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

Optimal. Leaf size=142 \[ \frac{(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^2 b^{5/2}}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2}-\frac{d \sqrt{c+d x} (b c-3 a d)}{a b^2}+\frac{(c+d x)^{3/2} (b c-a d)}{a b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.156058, antiderivative size = 142, 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, 154, 156, 63, 208} \[ \frac{(b c-a d)^{3/2} (3 a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^2 b^{5/2}}-\frac{2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2}-\frac{d \sqrt{c+d x} (b c-3 a d)}{a b^2}+\frac{(c+d x)^{3/2} (b c-a d)}{a b (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

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

Mathematica [A]  time = 0.217281, size = 145, normalized size = 1.02 \[ \frac{\frac{a \sqrt{c+d x} \left (3 a^2 d^2+2 a b d (d x-c)+b^2 c^2\right )}{b^2 (a+b x)}+\frac{\sqrt{b c-a d} \left (-3 a^2 d^2+a b c d+2 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{b^{5/2}}-2 c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 284, normalized size = 2. \begin{align*} 2\,{\frac{{d}^{2}\sqrt{dx+c}}{{b}^{2}}}-2\,{\frac{{c}^{5/2}}{{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+{\frac{{d}^{3}a}{{b}^{2} \left ( bdx+ad \right ) }\sqrt{dx+c}}-2\,{\frac{{d}^{2}\sqrt{dx+c}c}{b \left ( bdx+ad \right ) }}+{\frac{d{c}^{2}}{a \left ( bdx+ad \right ) }\sqrt{dx+c}}-3\,{\frac{{d}^{3}a}{{b}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+4\,{\frac{{d}^{2}c}{b\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+{\frac{d{c}^{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}}}}-2\,{\frac{b{c}^{3}}{{a}^{2}\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/(b*x+a)^2,x)

[Out]

2*d^2/b^2*(d*x+c)^(1/2)-2*c^(5/2)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^2+d^3/b^2*a*(d*x+c)^(1/2)/(b*d*x+a*d)-2*d^2
/b*(d*x+c)^(1/2)/(b*d*x+a*d)*c+d/a*(d*x+c)^(1/2)/(b*d*x+a*d)*c^2-3*d^3/b^2*a/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x
+c)^(1/2)/((a*d-b*c)*b)^(1/2))+4*d^2/b/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c+d/a/(
(a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))*c^2-2*b/a^2/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c
)^(1/2)/((a*d-b*c)*b)^(1/2))*c^3

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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/(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.06275, size = 1854, normalized size = 13.06 \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/(b*x+a)^2,x, algorithm="fricas")

[Out]

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

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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/(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.25356, size = 261, normalized size = 1.84 \begin{align*} \frac{2 \, c^{3} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c}} + \frac{2 \, \sqrt{d x + c} d^{2}}{b^{2}} - \frac{{\left (2 \, b^{3} c^{3} - a b^{2} c^{2} d - 4 \, a^{2} b c d^{2} + 3 \, 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^{2} b^{2}} + \frac{\sqrt{d x + c} b^{2} c^{2} d - 2 \, \sqrt{d x + c} a b c d^{2} + \sqrt{d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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