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

Optimal. Leaf size=133 \[ -\frac{(a+b x)^{5/2} (4 a B+3 A b)}{4 a x}+\frac{5 b (a+b x)^{3/2} (4 a B+3 A b)}{12 a}+\frac{5}{4} b \sqrt{a+b x} (4 a B+3 A b)-\frac{5}{4} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{A (a+b x)^{7/2}}{2 a x^2} \]

[Out]

(5*b*(3*A*b + 4*a*B)*Sqrt[a + b*x])/4 + (5*b*(3*A*b + 4*a*B)*(a + b*x)^(3/2))/(12*a) - ((3*A*b + 4*a*B)*(a + b
*x)^(5/2))/(4*a*x) - (A*(a + b*x)^(7/2))/(2*a*x^2) - (5*Sqrt[a]*b*(3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a
]])/4

________________________________________________________________________________________

Rubi [A]  time = 0.0552384, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {78, 47, 50, 63, 208} \[ -\frac{(a+b x)^{5/2} (4 a B+3 A b)}{4 a x}+\frac{5 b (a+b x)^{3/2} (4 a B+3 A b)}{12 a}+\frac{5}{4} b \sqrt{a+b x} (4 a B+3 A b)-\frac{5}{4} \sqrt{a} b (4 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )-\frac{A (a+b x)^{7/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*b*(3*A*b + 4*a*B)*Sqrt[a + b*x])/4 + (5*b*(3*A*b + 4*a*B)*(a + b*x)^(3/2))/(12*a) - ((3*A*b + 4*a*B)*(a + b
*x)^(5/2))/(4*a*x) - (A*(a + b*x)^(7/2))/(2*a*x^2) - (5*Sqrt[a]*b*(3*A*b + 4*a*B)*ArcTanh[Sqrt[a + b*x]/Sqrt[a
]])/4

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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^3} \, dx &=-\frac{A (a+b x)^{7/2}}{2 a x^2}+\frac{\left (\frac{3 A b}{2}+2 a B\right ) \int \frac{(a+b x)^{5/2}}{x^2} \, dx}{2 a}\\ &=-\frac{(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac{A (a+b x)^{7/2}}{2 a x^2}+\frac{(5 b (3 A b+4 a B)) \int \frac{(a+b x)^{3/2}}{x} \, dx}{8 a}\\ &=\frac{5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac{(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac{A (a+b x)^{7/2}}{2 a x^2}+\frac{1}{8} (5 b (3 A b+4 a B)) \int \frac{\sqrt{a+b x}}{x} \, dx\\ &=\frac{5}{4} b (3 A b+4 a B) \sqrt{a+b x}+\frac{5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac{(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac{A (a+b x)^{7/2}}{2 a x^2}+\frac{1}{8} (5 a b (3 A b+4 a B)) \int \frac{1}{x \sqrt{a+b x}} \, dx\\ &=\frac{5}{4} b (3 A b+4 a B) \sqrt{a+b x}+\frac{5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac{(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac{A (a+b x)^{7/2}}{2 a x^2}+\frac{1}{4} (5 a (3 A b+4 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )\\ &=\frac{5}{4} b (3 A b+4 a B) \sqrt{a+b x}+\frac{5 b (3 A b+4 a B) (a+b x)^{3/2}}{12 a}-\frac{(3 A b+4 a B) (a+b x)^{5/2}}{4 a x}-\frac{A (a+b x)^{7/2}}{2 a x^2}-\frac{5}{4} \sqrt{a} b (3 A b+4 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0258096, size = 56, normalized size = 0.42 \[ \frac{(a+b x)^{7/2} \left (b x^2 (4 a B+3 A b) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};\frac{b x}{a}+1\right )-7 a^2 A\right )}{14 a^3 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(7/2)*(-7*a^2*A + b*(3*A*b + 4*a*B)*x^2*Hypergeometric2F1[2, 7/2, 9/2, 1 + (b*x)/a]))/(14*a^3*x^2)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 110, normalized size = 0.8 \begin{align*} 2\,b \left ( 1/3\,B \left ( bx+a \right ) ^{3/2}+Ab\sqrt{bx+a}+2\,Ba\sqrt{bx+a}+a \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( \left ( -{\frac{9\,Ab}{8}}-1/2\,Ba \right ) \left ( bx+a \right ) ^{3/2}+ \left ({\frac{7\,Aba}{8}}+1/2\,B{a}^{2} \right ) \sqrt{bx+a} \right ) }-5/8\,{\frac{3\,Ab+4\,Ba}{\sqrt{a}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \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^3,x)

[Out]

2*b*(1/3*B*(b*x+a)^(3/2)+A*b*(b*x+a)^(1/2)+2*B*a*(b*x+a)^(1/2)+a*(((-9/8*A*b-1/2*B*a)*(b*x+a)^(3/2)+(7/8*A*b*a
+1/2*B*a^2)*(b*x+a)^(1/2))/b^2/x^2-5/8*(3*A*b+4*B*a)/a^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))))

________________________________________________________________________________________

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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.46324, size = 495, normalized size = 3.72 \begin{align*} \left [\frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{a} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt{b x + a}}{24 \, x^{2}}, \frac{15 \,{\left (4 \, B a b + 3 \, A b^{2}\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (8 \, B b^{2} x^{3} - 6 \, A a^{2} + 8 \,{\left (7 \, B a b + 3 \, A b^{2}\right )} x^{2} - 3 \,{\left (4 \, B a^{2} + 9 \, A a b\right )} x\right )} \sqrt{b x + a}}{12 \, x^{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^3,x, algorithm="fricas")

[Out]

[1/24*(15*(4*B*a*b + 3*A*b^2)*sqrt(a)*x^2*log((b*x - 2*sqrt(b*x + a)*sqrt(a) + 2*a)/x) + 2*(8*B*b^2*x^3 - 6*A*
a^2 + 8*(7*B*a*b + 3*A*b^2)*x^2 - 3*(4*B*a^2 + 9*A*a*b)*x)*sqrt(b*x + a))/x^2, 1/12*(15*(4*B*a*b + 3*A*b^2)*sq
rt(-a)*x^2*arctan(sqrt(b*x + a)*sqrt(-a)/a) + (8*B*b^2*x^3 - 6*A*a^2 + 8*(7*B*a*b + 3*A*b^2)*x^2 - 3*(4*B*a^2
+ 9*A*a*b)*x)*sqrt(b*x + a))/x^2]

________________________________________________________________________________________

Sympy [A]  time = 47.2451, size = 488, normalized size = 3.67 \begin{align*} - \frac{10 A a^{4} b^{2} \sqrt{a + b x}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac{6 A a^{3} b^{2} \left (a + b x\right )^{\frac{3}{2}}}{- 8 a^{4} - 16 a^{3} b x + 8 a^{2} \left (a + b x\right )^{2}} + \frac{3 A a^{3} b^{2} \sqrt{\frac{1}{a^{5}}} \log{\left (- a^{3} \sqrt{\frac{1}{a^{5}}} + \sqrt{a + b x} \right )}}{8} - \frac{3 A a^{3} b^{2} \sqrt{\frac{1}{a^{5}}} \log{\left (a^{3} \sqrt{\frac{1}{a^{5}}} + \sqrt{a + b x} \right )}}{8} - \frac{3 A a^{2} b^{2} \sqrt{\frac{1}{a^{3}}} \log{\left (- a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{3 A a^{2} b^{2} \sqrt{\frac{1}{a^{3}}} \log{\left (a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{6 A a b^{2} \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - \frac{3 A a b \sqrt{a + b x}}{x} + 2 A b^{2} \sqrt{a + b x} - \frac{B a^{3} b \sqrt{\frac{1}{a^{3}}} \log{\left (- a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{B 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 B a^{2} b \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - \frac{B a^{2} \sqrt{a + b x}}{x} + 4 B a b \sqrt{a + b x} + B 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*A*a**4*b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*A*a**3*b**2*(a + b*x)**(3/2)/(
-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 3*A*a**3*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a +
b*x))/8 - 3*A*a**3*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - 3*A*a**2*b**2*sqrt(a**(-3))*
log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 3*A*a**2*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x
))/2 + 6*A*a*b**2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 3*A*a*b*sqrt(a + b*x)/x + 2*A*b**2*sqrt(a + b*x) - B
*a**3*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + B*a**3*b*sqrt(a**(-3))*log(a**2*sqrt(a**(-3
)) + sqrt(a + b*x))/2 + 6*B*a**2*b*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - B*a**2*sqrt(a + b*x)/x + 4*B*a*b*sq
rt(a + b*x) + B*b**2*Piecewise((sqrt(a)*x, Eq(b, 0)), (2*(a + b*x)**(3/2)/(3*b), True))

________________________________________________________________________________________

Giac [A]  time = 1.25583, size = 209, normalized size = 1.57 \begin{align*} \frac{8 \,{\left (b x + a\right )}^{\frac{3}{2}} B b^{2} + 48 \, \sqrt{b x + a} B a b^{2} + 24 \, \sqrt{b x + a} A b^{3} + \frac{15 \,{\left (4 \, B a^{2} b^{2} + 3 \, A a b^{3}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{3 \,{\left (4 \,{\left (b x + a\right )}^{\frac{3}{2}} B a^{2} b^{2} - 4 \, \sqrt{b x + a} B a^{3} b^{2} + 9 \,{\left (b x + a\right )}^{\frac{3}{2}} A a b^{3} - 7 \, \sqrt{b x + a} A a^{2} b^{3}\right )}}{b^{2} x^{2}}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(8*(b*x + a)^(3/2)*B*b^2 + 48*sqrt(b*x + a)*B*a*b^2 + 24*sqrt(b*x + a)*A*b^3 + 15*(4*B*a^2*b^2 + 3*A*a*b^
3)*arctan(sqrt(b*x + a)/sqrt(-a))/sqrt(-a) - 3*(4*(b*x + a)^(3/2)*B*a^2*b^2 - 4*sqrt(b*x + a)*B*a^3*b^2 + 9*(b
*x + a)^(3/2)*A*a*b^3 - 7*sqrt(b*x + a)*A*a^2*b^3)/(b^2*x^2))/b