Integrand size = 20, antiderivative size = 219 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\frac {\left (12 b^2 c^2-17 a b c d+4 a^2 d^2\right ) \sqrt {c+d x}}{4 a^3 (a+b x)}+\frac {c (6 b c-7 a d) \sqrt {c+d x}}{4 a^2 x (a+b x)}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}-\frac {\sqrt {c} \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4}+\frac {(b c-a d)^{3/2} (6 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^4 \sqrt {b}} \]
-1/2*c*(d*x+c)^(3/2)/a/x^2/(b*x+a)+(-a*d+b*c)^(3/2)*(-a*d+6*b*c)*arctanh(b ^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))/a^4/b^(1/2)-1/4*(15*a^2*d^2-40*a*b* c*d+24*b^2*c^2)*arctanh((d*x+c)^(1/2)/c^(1/2))*c^(1/2)/a^4+1/4*(4*a^2*d^2- 17*a*b*c*d+12*b^2*c^2)*(d*x+c)^(1/2)/a^3/(b*x+a)+1/4*c*(-7*a*d+6*b*c)*(d*x +c)^(1/2)/a^2/x/(b*x+a)
Time = 0.68 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\frac {\frac {a \sqrt {c+d x} \left (12 b^2 c^2 x^2+a b c x (6 c-17 d x)+a^2 \left (-2 c^2-9 c d x+4 d^2 x^2\right )\right )}{x^2 (a+b x)}-\frac {4 (6 b c-a d) (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{\sqrt {b}}-\sqrt {c} \left (24 b^2 c^2-40 a b c d+15 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{4 a^4} \]
((a*Sqrt[c + d*x]*(12*b^2*c^2*x^2 + a*b*c*x*(6*c - 17*d*x) + a^2*(-2*c^2 - 9*c*d*x + 4*d^2*x^2)))/(x^2*(a + b*x)) - (4*(6*b*c - a*d)*(-(b*c) + a*d)^ (3/2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/Sqrt[b] - Sqrt[c ]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/( 4*a^4)
Time = 0.42 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {109, 27, 166, 27, 168, 27, 174, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (6 b c-7 a d)+d (3 b c-4 a d) x)}{2 x^2 (a+b x)^2}dx}{2 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (6 b c-7 a d)+d (3 b c-4 a d) x)}{x^2 (a+b x)^2}dx}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle -\frac {\frac {\int -\frac {c \left (24 b^2 c^2-40 a b d c+15 a^2 d^2\right )+d \left (18 b^2 c^2-27 a b d c+8 a^2 d^2\right ) x}{2 x (a+b x)^2 \sqrt {c+d x}}dx}{a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {c \left (24 b^2 c^2-40 a b d c+15 a^2 d^2\right )+d \left (18 b^2 c^2-27 a b d c+8 a^2 d^2\right ) x}{x (a+b x)^2 \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {(b c-a d) \left (c \left (24 b^2 c^2-40 a b d c+15 a^2 d^2\right )+d \left (12 b^2 c^2-17 a b d c+4 a^2 d^2\right ) x\right )}{x (a+b x) \sqrt {c+d x}}dx}{a (b c-a d)}+\frac {2 \sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{a (a+b x)}}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\frac {\int \frac {c \left (24 b^2 c^2-40 a b d c+15 a^2 d^2\right )+d \left (12 b^2 c^2-17 a b d c+4 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{a}+\frac {2 \sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{a (a+b x)}}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {c \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {4 (b c-a d)^2 (6 b c-a d) \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{a}+\frac {2 \sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{a (a+b x)}}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {2 c \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {8 (b c-a d)^2 (6 b c-a d) \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{a}+\frac {2 \sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{a (a+b x)}}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {-\frac {\frac {\frac {8 (b c-a d)^{3/2} (6 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a \sqrt {b}}-\frac {2 \sqrt {c} \left (15 a^2 d^2-40 a b c d+24 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a}}{a}+\frac {2 \sqrt {c+d x} \left (4 a^2 d^2-17 a b c d+12 b^2 c^2\right )}{a (a+b x)}}{2 a}-\frac {c \sqrt {c+d x} (6 b c-7 a d)}{a x (a+b x)}}{4 a}-\frac {c (c+d x)^{3/2}}{2 a x^2 (a+b x)}\) |
-1/2*(c*(c + d*x)^(3/2))/(a*x^2*(a + b*x)) - (-((c*(6*b*c - 7*a*d)*Sqrt[c + d*x])/(a*x*(a + b*x))) - ((2*(12*b^2*c^2 - 17*a*b*c*d + 4*a^2*d^2)*Sqrt[ c + d*x])/(a*(a + b*x)) + ((-2*Sqrt[c]*(24*b^2*c^2 - 40*a*b*c*d + 15*a^2*d ^2)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/a + (8*(b*c - a*d)^(3/2)*(6*b*c - a*d) *ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/(a*Sqrt[b]))/a)/(2*a))/ (4*a)
3.5.74.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] - Simp[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] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> 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] + S imp[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] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(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]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Time = 1.61 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.92
| method | result | size |
| pseudoelliptic | \(-\frac {6 \left (x^{2} \left (b c -\frac {a d}{6}\right ) \left (-a d +b c \right )^{2} \sqrt {c}\, \left (b x +a \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\frac {\left (\frac {15 x^{2} \left (a^{2} d^{2}-\frac {8}{3} a b c d +\frac {8}{5} b^{2} c^{2}\right ) \left (b x +a \right ) c \,\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{2}+\sqrt {d x +c}\, \sqrt {c}\, a \left (-6 b^{2} c^{2} x^{2}-3 \left (-\frac {17 d x}{6}+c \right ) x c a b +a^{2} \left (c^{2}+\frac {9}{2} c d x -2 d^{2} x^{2}\right )\right )\right ) \sqrt {\left (a d -b c \right ) b}}{12}\right )}{\sqrt {c}\, \sqrt {\left (a d -b c \right ) b}\, a^{4} \left (b x +a \right ) x^{2}}\) | \(201\) |
| derivativedivides | \(2 d^{4} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-40 a b c d +24 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{4} d^{4}}+\frac {\left (a d -b c \right )^{2} \left (\frac {\sqrt {d x +c}\, a d}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {\left (a d -6 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4}}\right )\) | \(203\) |
| default | \(2 d^{4} \left (-\frac {c \left (\frac {\left (\frac {9}{8} a^{2} d^{2}-a b c d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{8} c \,a^{2} d^{2}+b \,c^{2} d a \right ) \sqrt {d x +c}}{d^{2} x^{2}}+\frac {\left (15 a^{2} d^{2}-40 a b c d +24 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8 \sqrt {c}}\right )}{a^{4} d^{4}}+\frac {\left (a d -b c \right )^{2} \left (\frac {\sqrt {d x +c}\, a d}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {\left (a d -6 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{a^{4} d^{4}}\right )\) | \(203\) |
| risch | \(-\frac {c \sqrt {d x +c}\, \left (9 a d x -8 b c x +2 a c \right )}{4 a^{3} x^{2}}-\frac {d \left (\frac {\frac {8 \left (-\frac {1}{2} a^{3} d^{3}+a^{2} b c \,d^{2}-\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}-\frac {4 \left (a^{3} d^{3}-8 a^{2} b c \,d^{2}+13 a \,b^{2} c^{2} d -6 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{a d}+\frac {\sqrt {c}\, \left (15 a^{2} d^{2}-40 a b c d +24 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d}\right )}{4 a^{3}}\) | \(219\) |
-6*(x^2*(b*c-1/6*a*d)*(-a*d+b*c)^2*c^(1/2)*(b*x+a)*arctan(b*(d*x+c)^(1/2)/ ((a*d-b*c)*b)^(1/2))+1/12*(15/2*x^2*(a^2*d^2-8/3*a*b*c*d+8/5*b^2*c^2)*(b*x +a)*c*arctanh((d*x+c)^(1/2)/c^(1/2))+(d*x+c)^(1/2)*c^(1/2)*a*(-6*b^2*c^2*x ^2-3*(-17/6*d*x+c)*x*c*a*b+a^2*(c^2+9/2*c*d*x-2*d^2*x^2)))*((a*d-b*c)*b)^( 1/2))/c^(1/2)/((a*d-b*c)*b)^(1/2)/a^4/(b*x+a)/x^2
Time = 0.32 (sec) , antiderivative size = 1173, normalized size of antiderivative = 5.36 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\text {Too large to display} \]
[1/8*(4*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2* b*c*d + a^3*d^2)*x^2)*sqrt((b*c - a*d)/b)*log((b*d*x + 2*b*c - a*d + 2*sqr t(d*x + c)*b*sqrt((b*c - a*d)/b))/(b*x + a)) + ((24*b^3*c^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^2)*x^2)*sqr t(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2*a^3*c^2 - (12*a*b ^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sq rt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/8*(8*((6*b^3*c^2 - 7*a*b^2*c*d + a^2 *b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*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)) + ((24*b^3*c ^2 - 40*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15* a^3*d^2)*x^2)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(2* a^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)*x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(((24*b^3*c^2 - 4 0*a*b^2*c*d + 15*a^2*b*d^2)*x^3 + (24*a*b^2*c^2 - 40*a^2*b*c*d + 15*a^3*d^ 2)*x^2)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 2*((6*b^3*c^2 - 7*a*b^ 2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a^2*b*c*d + a^3*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^3*c^2 - (12*a*b^2*c^2 - 17*a^2*b*c*d + 4*a^3*d^2)* x^2 - 3*(2*a^2*b*c^2 - 3*a^3*c*d)*x)*sqrt(d*x + c))/(a^4*b*x^3 + a^5*x^2), 1/4*(4*((6*b^3*c^2 - 7*a*b^2*c*d + a^2*b*d^2)*x^3 + (6*a*b^2*c^2 - 7*a...
Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m ore detail
Time = 0.28 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.21 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=-\frac {{\left (6 \, b^{3} c^{3} - 13 \, a b^{2} c^{2} d + 8 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{4}} + \frac {{\left (24 \, b^{2} c^{3} - 40 \, a b c^{2} d + 15 \, a^{2} c d^{2}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{4 \, a^{4} \sqrt {-c}} + \frac {\sqrt {d x + c} b^{2} c^{2} d - 2 \, \sqrt {d x + c} a b c d^{2} + \sqrt {d x + c} a^{2} d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a^{3}} + \frac {8 \, {\left (d x + c\right )}^{\frac {3}{2}} b c^{2} d - 8 \, \sqrt {d x + c} b c^{3} d - 9 \, {\left (d x + c\right )}^{\frac {3}{2}} a c d^{2} + 7 \, \sqrt {d x + c} a c^{2} d^{2}}{4 \, a^{3} d^{2} x^{2}} \]
-(6*b^3*c^3 - 13*a*b^2*c^2*d + 8*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^4) + 1/4*(24*b^2*c^3 - 40*a*b*c^2*d + 15*a^2*c*d^2)*arctan(sqrt(d*x + c)/sqrt(-c))/(a^4*sqrt(-c)) + (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^3) + 1/4*(8*(d*x + c)^(3/2)*b*c^2*d - 8*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^3*d^2*x^2)
Time = 0.93 (sec) , antiderivative size = 670, normalized size of antiderivative = 3.06 \[ \int \frac {(c+d x)^{5/2}}{x^3 (a+b x)^2} \, dx=\frac {\frac {\sqrt {c+d\,x}\,\left (11\,a^2\,c^2\,d^3-23\,a\,b\,c^3\,d^2+12\,b^2\,c^4\,d\right )}{4\,a^3}-\frac {{\left (c+d\,x\right )}^{3/2}\,\left (17\,a^2\,c\,d^3-40\,a\,b\,c^2\,d^2+24\,b^2\,c^3\,d\right )}{4\,a^3}+\frac {d\,{\left (c+d\,x\right )}^{5/2}\,\left (4\,a^2\,d^2-17\,a\,b\,c\,d+12\,b^2\,c^2\right )}{4\,a^3}}{b\,{\left (c+d\,x\right )}^3+\left (a\,d-3\,b\,c\right )\,{\left (c+d\,x\right )}^2-b\,c^3+\left (3\,b\,c^2-2\,a\,c\,d\right )\,\left (c+d\,x\right )+a\,c^2\,d}+\frac {\sqrt {c}\,\ln \left (\sqrt {c+d\,x}-\sqrt {c}\right )\,\left (\frac {15\,a^2\,d^2}{8}-5\,a\,b\,c\,d+3\,b^2\,c^2\right )}{a^4}-\frac {\sqrt {c}\,\ln \left (\sqrt {c+d\,x}+\sqrt {c}\right )\,\left (15\,a^2\,d^2-40\,a\,b\,c\,d+24\,b^2\,c^2\right )}{8\,a^4}+\frac {\mathrm {atan}\left (-\frac {b^2\,c^2\,d^7\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,111{}\mathrm {i}}{8\,\left (41\,a\,b^3\,c^3\,d^8-\frac {291\,b^4\,c^4\,d^7}{8}-\frac {143\,a^2\,b^2\,c^2\,d^9}{8}+\frac {45\,b^5\,c^5\,d^6}{4\,a}+2\,a^3\,b\,c\,d^{10}\right )}+\frac {b^3\,c^3\,d^6\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,45{}\mathrm {i}}{4\,\left (2\,a^4\,b\,c\,d^{10}-\frac {143\,a^3\,b^2\,c^2\,d^9}{8}+41\,a^2\,b^3\,c^3\,d^8-\frac {291\,a\,b^4\,c^4\,d^7}{8}+\frac {45\,b^5\,c^5\,d^6}{4}\right )}+\frac {b\,c\,d^8\,\sqrt {c+d\,x}\,\sqrt {-a^3\,b\,d^3+3\,a^2\,b^2\,c\,d^2-3\,a\,b^3\,c^2\,d+b^4\,c^3}\,2{}\mathrm {i}}{41\,b^3\,c^3\,d^8-\frac {143\,a\,b^2\,c^2\,d^9}{8}-\frac {291\,b^4\,c^4\,d^7}{8\,a}+\frac {45\,b^5\,c^5\,d^6}{4\,a^2}+2\,a^2\,b\,c\,d^{10}}\right )\,\left (a\,d-6\,b\,c\right )\,\sqrt {-b\,{\left (a\,d-b\,c\right )}^3}\,1{}\mathrm {i}}{a^4\,b} \]
(((c + d*x)^(1/2)*(12*b^2*c^4*d + 11*a^2*c^2*d^3 - 23*a*b*c^3*d^2))/(4*a^3 ) - ((c + d*x)^(3/2)*(17*a^2*c*d^3 + 24*b^2*c^3*d - 40*a*b*c^2*d^2))/(4*a^ 3) + (d*(c + d*x)^(5/2)*(4*a^2*d^2 + 12*b^2*c^2 - 17*a*b*c*d))/(4*a^3))/(b *(c + d*x)^3 + (a*d - 3*b*c)*(c + d*x)^2 - b*c^3 + (3*b*c^2 - 2*a*c*d)*(c + d*x) + a*c^2*d) + (c^(1/2)*log((c + d*x)^(1/2) - c^(1/2))*((15*a^2*d^2)/ 8 + 3*b^2*c^2 - 5*a*b*c*d))/a^4 - (c^(1/2)*log((c + d*x)^(1/2) + c^(1/2))* (15*a^2*d^2 + 24*b^2*c^2 - 40*a*b*c*d))/(8*a^4) + (atan((b^3*c^3*d^6*(c + d*x)^(1/2)*(b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2)*4 5i)/(4*((45*b^5*c^5*d^6)/4 - (291*a*b^4*c^4*d^7)/8 + 41*a^2*b^3*c^3*d^8 - (143*a^3*b^2*c^2*d^9)/8 + 2*a^4*b*c*d^10)) - (b^2*c^2*d^7*(c + d*x)^(1/2)* (b^4*c^3 - a^3*b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2)*111i)/(8*(41 *a*b^3*c^3*d^8 - (291*b^4*c^4*d^7)/8 - (143*a^2*b^2*c^2*d^9)/8 + (45*b^5*c ^5*d^6)/(4*a) + 2*a^3*b*c*d^10)) + (b*c*d^8*(c + d*x)^(1/2)*(b^4*c^3 - a^3 *b*d^3 + 3*a^2*b^2*c*d^2 - 3*a*b^3*c^2*d)^(1/2)*2i)/(41*b^3*c^3*d^8 - (143 *a*b^2*c^2*d^9)/8 - (291*b^4*c^4*d^7)/(8*a) + (45*b^5*c^5*d^6)/(4*a^2) + 2 *a^2*b*c*d^10))*(a*d - 6*b*c)*(-b*(a*d - b*c)^3)^(1/2)*1i)/(a^4*b)