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

Optimal. Leaf size=107 \[ -\frac{(a+b x)^{3/2} (4 a B+A b)}{4 a x}+\frac{3 b \sqrt{a+b x} (4 a B+A b)}{4 a}-\frac{3 b (4 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{A (a+b x)^{5/2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.04583, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ -\frac{(a+b x)^{3/2} (4 a B+A b)}{4 a x}+\frac{3 b \sqrt{a+b x} (4 a B+A b)}{4 a}-\frac{3 b (4 a B+A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 \sqrt{a}}-\frac{A (a+b x)^{5/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.0232135, size = 55, normalized size = 0.51 \[ \frac{(a+b x)^{5/2} \left (b x^2 (4 a B+A b) \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};\frac{b x}{a}+1\right )-5 a^2 A\right )}{10 a^3 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(5/2)*(-5*a^2*A + b*(A*b + 4*a*B)*x^2*Hypergeometric2F1[2, 5/2, 7/2, 1 + (b*x)/a]))/(10*a^3*x^2)

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Maple [A]  time = 0.012, size = 84, normalized size = 0.8 \begin{align*} 2\,b \left ( B\sqrt{bx+a}+{\frac{ \left ( -5/8\,Ab-1/2\,Ba \right ) \left ( bx+a \right ) ^{3/2}+ \left ( 1/2\,B{a}^{2}+3/8\,Aba \right ) \sqrt{bx+a}}{{b}^{2}{x}^{2}}}-3/8\,{\frac{Ab+4\,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)^(3/2)*(B*x+A)/x^3,x)

[Out]

2*b*(B*(b*x+a)^(1/2)+((-5/8*A*b-1/2*B*a)*(b*x+a)^(3/2)+(1/2*B*a^2+3/8*A*b*a)*(b*x+a)^(1/2))/b^2/x^2-3/8*(A*b+4
*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)^(3/2)*(B*x+A)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.44013, size = 414, normalized size = 3.87 \begin{align*} \left [\frac{3 \,{\left (4 \, B a b + 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 a b x^{2} - 2 \, A a^{2} -{\left (4 \, B a^{2} + 5 \, A a b\right )} x\right )} \sqrt{b x + a}}{8 \, a x^{2}}, \frac{3 \,{\left (4 \, B a b + A b^{2}\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) +{\left (8 \, B a b x^{2} - 2 \, A a^{2} -{\left (4 \, B a^{2} + 5 \, A a b\right )} x\right )} \sqrt{b x + a}}{4 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 44.2251, size = 428, normalized size = 4. \begin{align*} - \frac{10 A a^{3} 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^{2} 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^{2} 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^{5}}} \log{\left (a^{3} \sqrt{\frac{1}{a^{5}}} + \sqrt{a + b x} \right )}}{8} - A a b^{2} \sqrt{\frac{1}{a^{3}}} \log{\left (- a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )} + A a b^{2} \sqrt{\frac{1}{a^{3}}} \log{\left (a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )} + \frac{2 A b^{2} \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - \frac{2 A b \sqrt{a + b x}}{x} - \frac{B a^{2} b \sqrt{\frac{1}{a^{3}}} \log{\left (- a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{B a^{2} b \sqrt{\frac{1}{a^{3}}} \log{\left (a^{2} \sqrt{\frac{1}{a^{3}}} + \sqrt{a + b x} \right )}}{2} + \frac{4 B a b \operatorname{atan}{\left (\frac{\sqrt{a + b x}}{\sqrt{- a}} \right )}}{\sqrt{- a}} - \frac{B a \sqrt{a + b x}}{x} + 2 B b \sqrt{a + b x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-10*A*a**3*b**2*sqrt(a + b*x)/(-8*a**4 - 16*a**3*b*x + 8*a**2*(a + b*x)**2) + 6*A*a**2*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**2*b**2*sqrt(a**(-5))*log(-a**3*sqrt(a**(-5)) + sqrt(a +
b*x))/8 - 3*A*a**2*b**2*sqrt(a**(-5))*log(a**3*sqrt(a**(-5)) + sqrt(a + b*x))/8 - A*a*b**2*sqrt(a**(-3))*log(-
a**2*sqrt(a**(-3)) + sqrt(a + b*x)) + A*a*b**2*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x)) + 2*A*b**
2*atan(sqrt(a + b*x)/sqrt(-a))/sqrt(-a) - 2*A*b*sqrt(a + b*x)/x - B*a**2*b*sqrt(a**(-3))*log(-a**2*sqrt(a**(-3
)) + sqrt(a + b*x))/2 + B*a**2*b*sqrt(a**(-3))*log(a**2*sqrt(a**(-3)) + sqrt(a + b*x))/2 + 4*B*a*b*atan(sqrt(a
+ b*x)/sqrt(-a))/sqrt(-a) - B*a*sqrt(a + b*x)/x + 2*B*b*sqrt(a + b*x)

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Giac [A]  time = 1.2423, size = 161, normalized size = 1.5 \begin{align*} \frac{8 \, \sqrt{b x + a} B b^{2} + \frac{3 \,{\left (4 \, B a b^{2} + A b^{3}\right )} \arctan \left (\frac{\sqrt{b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{4 \,{\left (b x + a\right )}^{\frac{3}{2}} B a b^{2} - 4 \, \sqrt{b x + a} B a^{2} b^{2} + 5 \,{\left (b x + a\right )}^{\frac{3}{2}} A b^{3} - 3 \, \sqrt{b x + a} A a b^{3}}{b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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