Integrand size = 20, antiderivative size = 284 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=-\frac {b \left (32 b^2 c^2-52 a b c d+19 a^2 d^2\right ) \sqrt {c+d x}}{8 a^4 (a+b x)}+\frac {c (8 b c-9 a d) \sqrt {c+d x}}{12 a^2 x^2 (a+b x)}-\frac {\left (48 b^2 c^2-82 a b c d+33 a^2 d^2\right ) \sqrt {c+d x}}{24 a^3 x (a+b x)}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}+\frac {\left (64 b^3 c^3-120 a b^2 c^2 d+60 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{8 a^5 \sqrt {c}}-\frac {\sqrt {b} (8 b c-3 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a^5} \]
-1/3*c*(d*x+c)^(3/2)/a/x^3/(b*x+a)-(-3*a*d+8*b*c)*(-a*d+b*c)^(3/2)*arctanh (b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)/a^5+1/8*(-5*a^3*d^3+60*a^ 2*b*c*d^2-120*a*b^2*c^2*d+64*b^3*c^3)*arctanh((d*x+c)^(1/2)/c^(1/2))/a^5/c ^(1/2)-1/8*b*(19*a^2*d^2-52*a*b*c*d+32*b^2*c^2)*(d*x+c)^(1/2)/a^4/(b*x+a)+ 1/12*c*(-9*a*d+8*b*c)*(d*x+c)^(1/2)/a^2/x^2/(b*x+a)-1/24*(33*a^2*d^2-82*a* b*c*d+48*b^2*c^2)*(d*x+c)^(1/2)/a^3/x/(b*x+a)
Time = 0.81 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.79 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\frac {-\frac {a \sqrt {c+d x} \left (96 b^3 c^2 x^3+12 a b^2 c x^2 (4 c-13 d x)+a^3 \left (8 c^2+26 c d x+33 d^2 x^2\right )+a^2 b x \left (-16 c^2-82 c d x+57 d^2 x^2\right )\right )}{x^3 (a+b x)}+24 \sqrt {b} (8 b c-3 a d) (-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )+\frac {3 \left (64 b^3 c^3-120 a b^2 c^2 d+60 a^2 b c d^2-5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{\sqrt {c}}}{24 a^5} \]
(-((a*Sqrt[c + d*x]*(96*b^3*c^2*x^3 + 12*a*b^2*c*x^2*(4*c - 13*d*x) + a^3* (8*c^2 + 26*c*d*x + 33*d^2*x^2) + a^2*b*x*(-16*c^2 - 82*c*d*x + 57*d^2*x^2 )))/(x^3*(a + b*x))) + 24*Sqrt[b]*(8*b*c - 3*a*d)*(-(b*c) + a*d)^(3/2)*Arc Tan[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]] + (3*(64*b^3*c^3 - 120*a*b ^2*c^2*d + 60*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/Sqr t[c])/(24*a^5)
Time = 0.52 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.08, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {109, 27, 166, 27, 168, 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.
| \(\displaystyle \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (8 b c-9 a d)+d (5 b c-6 a d) x)}{2 x^3 (a+b x)^2}dx}{3 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\sqrt {c+d x} (c (8 b c-9 a d)+d (5 b c-6 a d) x)}{x^3 (a+b x)^2}dx}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle -\frac {\frac {\int -\frac {c \left (48 b^2 c^2-82 a b d c+33 a^2 d^2\right )+d \left (40 b^2 c^2-65 a b d c+24 a^2 d^2\right ) x}{2 x^2 (a+b x)^2 \sqrt {c+d x}}dx}{2 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {c \left (48 b^2 c^2-82 a b d c+33 a^2 d^2\right )+d \left (40 b^2 c^2-65 a b d c+24 a^2 d^2\right ) x}{x^2 (a+b x)^2 \sqrt {c+d x}}dx}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {3 c \left (64 b^3 c^3-120 a b^2 d c^2+60 a^2 b d^2 c-5 a^3 d^3+b d \left (48 b^2 c^2-82 a b d c+33 a^2 d^2\right ) x\right )}{2 x (a+b x)^2 \sqrt {c+d x}}dx}{a c}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {-\frac {3 \int \frac {64 b^3 c^3-120 a b^2 d c^2+60 a^2 b d^2 c-5 a^3 d^3+b d \left (48 b^2 c^2-82 a b d c+33 a^2 d^2\right ) x}{x (a+b x)^2 \sqrt {c+d x}}dx}{2 a}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle -\frac {-\frac {-\frac {3 \left (\frac {\int \frac {(b c-a d) \left (64 b^3 c^3-120 a b^2 d c^2+60 a^2 b d^2 c-5 a^3 d^3+b d \left (32 b^2 c^2-52 a b d c+19 a^2 d^2\right ) x\right )}{x (a+b x) \sqrt {c+d x}}dx}{a (b c-a d)}+\frac {2 b \sqrt {c+d x} \left (19 a^2 d^2-52 a b c d+32 b^2 c^2\right )}{a (a+b x)}\right )}{2 a}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {-\frac {3 \left (\frac {\int \frac {64 b^3 c^3-120 a b^2 d c^2+60 a^2 b d^2 c-5 a^3 d^3+b d \left (32 b^2 c^2-52 a b d c+19 a^2 d^2\right ) x}{x (a+b x) \sqrt {c+d x}}dx}{a}+\frac {2 b \sqrt {c+d x} \left (19 a^2 d^2-52 a b c d+32 b^2 c^2\right )}{a (a+b x)}\right )}{2 a}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle -\frac {-\frac {-\frac {3 \left (\frac {\frac {\left (-5 a^3 d^3+60 a^2 b c d^2-120 a b^2 c^2 d+64 b^3 c^3\right ) \int \frac {1}{x \sqrt {c+d x}}dx}{a}-\frac {8 b (8 b c-3 a d) (b c-a d)^2 \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{a}}{a}+\frac {2 b \sqrt {c+d x} \left (19 a^2 d^2-52 a b c d+32 b^2 c^2\right )}{a (a+b x)}\right )}{2 a}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {-\frac {-\frac {3 \left (\frac {\frac {2 \left (-5 a^3 d^3+60 a^2 b c d^2-120 a b^2 c^2 d+64 b^3 c^3\right ) \int \frac {1}{\frac {c+d x}{d}-\frac {c}{d}}d\sqrt {c+d x}}{a d}-\frac {16 b (8 b c-3 a d) (b c-a d)^2 \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{a d}}{a}+\frac {2 b \sqrt {c+d x} \left (19 a^2 d^2-52 a b c d+32 b^2 c^2\right )}{a (a+b x)}\right )}{2 a}-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {-\frac {-\frac {\sqrt {c+d x} \left (33 a^2 d^2-82 a b c d+48 b^2 c^2\right )}{a x (a+b x)}-\frac {3 \left (\frac {2 b \sqrt {c+d x} \left (19 a^2 d^2-52 a b c d+32 b^2 c^2\right )}{a (a+b x)}+\frac {\frac {16 \sqrt {b} (8 b c-3 a d) (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{a}-\frac {2 \left (-5 a^3 d^3+60 a^2 b c d^2-120 a b^2 c^2 d+64 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c}}\right )}{a \sqrt {c}}}{a}\right )}{2 a}}{4 a}-\frac {c \sqrt {c+d x} (8 b c-9 a d)}{2 a x^2 (a+b x)}}{6 a}-\frac {c (c+d x)^{3/2}}{3 a x^3 (a+b x)}\) |
-1/3*(c*(c + d*x)^(3/2))/(a*x^3*(a + b*x)) - (-1/2*(c*(8*b*c - 9*a*d)*Sqrt [c + d*x])/(a*x^2*(a + b*x)) - (-(((48*b^2*c^2 - 82*a*b*c*d + 33*a^2*d^2)* Sqrt[c + d*x])/(a*x*(a + b*x))) - (3*((2*b*(32*b^2*c^2 - 52*a*b*c*d + 19*a ^2*d^2)*Sqrt[c + d*x])/(a*(a + b*x)) + ((-2*(64*b^3*c^3 - 120*a*b^2*c^2*d + 60*a^2*b*c*d^2 - 5*a^3*d^3)*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(a*Sqrt[c]) + (16*Sqrt[b]*(8*b*c - 3*a*d)*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/Sqrt[b*c - a*d]])/a)/a))/(2*a))/(4*a))/(6*a)
3.5.75.3.1 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 = 0.79 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.86
| method | result | size |
| pseudoelliptic | \(-\frac {-24 x^{3} \left (-a d +b c \right )^{2} \left (b c -\frac {3 a d}{8}\right ) b \left (b x +a \right ) \sqrt {c}\, \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {15 x^{3} \left (a^{3} d^{3}-12 a^{2} b c \,d^{2}+24 a \,b^{2} c^{2} d -\frac {64}{5} b^{3} c^{3}\right ) \left (b x +a \right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{8}+\left (12 x^{3} b^{3} c^{2}+6 x^{2} \left (-\frac {13 d x}{4}+c \right ) c a \,b^{2}-2 x \left (-\frac {57}{16} d^{2} x^{2}+\frac {41}{8} c d x +c^{2}\right ) a^{2} b +a^{3} \left (c^{2}+\frac {13}{4} c d x +\frac {33}{8} d^{2} x^{2}\right )\right ) \sqrt {d x +c}\, a \sqrt {c}\right ) \sqrt {\left (a d -b c \right ) b}}{3 \sqrt {c}\, \sqrt {\left (a d -b c \right ) b}\, a^{5} \left (b x +a \right ) x^{3}}\) | \(243\) |
| risch | \(-\frac {\sqrt {d x +c}\, \left (33 a^{2} d^{2} x^{2}-108 a b c d \,x^{2}+72 b^{2} c^{2} x^{2}+26 a^{2} c d x -24 a b \,c^{2} x +8 a^{2} c^{2}\right )}{24 a^{4} x^{3}}-\frac {d \left (\frac {16 b \left (\frac {\left (\frac {1}{2} a^{3} d^{3}-a^{2} b c \,d^{2}+\frac {1}{2} a \,b^{2} c^{2} d \right ) \sqrt {d x +c}}{\left (d x +c \right ) b +a d -b c}+\frac {\left (3 a^{3} d^{3}-14 a^{2} b c \,d^{2}+19 a \,b^{2} c^{2} d -8 b^{3} c^{3}\right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{a d}-\frac {\left (-5 a^{3} d^{3}+60 a^{2} b c \,d^{2}-120 a \,b^{2} c^{2} d +64 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{a d \sqrt {c}}\right )}{8 a^{4}}\) | \(277\) |
| derivativedivides | \(2 d^{5} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{4} a^{2} b c \,d^{2}+\frac {3}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+4 a^{2} d^{2} b \,c^{2}-3 a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{4} a^{2} b \,c^{3} d^{2}+\frac {3}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-60 a^{2} b c \,d^{2}+120 a \,b^{2} c^{2} d -64 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{5} d^{5}}-\frac {\left (a d -b c \right )^{2} b \left (\frac {\sqrt {d x +c}\, a d}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {\left (3 a d -8 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{a^{5} d^{5}}\right )\) | \(290\) |
| default | \(2 d^{5} \left (\frac {-\frac {\left (\frac {11}{16} a^{3} d^{3}-\frac {9}{4} a^{2} b c \,d^{2}+\frac {3}{2} a \,b^{2} c^{2} d \right ) \left (d x +c \right )^{\frac {5}{2}}+\left (-\frac {5}{6} a^{3} c \,d^{3}+4 a^{2} d^{2} b \,c^{2}-3 a \,b^{2} c^{3} d \right ) \left (d x +c \right )^{\frac {3}{2}}+\left (-\frac {7}{4} a^{2} b \,c^{3} d^{2}+\frac {3}{2} a \,b^{2} c^{4} d +\frac {5}{16} a^{3} c^{2} d^{3}\right ) \sqrt {d x +c}}{d^{3} x^{3}}-\frac {\left (5 a^{3} d^{3}-60 a^{2} b c \,d^{2}+120 a \,b^{2} c^{2} d -64 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c}}\right )}{16 \sqrt {c}}}{a^{5} d^{5}}-\frac {\left (a d -b c \right )^{2} b \left (\frac {\sqrt {d x +c}\, a d}{2 \left (d x +c \right ) b +2 a d -2 b c}+\frac {\left (3 a d -8 b c \right ) \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{2 \sqrt {\left (a d -b c \right ) b}}\right )}{a^{5} d^{5}}\right )\) | \(290\) |
-1/3*(-24*x^3*(-a*d+b*c)^2*(b*c-3/8*a*d)*b*(b*x+a)*c^(1/2)*arctan(b*(d*x+c )^(1/2)/((a*d-b*c)*b)^(1/2))+(15/8*x^3*(a^3*d^3-12*a^2*b*c*d^2+24*a*b^2*c^ 2*d-64/5*b^3*c^3)*(b*x+a)*arctanh((d*x+c)^(1/2)/c^(1/2))+(12*x^3*b^3*c^2+6 *x^2*(-13/4*d*x+c)*c*a*b^2-2*x*(-57/16*d^2*x^2+41/8*c*d*x+c^2)*a^2*b+a^3*( c^2+13/4*c*d*x+33/8*d^2*x^2))*(d*x+c)^(1/2)*a*c^(1/2))*((a*d-b*c)*b)^(1/2) )/c^(1/2)/((a*d-b*c)*b)^(1/2)/a^5/(b*x+a)/x^3
Time = 0.39 (sec) , antiderivative size = 1482, normalized size of antiderivative = 5.22 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\text {Too large to display} \]
[1/48*(24*((8*b^3*c^3 - 11*a*b^2*c^2*d + 3*a^2*b*c*d^2)*x^4 + (8*a*b^2*c^3 - 11*a^2*b*c^2*d + 3*a^3*c*d^2)*x^3)*sqrt(b^2*c - a*b*d)*log((b*d*x + 2*b *c - a*d - 2*sqrt(b^2*c - a*b*d)*sqrt(d*x + c))/(b*x + a)) - 3*((64*b^4*c^ 3 - 120*a*b^3*c^2*d + 60*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*x^4 + (64*a*b^3*c^3 - 120*a^2*b^2*c^2*d + 60*a^3*b*c*d^2 - 5*a^4*d^3)*x^3)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^4*c^3 + 3*(32*a*b^3*c^3 - 52*a^ 2*b^2*c^2*d + 19*a^3*b*c*d^2)*x^3 + (48*a^2*b^2*c^3 - 82*a^3*b*c^2*d + 33* a^4*c*d^2)*x^2 - 2*(8*a^3*b*c^3 - 13*a^4*c^2*d)*x)*sqrt(d*x + c))/(a^5*b*c *x^4 + a^6*c*x^3), 1/48*(48*((8*b^3*c^3 - 11*a*b^2*c^2*d + 3*a^2*b*c*d^2)* x^4 + (8*a*b^2*c^3 - 11*a^2*b*c^2*d + 3*a^3*c*d^2)*x^3)*sqrt(-b^2*c + a*b* d)*arctan(sqrt(-b^2*c + a*b*d)*sqrt(d*x + c)/(b*d*x + b*c)) - 3*((64*b^4*c ^3 - 120*a*b^3*c^2*d + 60*a^2*b^2*c*d^2 - 5*a^3*b*d^3)*x^4 + (64*a*b^3*c^3 - 120*a^2*b^2*c^2*d + 60*a^3*b*c*d^2 - 5*a^4*d^3)*x^3)*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) - 2*(8*a^4*c^3 + 3*(32*a*b^3*c^3 - 52*a ^2*b^2*c^2*d + 19*a^3*b*c*d^2)*x^3 + (48*a^2*b^2*c^3 - 82*a^3*b*c^2*d + 33 *a^4*c*d^2)*x^2 - 2*(8*a^3*b*c^3 - 13*a^4*c^2*d)*x)*sqrt(d*x + c))/(a^5*b* c*x^4 + a^6*c*x^3), -1/24*(3*((64*b^4*c^3 - 120*a*b^3*c^2*d + 60*a^2*b^2*c *d^2 - 5*a^3*b*d^3)*x^4 + (64*a*b^3*c^3 - 120*a^2*b^2*c^2*d + 60*a^3*b*c*d ^2 - 5*a^4*d^3)*x^3)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) - 12*((8*b^ 3*c^3 - 11*a*b^2*c^2*d + 3*a^2*b*c*d^2)*x^4 + (8*a*b^2*c^3 - 11*a^2*b*c...
Timed out. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.30 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\frac {{\left (8 \, b^{4} c^{3} - 19 \, a b^{3} c^{2} d + 14 \, a^{2} b^{2} c d^{2} - 3 \, a^{3} b d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{\sqrt {-b^{2} c + a b d} a^{5}} - \frac {{\left (64 \, b^{3} c^{3} - 120 \, a b^{2} c^{2} d + 60 \, a^{2} b c d^{2} - 5 \, a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c}}\right )}{8 \, a^{5} \sqrt {-c}} - \frac {\sqrt {d x + c} b^{3} c^{2} d - 2 \, \sqrt {d x + c} a b^{2} c d^{2} + \sqrt {d x + c} a^{2} b d^{3}}{{\left ({\left (d x + c\right )} b - b c + a d\right )} a^{4}} - \frac {72 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{2} c^{2} d - 144 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{2} c^{3} d + 72 \, \sqrt {d x + c} b^{2} c^{4} d - 108 \, {\left (d x + c\right )}^{\frac {5}{2}} a b c d^{2} + 192 \, {\left (d x + c\right )}^{\frac {3}{2}} a b c^{2} d^{2} - 84 \, \sqrt {d x + c} a b c^{3} d^{2} + 33 \, {\left (d x + c\right )}^{\frac {5}{2}} a^{2} d^{3} - 40 \, {\left (d x + c\right )}^{\frac {3}{2}} a^{2} c d^{3} + 15 \, \sqrt {d x + c} a^{2} c^{2} d^{3}}{24 \, a^{4} d^{3} x^{3}} \]
(8*b^4*c^3 - 19*a*b^3*c^2*d + 14*a^2*b^2*c*d^2 - 3*a^3*b*d^3)*arctan(sqrt( d*x + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^5) - 1/8*(64*b^3* c^3 - 120*a*b^2*c^2*d + 60*a^2*b*c*d^2 - 5*a^3*d^3)*arctan(sqrt(d*x + c)/s qrt(-c))/(a^5*sqrt(-c)) - (sqrt(d*x + c)*b^3*c^2*d - 2*sqrt(d*x + c)*a*b^2 *c*d^2 + sqrt(d*x + c)*a^2*b*d^3)/(((d*x + c)*b - b*c + a*d)*a^4) - 1/24*( 72*(d*x + c)^(5/2)*b^2*c^2*d - 144*(d*x + c)^(3/2)*b^2*c^3*d + 72*sqrt(d*x + c)*b^2*c^4*d - 108*(d*x + c)^(5/2)*a*b*c*d^2 + 192*(d*x + c)^(3/2)*a*b* c^2*d^2 - 84*sqrt(d*x + c)*a*b*c^3*d^2 + 33*(d*x + c)^(5/2)*a^2*d^3 - 40*( d*x + c)^(3/2)*a^2*c*d^3 + 15*sqrt(d*x + c)*a^2*c^2*d^3)/(a^4*d^3*x^3)
Time = 1.25 (sec) , antiderivative size = 2151, normalized size of antiderivative = 7.57 \[ \int \frac {(c+d x)^{5/2}}{x^4 (a+b x)^2} \, dx=\text {Too large to display} \]
(((c + d*x)^(3/2)*(40*a^3*c*d^4 - 288*b^3*c^4*d + 564*a*b^2*c^3*d^2 - 319* a^2*b*c^2*d^3))/(24*a^4) - ((c + d*x)^(5/2)*(33*a^3*d^4 - 288*b^3*c^3*d + 516*a*b^2*c^2*d^2 - 253*a^2*b*c*d^3))/(24*a^4) + ((c + d*x)^(1/2)*(32*b^3* c^5*d - 5*a^3*c^2*d^4 - 68*a*b^2*c^4*d^2 + 41*a^2*b*c^3*d^3))/(8*a^4) - (b *d*(c + d*x)^(7/2)*(19*a^2*d^2 + 32*b^2*c^2 - 52*a*b*c*d))/(8*a^4))/(b*(c + d*x)^4 - (4*b*c^3 - 3*a*c^2*d)*(c + d*x) + (6*b*c^2 - 3*a*c*d)*(c + d*x) ^2 + (a*d - 4*b*c)*(c + d*x)^3 + b*c^4 - a*c^3*d) + (atan((((((c + d*x)^(1 /2)*(601*a^6*b^3*d^8 + 8192*b^9*c^6*d^2 - 34816*a*b^8*c^5*d^3 - 5976*a^5*b ^4*c*d^7 + 59520*a^2*b^7*c^4*d^4 - 52160*a^3*b^6*c^3*d^5 + 24640*a^4*b^5*c ^2*d^6))/(32*a^8) - ((((5*a^13*b^2*d^6)/2 - (41*a^12*b^3*c*d^5)/2 - 16*a^1 0*b^5*c^3*d^3 + 34*a^11*b^4*c^2*d^4)/a^12 - ((256*a^11*b^2*d^3 - 512*a^10* b^3*c*d^2)*(c + d*x)^(1/2)*(5*a^3*d^3 - 64*b^3*c^3 + 120*a*b^2*c^2*d - 60* a^2*b*c*d^2))/(512*a^13*c^(1/2)))*(5*a^3*d^3 - 64*b^3*c^3 + 120*a*b^2*c^2* d - 60*a^2*b*c*d^2))/(16*a^5*c^(1/2)))*(5*a^3*d^3 - 64*b^3*c^3 + 120*a*b^2 *c^2*d - 60*a^2*b*c*d^2)*1i)/(16*a^5*c^(1/2)) + ((((c + d*x)^(1/2)*(601*a^ 6*b^3*d^8 + 8192*b^9*c^6*d^2 - 34816*a*b^8*c^5*d^3 - 5976*a^5*b^4*c*d^7 + 59520*a^2*b^7*c^4*d^4 - 52160*a^3*b^6*c^3*d^5 + 24640*a^4*b^5*c^2*d^6))/(3 2*a^8) + ((((5*a^13*b^2*d^6)/2 - (41*a^12*b^3*c*d^5)/2 - 16*a^10*b^5*c^3*d ^3 + 34*a^11*b^4*c^2*d^4)/a^12 + ((256*a^11*b^2*d^3 - 512*a^10*b^3*c*d^2)* (c + d*x)^(1/2)*(5*a^3*d^3 - 64*b^3*c^3 + 120*a*b^2*c^2*d - 60*a^2*b*c*...