*x) + 2*B*a*b*Piecewise((sqrt(a)*x, Eq(b, 0)), (2*(a + b*x)*
*(3/2)/(3*b), True)) - 2*B*a*(a + b*x)**(3/2)/3 + 2*B*(a + b*x)**(5/2)/5

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Giac [A]  time = 1.18419, size = 169, normalized size = 1.43 \begin{align*} \frac{6 \,{\left (b x + a\right )}^{\frac{5}{2}} B b + 10 \,{\left (b x + a\right )}^{\frac{3}{2}} B a b + 30 \, \sqrt{b x + a} B a^{2} b + 10 \,{\left (b x + a\right )}^{\frac{3}{2}} A b^{2} + 60 \, \sqrt{b x + a} A a b^{2} - \frac{15 \, \sqrt{b x + a} A a^{2} b}{x} + \frac{15 \,{\left (2 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(6*(b*x + a)^(5/2)*B*b + 10*(b*x + a)^(3/2)*B*a*b + 30*sqrt(b*x + a)*B*a^2*b + 10*(b*x + a)^(3/2)*A*b^2 +
60*sqrt(b*x + a)*A*a*b^2 - 15*sqrt(b*x + a)*A*a^2*b/x + 15*(2*B*a^3*b + 5*A*a^2*b^2)*arctan(sqrt(b*x + a)/sqr
t(-a))/sqrt(-a))/b