3.461 \(\int \frac{(c+d x)^{5/2}}{x^3 (a+b x)} \, dx\)

Optimal. Leaf size=151 \[ -\frac{\sqrt{c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^3}+\frac{c \sqrt{c+d x} (4 b c-7 a d)}{4 a^2 x}+\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 \sqrt{b}}-\frac{c (c+d x)^{3/2}}{2 a x^2} \]

[Out]

(c*(4*b*c - 7*a*d)*Sqrt[c + d*x])/(4*a^2*x) - (c*(c + d*x)^(3/2))/(2*a*x^2) - (Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d
+ 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^3) + (2*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/
Sqrt[b*c - a*d]])/(a^3*Sqrt[b])

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Rubi [A]  time = 0.14735, antiderivative size = 151, 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, 149, 156, 63, 208} \[ -\frac{\sqrt{c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^3}+\frac{c \sqrt{c+d x} (4 b c-7 a d)}{4 a^2 x}+\frac{2 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{a^3 \sqrt{b}}-\frac{c (c+d x)^{3/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(4*b*c - 7*a*d)*Sqrt[c + d*x])/(4*a^2*x) - (c*(c + d*x)^(3/2))/(2*a*x^2) - (Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d
+ 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(4*a^3) + (2*(b*c - a*d)^(5/2)*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 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^3 (a+b x)} \, dx &=-\frac{c (c+d x)^{3/2}}{2 a x^2}-\frac{\int \frac{\sqrt{c+d x} \left (\frac{1}{2} c (4 b c-7 a d)+\frac{1}{2} d (b c-4 a d) x\right )}{x^2 (a+b x)} \, dx}{2 a}\\ &=\frac{c (4 b c-7 a d) \sqrt{c+d x}}{4 a^2 x}-\frac{c (c+d x)^{3/2}}{2 a x^2}-\frac{\int \frac{-\frac{1}{4} c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )-\frac{1}{4} d \left (4 b^2 c^2-9 a b c d+8 a^2 d^2\right ) x}{x (a+b x) \sqrt{c+d x}} \, dx}{2 a^2}\\ &=\frac{c (4 b c-7 a d) \sqrt{c+d x}}{4 a^2 x}-\frac{c (c+d x)^{3/2}}{2 a x^2}-\frac{(b c-a d)^3 \int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx}{a^3}+\frac{\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \int \frac{1}{x \sqrt{c+d x}} \, dx}{8 a^3}\\ &=\frac{c (4 b c-7 a d) \sqrt{c+d x}}{4 a^2 x}-\frac{c (c+d x)^{3/2}}{2 a x^2}-\frac{\left (2 (b c-a d)^3\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 d}+\frac{\left (c \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x}\right )}{4 a^3 d}\\ &=\frac{c (4 b c-7 a d) \sqrt{c+d x}}{4 a^2 x}-\frac{c (c+d x)^{3/2}}{2 a x^2}-\frac{\sqrt{c} \left (8 b^2 c^2-20 a b c d+15 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{4 a^3}+\frac{2 (b c-a d)^{5/2} \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 = 0.146113, size = 131, normalized size = 0.87 \[ \frac{-\sqrt{c} \left (15 a^2 d^2-20 a b c d+8 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )+\frac{a c \sqrt{c+d x} (-2 a c-9 a d x+4 b c x)}{x^2}+\frac{8 (b c-a d)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x}}{\sqrt{b c-a d}}\right )}{\sqrt{b}}}{4 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*c*Sqrt[c + d*x]*(-2*a*c + 4*b*c*x - 9*a*d*x))/x^2 - Sqrt[c]*(8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*ArcTanh[
Sqrt[c + d*x]/Sqrt[c]] + (8*(b*c - a*d)^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/Sqrt[b])/(4*a^
3)

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Maple [B]  time = 0.013, size = 321, normalized size = 2.1 \begin{align*} -{\frac{9\,c}{4\,a{x}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{{c}^{2}b}{d{a}^{2}{x}^{2}} \left ( dx+c \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{c}^{2}}{4\,a{x}^{2}}\sqrt{dx+c}}-{\frac{{c}^{3}b}{d{a}^{2}{x}^{2}}\sqrt{dx+c}}-{\frac{15\,{d}^{2}}{4\,a}\sqrt{c}{\it Artanh} \left ({\sqrt{dx+c}{\frac{1}{\sqrt{c}}}} \right ) }+5\,{\frac{d{c}^{3/2}b}{{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }-2\,{\frac{{c}^{5/2}{b}^{2}}{{a}^{3}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) }+2\,{\frac{{d}^{3}}{\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-6\,{\frac{{d}^{2}cb}{a\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }+6\,{\frac{{b}^{2}d{c}^{2}}{{a}^{2}\sqrt{ \left ( ad-bc \right ) b}}\arctan \left ({\frac{b\sqrt{dx+c}}{\sqrt{ \left ( ad-bc \right ) b}}} \right ) }-2\,{\frac{{b}^{3}{c}^{3}}{{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^3/(b*x+a),x)

[Out]

-9/4*c*(d*x+c)^(3/2)/a/x^2+1/d*c^2/a^2/x^2*(d*x+c)^(3/2)*b+7/4*c^2/a/x^2*(d*x+c)^(1/2)-1/d*c^3/a^2/x^2*(d*x+c)
^(1/2)*b-15/4*d^2*c^(1/2)/a*arctanh((d*x+c)^(1/2)/c^(1/2))+5*d*c^(3/2)/a^2*arctanh((d*x+c)^(1/2)/c^(1/2))*b-2*
c^(5/2)/a^3*arctanh((d*x+c)^(1/2)/c^(1/2))*b^2+2*d^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*c)*b)^
(1/2))-6*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*b+6*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^2*c^2-2/a^3/((a*d-b*c)*b)^(1/2)*arctan(b*(d*x+c)^(1/2)/((a*d-b*
c)*b)^(1/2))*b^3*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^3/(b*x+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.17456, size = 1586, normalized size = 10.5 \begin{align*} \left [\frac{8 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) +{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{c} x^{2} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) - 2 \,{\left (2 \, a^{2} c^{2} -{\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{d x + c}}{8 \, a^{3} x^{2}}, \frac{16 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{c} x^{2} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) - 2 \,{\left (2 \, a^{2} c^{2} -{\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{d x + c}}{8 \, a^{3} x^{2}}, \frac{{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) + 4 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x + 2 \, b c - a d + 2 \, \sqrt{d x + c} b \sqrt{\frac{b c - a d}{b}}}{b x + a}\right ) -{\left (2 \, a^{2} c^{2} -{\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{d x + c}}{4 \, a^{3} x^{2}}, \frac{8 \,{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} x^{2} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) +{\left (8 \, b^{2} c^{2} - 20 \, a b c d + 15 \, a^{2} d^{2}\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) -{\left (2 \, a^{2} c^{2} -{\left (4 \, a b c^{2} - 9 \, a^{2} c d\right )} x\right )} \sqrt{d x + c}}{4 \, a^{3} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(5/2)/x^3/(b*x+a),x, algorithm="fricas")

[Out]

[1/8*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqrt(d*x + c)*b*s
qrt((b*c - a*d)/b))/(b*x + a)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*
sqrt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2), 1/8*(16*(b^2*c^2 - 2*a
*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (8*b^2*
c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(c)*x^2*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^2*c^2 - (4*a*
b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^2), 1/4*((8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan
(sqrt(d*x + c)*sqrt(-c)/c) + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*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 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))
/(a^3*x^2), 1/4*(8*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*x^2*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x + c)*b*sqrt(-(b*c
- a*d)/b)/(b*c - a*d)) + (8*b^2*c^2 - 20*a*b*c*d + 15*a^2*d^2)*sqrt(-c)*x^2*arctan(sqrt(d*x + c)*sqrt(-c)/c)
- (2*a^2*c^2 - (4*a*b*c^2 - 9*a^2*c*d)*x)*sqrt(d*x + c))/(a^3*x^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**3/(b*x+a),x)

[Out]

Timed out

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Giac [A]  time = 1.19982, size = 267, normalized size = 1.77 \begin{align*} -\frac{2 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, 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}} + \frac{{\left (8 \, b^{2} c^{3} - 20 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{4 \, a^{3} \sqrt{-c}} + \frac{4 \,{\left (d x + c\right )}^{\frac{3}{2}} b c^{2} d - 4 \, \sqrt{d x + c} b c^{3} d - 9 \,{\left (d x + c\right )}^{\frac{3}{2}} a c d^{2} + 7 \, \sqrt{d x + c} a c^{2} d^{2}}{4 \, a^{2} d^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*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) + 1/4*(8*b^2*c^3 - 20*a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^3*sqrt(-c))
+ 1/4*(4*(d*x + c)^(3/2)*b*c^2*d - 4*sqrt(d*x + c)*b*c^3*d - 9*(d*x + c)^(3/2)*a*c*d^2 + 7*sqrt(d*x + c)*a*c^
2*d^2)/(a^2*d^2*x^2)