3.473 \(\int \frac{(c+d x)^{5/2}}{x^2 (a+b x)^2} \, dx\)

Optimal. Leaf size=147 \[ -\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 b^{3/2}}+\frac{c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{\sqrt{c+d x} (b c-a d)^2}{a^2 b (a+b x)}-\frac{c^2 \sqrt{c+d x}}{a^2 x} \]

[Out]

-((c^2*Sqrt[c + d*x])/(a^2*x)) - ((b*c - a*d)^2*Sqrt[c + d*x])/(a^2*b*(a + b*x)) + (c^(3/2)*(4*b*c - 5*a*d)*Ar
cTanh[Sqrt[c + d*x]/Sqrt[c]])/a^3 - ((b*c - a*d)^(3/2)*(4*b*c + a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c
- a*d]])/(a^3*b^(3/2))

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Rubi [A]  time = 0.197115, antiderivative size = 159, normalized size of antiderivative = 1.08, 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, 149, 156, 63, 208} \[ -\frac{(b c-a d)^{3/2} (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 b^{3/2}}+\frac{c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{\sqrt{c+d x} (b c-a d) (2 b c-a d)}{a^2 b (a+b x)}-\frac{c (c+d x)^{3/2}}{a x (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(5/2)/(x^2*(a + b*x)^2),x]

[Out]

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

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 149

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*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && 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)^{5/2}}{x^2 (a+b x)^2} \, dx &=-\frac{c (c+d x)^{3/2}}{a x (a+b x)}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{1}{2} c (4 b c-5 a d)+\frac{1}{2} d (b c-2 a d) x\right )}{x (a+b x)^2} \, dx}{a}\\ &=-\frac{(b c-a d) (2 b c-a d) \sqrt{c+d x}}{a^2 b (a+b x)}-\frac{c (c+d x)^{3/2}}{a x (a+b x)}+\frac{\int \frac{-\frac{1}{2} b c^2 (4 b c-5 a d)-\frac{1}{2} d \left (2 b^2 c^2-2 a b c d-a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx}{a^2 b}\\ &=-\frac{(b c-a d) (2 b c-a d) \sqrt{c+d x}}{a^2 b (a+b x)}-\frac{c (c+d x)^{3/2}}{a x (a+b x)}-\frac{\left (c^2 (4 b c-5 a d)\right ) \int \frac{1}{x \sqrt{c+d x}} \, dx}{2 a^3}+\frac{\left ((b c-a d)^2 (4 b c+a d)\right ) \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{2 a^3 b}\\ &=-\frac{(b c-a d) (2 b c-a d) \sqrt{c+d x}}{a^2 b (a+b x)}-\frac{c (c+d x)^{3/2}}{a x (a+b x)}-\frac{\left (c^2 (4 b c-5 a d)\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{a^3 d}+\frac{\left ((b c-a d)^2 (4 b c+a d)\right ) \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 b d}\\ &=-\frac{(b c-a d) (2 b c-a d) \sqrt{c+d x}}{a^2 b (a+b x)}-\frac{c (c+d x)^{3/2}}{a x (a+b x)}+\frac{c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3}-\frac{(b c-a d)^{3/2} (4 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.260707, size = 159, normalized size = 1.08 \[ \frac{-\frac{a \sqrt{c+d x} \left (a^2 d^2 x+a b c (c-2 d x)+2 b^2 c^2 x\right )}{b x (a+b x)}-\frac{\sqrt{b c-a d} \left (-a^2 d^2-3 a b c d+4 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{3/2}}+c^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(5/2)/(x^2*(a + b*x)^2),x]

[Out]

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

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Maple [B]  time = 0.017, size = 313, normalized size = 2.1 \begin{align*} -{\frac{{c}^{2}}{{a}^{2}x}\sqrt{dx+c}}-5\,{\frac{d{c}^{3/2}}{{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+4\,{\frac{{c}^{5/2}b}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-{\frac{{d}^{3}}{b \left ( bdx+ad \right ) }\sqrt{dx+c}}+2\,{\frac{{d}^{2}\sqrt{dx+c}c}{a \left ( bdx+ad \right ) }}-{\frac{bd{c}^{2}}{{a}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}+{\frac{{d}^{3}}{b}\arctan \left ({b\sqrt{dx+c}{\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ad-bc \right ) b}}}}+2\,{\frac{{d}^{2}c}{a\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-7\,{\frac{bd{c}^{2}}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+4\,{\frac{{c}^{3}{b}^{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)^(5/2)/x^2/(b*x+a)^2,x)

[Out]

-c^2*(d*x+c)^(1/2)/a^2/x-5*d*c^(3/2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))+4*c^(5/2)/a^3*arctanh((d*x+c)^(1/2)/c^
(1/2))*b-d^3/b*(d*x+c)^(1/2)/(b*d*x+a*d)+2*d^2/a*(d*x+c)^(1/2)/(b*d*x+a*d)*c-d/a^2*b*(d*x+c)^(1/2)/(b*d*x+a*d)
*c^2+d^3/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))+2*d^2/a/((a*d-b*c)*b)^(1/2)*arctan(
b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c-7*d/a^2*b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2
))*c^2+4/a^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^3*b^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)^(5/2)/x^2/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.87942, size = 2122, normalized size = 14.44 \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)^(5/2)/x^2/(b*x+a)^2,x, algorithm="fricas")

[Out]

[-1/2*(((4*b^3*c^2 - 3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*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^3*c^2 - 5*a*b^2*c*d)*x
^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(a^2*b*c^2 + (2*a
*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/(a^3*b^2*x^2 + a^4*b*x), -1/2*(2*((4*b^3*c^2 - 3*a*b^2*c*d
- a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*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^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(c)*log((d
*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*(a^2*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c)
)/(a^3*b^2*x^2 + a^4*b*x), -1/2*(2*((4*b^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqrt(-c)*ar
ctan(sqrt(d*x + c)*sqrt(-c)/c) + ((4*b^3*c^2 - 3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3
*d^2)*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*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/(a^3*b^2*x^2 + a^4*b*x), -(((4*b^3*c^2 -
3*a*b^2*c*d - a^2*b*d^2)*x^2 + (4*a*b^2*c^2 - 3*a^2*b*c*d - a^3*d^2)*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^3*c^2 - 5*a*b^2*c*d)*x^2 + (4*a*b^2*c^2 - 5*a^2*b*c*d)*x)*sqr
t(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + (a^2*b*c^2 + (2*a*b^2*c^2 - 2*a^2*b*c*d + a^3*d^2)*x)*sqrt(d*x + c))/
(a^3*b^2*x^2 + a^4*b*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)**(5/2)/x**2/(b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.20121, size = 351, normalized size = 2.39 \begin{align*} -\frac{{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{a^{3} \sqrt{-c}} + \frac{{\left (4 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d + 2 \, a^{2} b c d^{2} + a^{3} d^{3}\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} b} - \frac{2 \,{\left (d x + c\right )}^{\frac{3}{2}} b^{2} c^{2} d - 2 \, \sqrt{d x + c} b^{2} c^{3} d - 2 \,{\left (d x + c\right )}^{\frac{3}{2}} a b c d^{2} + 3 \, \sqrt{d x + c} a b c^{2} d^{2} +{\left (d x + c\right )}^{\frac{3}{2}} a^{2} d^{3} - \sqrt{d x + c} a^{2} c d^{3}}{{\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} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*b*c^3 - 5*a*c^2*d)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c)) + (4*b^3*c^3 - 7*a*b^2*c^2*d + 2*a^2*b*c*
d^2 + a^3*d^3)*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3*b) - (2*(d*x + c)^(3/2)*
b^2*c^2*d - 2*sqrt(d*x + c)*b^2*c^3*d - 2*(d*x + c)^(3/2)*a*b*c*d^2 + 3*sqrt(d*x + c)*a*b*c^2*d^2 + (d*x + c)^
(3/2)*a^2*d^3 - sqrt(d*x + c)*a^2*c*d^3)/(((d*x + c)^2*b - 2*(d*x + c)*b*c + b*c^2 + (d*x + c)*a*d - a*c*d)*a^
2*b)