3.467 \(\int \frac{(c+d x)^{3/2}}{x^2 (a+b x)^2} \, dx\)

Optimal. Leaf size=149 \[ -\frac{\sqrt{c+d x} (2 b c-a d)}{a^2 (a+b x)}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 \sqrt{b}}-\frac{c \sqrt{c+d x}}{a x (a+b x)} \]

[Out]

-(((2*b*c - a*d)*Sqrt[c + d*x])/(a^2*(a + b*x))) - (c*Sqrt[c + d*x])/(a*x*(a + b*x)) + (Sqrt[c]*(4*b*c - 3*a*d
)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^3 - (Sqrt[b*c - a*d]*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*
c - a*d]])/(a^3*Sqrt[b])

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Rubi [A]  time = 0.195907, antiderivative size = 149, 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 = {98, 151, 156, 63, 208} \[ -\frac{\sqrt{c+d x} (2 b c-a d)}{a^2 (a+b x)}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 \sqrt{b}}-\frac{c \sqrt{c+d x}}{a x (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(3/2)/(x^2*(a + b*x)^2),x]

[Out]

-(((2*b*c - a*d)*Sqrt[c + d*x])/(a^2*(a + b*x))) - (c*Sqrt[c + d*x])/(a*x*(a + b*x)) + (Sqrt[c]*(4*b*c - 3*a*d
)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a^3 - (Sqrt[b*c - a*d]*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*
c - a*d]])/(a^3*Sqrt[b])

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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{(c+d x)^{3/2}}{x^2 (a+b x)^2} \, dx &=-\frac{c \sqrt{c+d x}}{a x (a+b x)}-\frac{\int \frac{\frac{1}{2} c (4 b c-3 a d)+\frac{1}{2} d (3 b c-2 a d) x}{x (a+b x)^2 \sqrt{c+d x}} \, dx}{a}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x}}{a^2 (a+b x)}-\frac{c \sqrt{c+d x}}{a x (a+b x)}-\frac{\int \frac{\frac{1}{2} c (4 b c-3 a d) (b c-a d)+\frac{1}{2} d (b c-a d) (2 b c-a d) x}{x (a+b x) \sqrt{c+d x}} \, dx}{a^2 (b c-a d)}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x}}{a^2 (a+b x)}-\frac{c \sqrt{c+d x}}{a x (a+b x)}-\frac{(c (4 b c-3 a d)) \int \frac{1}{x \sqrt{c+d x}} \, dx}{2 a^3}+\frac{((b c-a d) (4 b c-a d)) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 a^3}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x}}{a^2 (a+b x)}-\frac{c \sqrt{c+d x}}{a x (a+b x)}-\frac{(c (4 b c-3 a d)) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^3 d}+\frac{((b c-a d) (4 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^3 d}\\ &=-\frac{(2 b c-a d) \sqrt{c+d x}}{a^2 (a+b x)}-\frac{c \sqrt{c+d x}}{a x (a+b x)}+\frac{\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.206961, size = 127, normalized size = 0.85 \[ \frac{\frac{a \sqrt{c+d x} (-a c+a d x-2 b c x)}{x (a+b x)}+\sqrt{c} (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )-\frac{\sqrt{b c-a d} (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(3/2)/(x^2*(a + b*x)^2),x]

[Out]

((a*Sqrt[c + d*x]*(-(a*c) - 2*b*c*x + a*d*x))/(x*(a + b*x)) + Sqrt[c]*(4*b*c - 3*a*d)*ArcTanh[Sqrt[c + d*x]/Sq
rt[c]] - (Sqrt[b*c - a*d]*(4*b*c - a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b])/a^3

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Maple [A]  time = 0.016, size = 237, normalized size = 1.6 \begin{align*} -{\frac{c}{{a}^{2}x}\sqrt{dx+c}}-3\,{\frac{d\sqrt{c}}{{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+4\,{\frac{{c}^{3/2}b}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+{\frac{{d}^{2}}{a \left ( bdx+ad \right ) }\sqrt{dx+c}}-{\frac{bdc}{{a}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{{d}^{2}}{a}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}-5\,{\frac{bdc}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+4\,{\frac{{b}^{2}{c}^{2}}{{a}^{3}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(3/2)/x^2/(b*x+a)^2,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.39628, size = 1679, normalized size = 11.27 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((4*b^2*c - a*b*d)*x^2 + (4*a*b*c - a^2*d)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x
+ c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + ((4*b^2*c - 3*a*b*d)*x^2 + (4*a*b*c - 3*a^2*d)*x)*sqrt(c)*log((d*x -
2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(a^2*c + (2*a*b*c - a^2*d)*x)*sqrt(d*x + c))/(a^3*b*x^2 + a^4*x), -1/2*
(2*((4*b^2*c - a*b*d)*x^2 + (4*a*b*c - a^2*d)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d
)/b)/(b*c - a*d)) + ((4*b^2*c - 3*a*b*d)*x^2 + (4*a*b*c - 3*a^2*d)*x)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(
c) + 2*c)/x) + 2*(a^2*c + (2*a*b*c - a^2*d)*x)*sqrt(d*x + c))/(a^3*b*x^2 + a^4*x), -1/2*(2*((4*b^2*c - 3*a*b*d
)*x^2 + (4*a*b*c - 3*a^2*d)*x)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + ((4*b^2*c - a*b*d)*x^2 + (4*a*b*c -
a^2*d)*x)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) +
2*(a^2*c + (2*a*b*c - a^2*d)*x)*sqrt(d*x + c))/(a^3*b*x^2 + a^4*x), -(((4*b^2*c - a*b*d)*x^2 + (4*a*b*c - a^2*
d)*x)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + ((4*b^2*c - 3*a*b*d)*x^
2 + (4*a*b*c - 3*a^2*d)*x)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (a^2*c + (2*a*b*c - a^2*d)*x)*sqrt(d*x
+ c))/(a^3*b*x^2 + a^4*x)]

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Sympy [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((d*x+c)**(3/2)/x**2/(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.24907, size = 266, normalized size = 1.79 \begin{align*} \frac{{\left (4 \, b^{2} c^{2} - 5 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{3}} - \frac{{\left (4 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c}} - \frac{2 \,{\left (d x + c\right )}^{\frac{3}{2}} b c d - 2 \, \sqrt{d x + c} b c^{2} d -{\left (d x + c\right )}^{\frac{3}{2}} a d^{2} + 2 \, \sqrt{d x + c} a c d^{2}}{{\left ({\left (d x + c\right )}^{2} b - 2 \,{\left (d x + c\right )} b c + b c^{2} +{\left (d x + c\right )} a d - a c d\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3) - (4
*b*c^2 - 3*a*c*d)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c)) - (2*(d*x + c)^(3/2)*b*c*d - 2*sqrt(d*x + c)*b
*c^2*d - (d*x + c)^(3/2)*a*d^2 + 2*sqrt(d*x + c)*a*c*d^2)/(((d*x + c)^2*b - 2*(d*x + c)*b*c + b*c^2 + (d*x + c
)*a*d - a*c*d)*a^2)