Integrand size = 17, antiderivative size = 173 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=-\frac {35 d^3}{8 (b c-a d)^4 \sqrt {c+d x}}-\frac {1}{3 (b c-a d) (a+b x)^3 \sqrt {c+d x}}+\frac {7 d}{12 (b c-a d)^2 (a+b x)^2 \sqrt {c+d x}}-\frac {35 d^2}{24 (b c-a d)^3 (a+b x) \sqrt {c+d x}}+\frac {35 \sqrt {b} d^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{8 (b c-a d)^{9/2}} \]
35/8*d^3*arctanh(b^(1/2)*(d*x+c)^(1/2)/(-a*d+b*c)^(1/2))*b^(1/2)/(-a*d+b*c )^(9/2)-35/8*d^3/(-a*d+b*c)^4/(d*x+c)^(1/2)-1/3/(-a*d+b*c)/(b*x+a)^3/(d*x+ c)^(1/2)+7/12*d/(-a*d+b*c)^2/(b*x+a)^2/(d*x+c)^(1/2)-35/24*d^2/(-a*d+b*c)^ 3/(b*x+a)/(d*x+c)^(1/2)
Time = 0.64 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=\frac {-48 a^3 d^3-3 a^2 b d^2 (29 c+77 d x)-2 a b^2 d \left (-19 c^2+49 c d x+140 d^2 x^2\right )-b^3 \left (8 c^3-14 c^2 d x+35 c d^2 x^2+105 d^3 x^3\right )}{24 (b c-a d)^4 (a+b x)^3 \sqrt {c+d x}}-\frac {35 \sqrt {b} d^3 \arctan \left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {-b c+a d}}\right )}{8 (-b c+a d)^{9/2}} \]
(-48*a^3*d^3 - 3*a^2*b*d^2*(29*c + 77*d*x) - 2*a*b^2*d*(-19*c^2 + 49*c*d*x + 140*d^2*x^2) - b^3*(8*c^3 - 14*c^2*d*x + 35*c*d^2*x^2 + 105*d^3*x^3))/( 24*(b*c - a*d)^4*(a + b*x)^3*Sqrt[c + d*x]) - (35*Sqrt[b]*d^3*ArcTan[(Sqrt [b]*Sqrt[c + d*x])/Sqrt[-(b*c) + a*d]])/(8*(-(b*c) + a*d)^(9/2))
Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {52, 52, 52, 61, 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.
\(\displaystyle \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {7 d \int \frac {1}{(a+b x)^3 (c+d x)^{3/2}}dx}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {7 d \left (-\frac {5 d \int \frac {1}{(a+b x)^2 (c+d x)^{3/2}}dx}{4 (b c-a d)}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \int \frac {1}{(a+b x) (c+d x)^{3/2}}dx}{2 (b c-a d)}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}\right )}{4 (b c-a d)}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (\frac {b \int \frac {1}{(a+b x) \sqrt {c+d x}}dx}{b c-a d}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}\right )}{4 (b c-a d)}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (\frac {2 b \int \frac {1}{a+\frac {b (c+d x)}{d}-\frac {b c}{d}}d\sqrt {c+d x}}{d (b c-a d)}+\frac {2}{\sqrt {c+d x} (b c-a d)}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}\right )}{4 (b c-a d)}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {7 d \left (-\frac {5 d \left (-\frac {3 d \left (\frac {2}{\sqrt {c+d x} (b c-a d)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}\right )}{2 (b c-a d)}-\frac {1}{(a+b x) \sqrt {c+d x} (b c-a d)}\right )}{4 (b c-a d)}-\frac {1}{2 (a+b x)^2 \sqrt {c+d x} (b c-a d)}\right )}{6 (b c-a d)}-\frac {1}{3 (a+b x)^3 \sqrt {c+d x} (b c-a d)}\) |
-1/3*1/((b*c - a*d)*(a + b*x)^3*Sqrt[c + d*x]) - (7*d*(-1/2*1/((b*c - a*d) *(a + b*x)^2*Sqrt[c + d*x]) - (5*d*(-(1/((b*c - a*d)*(a + b*x)*Sqrt[c + d* x])) - (3*d*(2/((b*c - a*d)*Sqrt[c + d*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*S qrt[c + d*x])/Sqrt[b*c - a*d]])/(b*c - a*d)^(3/2)))/(2*(b*c - a*d))))/(4*( b*c - a*d))))/(6*(b*c - a*d))
3.15.33.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, 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_)^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 = 0.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(2 d^{3} \left (-\frac {1}{\left (a d -b c \right )^{4} \sqrt {d x +c}}-\frac {b \left (\frac {\frac {19 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{16}+\frac {17 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{6}+\left (\frac {29}{16} a^{2} d^{2}-\frac {29}{8} a b c d +\frac {29}{16} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {35 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{4}}\right )\) | \(156\) |
default | \(2 d^{3} \left (-\frac {1}{\left (a d -b c \right )^{4} \sqrt {d x +c}}-\frac {b \left (\frac {\frac {19 \left (d x +c \right )^{\frac {5}{2}} b^{2}}{16}+\frac {17 \left (a d -b c \right ) b \left (d x +c \right )^{\frac {3}{2}}}{6}+\left (\frac {29}{16} a^{2} d^{2}-\frac {29}{8} a b c d +\frac {29}{16} b^{2} c^{2}\right ) \sqrt {d x +c}}{\left (\left (d x +c \right ) b +a d -b c \right )^{3}}+\frac {35 \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16 \sqrt {\left (a d -b c \right ) b}}\right )}{\left (a d -b c \right )^{4}}\right )\) | \(156\) |
pseudoelliptic | \(-\frac {2 \left (\frac {35 b \,d^{3} \sqrt {d x +c}\, \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {d x +c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{16}+\sqrt {\left (a d -b c \right ) b}\, \left (\left (\frac {35}{16} b^{3} x^{3}+\frac {35}{6} a \,b^{2} x^{2}+\frac {77}{16} a^{2} b x +a^{3}\right ) d^{3}+\frac {29 b c \left (\frac {35}{87} b^{2} x^{2}+\frac {98}{87} a b x +a^{2}\right ) d^{2}}{16}-\frac {19 b^{2} c^{2} \left (\frac {7 b x}{19}+a \right ) d}{24}+\frac {b^{3} c^{3}}{6}\right )\right )}{\sqrt {d x +c}\, \sqrt {\left (a d -b c \right ) b}\, \left (b x +a \right )^{3} \left (a d -b c \right )^{4}}\) | \(175\) |
2*d^3*(-1/(a*d-b*c)^4/(d*x+c)^(1/2)-1/(a*d-b*c)^4*b*((19/16*(d*x+c)^(5/2)* b^2+17/6*(a*d-b*c)*b*(d*x+c)^(3/2)+(29/16*a^2*d^2-29/8*a*b*c*d+29/16*b^2*c ^2)*(d*x+c)^(1/2))/((d*x+c)*b+a*d-b*c)^3+35/16/((a*d-b*c)*b)^(1/2)*arctan( b*(d*x+c)^(1/2)/((a*d-b*c)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 597 vs. \(2 (145) = 290\).
Time = 0.27 (sec) , antiderivative size = 1204, normalized size of antiderivative = 6.96 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx =\text {Too large to display} \]
[1/48*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a *b^2*c*d^3 + a^2*b*d^4)*x^2 + (3*a^2*b*c*d^3 + a^3*d^4)*x)*sqrt(b/(b*c - a *d))*log((b*d*x + 2*b*c - a*d + 2*(b*c - a*d)*sqrt(d*x + c)*sqrt(b/(b*c - a*d)))/(b*x + a)) - 2*(105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a ^2*b*c*d^2 + 48*a^3*d^3 + 35*(b^3*c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3*c^2* d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3*b^4*c^5 - 4*a^4* b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^4 *b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b ^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2 *b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 - a^6 *b*c*d^4 + a^7*d^5)*x), 1/24*(105*(b^3*d^4*x^4 + a^3*c*d^3 + (b^3*c*d^3 + 3*a*b^2*d^4)*x^3 + 3*(a*b^2*c*d^3 + a^2*b*d^4)*x^2 + (3*a^2*b*c*d^3 + a^3* d^4)*x)*sqrt(-b/(b*c - a*d))*arctan(-(b*c - a*d)*sqrt(d*x + c)*sqrt(-b/(b* c - a*d))/(b*d*x + b*c)) - (105*b^3*d^3*x^3 + 8*b^3*c^3 - 38*a*b^2*c^2*d + 87*a^2*b*c*d^2 + 48*a^3*d^3 + 35*(b^3*c*d^2 + 8*a*b^2*d^3)*x^2 - 7*(2*b^3 *c^2*d - 14*a*b^2*c*d^2 - 33*a^2*b*d^3)*x)*sqrt(d*x + c))/(a^3*b^4*c^5 - 4 *a^4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^ 4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d...
Timed out. \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (145) = 290\).
Time = 0.34 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=-\frac {35 \, b d^{3} \arctan \left (\frac {\sqrt {d x + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {-b^{2} c + a b d}} - \frac {2 \, d^{3}}{{\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} \sqrt {d x + c}} - \frac {57 \, {\left (d x + c\right )}^{\frac {5}{2}} b^{3} d^{3} - 136 \, {\left (d x + c\right )}^{\frac {3}{2}} b^{3} c d^{3} + 87 \, \sqrt {d x + c} b^{3} c^{2} d^{3} + 136 \, {\left (d x + c\right )}^{\frac {3}{2}} a b^{2} d^{4} - 174 \, \sqrt {d x + c} a b^{2} c d^{4} + 87 \, \sqrt {d x + c} a^{2} b d^{5}}{24 \, {\left (b^{4} c^{4} - 4 \, a b^{3} c^{3} d + 6 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + a^{4} d^{4}\right )} {\left ({\left (d x + c\right )} b - b c + a d\right )}^{3}} \]
-35/8*b*d^3*arctan(sqrt(d*x + c)*b/sqrt(-b^2*c + a*b*d))/((b^4*c^4 - 4*a*b ^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(-b^2*c + a*b* d)) - 2*d^3/((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sqrt(d*x + c)) - 1/24*(57*(d*x + c)^(5/2)*b^3*d^3 - 136*(d*x + c)^(3/2)*b^3*c*d^3 + 87*sqrt(d*x + c)*b^3*c^2*d^3 + 136*(d*x + c)^(3/2)*a* b^2*d^4 - 174*sqrt(d*x + c)*a*b^2*c*d^4 + 87*sqrt(d*x + c)*a^2*b*d^5)/((b^ 4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*((d*x + c)*b - b*c + a*d)^3)
Time = 0.59 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=-\frac {\frac {2\,d^3}{a\,d-b\,c}+\frac {35\,b^2\,d^3\,{\left (c+d\,x\right )}^2}{3\,{\left (a\,d-b\,c\right )}^3}+\frac {35\,b^3\,d^3\,{\left (c+d\,x\right )}^3}{8\,{\left (a\,d-b\,c\right )}^4}+\frac {77\,b\,d^3\,\left (c+d\,x\right )}{8\,{\left (a\,d-b\,c\right )}^2}}{\sqrt {c+d\,x}\,\left (a^3\,d^3-3\,a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-b^3\,c^3\right )+b^3\,{\left (c+d\,x\right )}^{7/2}-\left (3\,b^3\,c-3\,a\,b^2\,d\right )\,{\left (c+d\,x\right )}^{5/2}+{\left (c+d\,x\right )}^{3/2}\,\left (3\,a^2\,b\,d^2-6\,a\,b^2\,c\,d+3\,b^3\,c^2\right )}-\frac {35\,\sqrt {b}\,d^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c+d\,x}\,\left (a^4\,d^4-4\,a^3\,b\,c\,d^3+6\,a^2\,b^2\,c^2\,d^2-4\,a\,b^3\,c^3\,d+b^4\,c^4\right )}{{\left (a\,d-b\,c\right )}^{9/2}}\right )}{8\,{\left (a\,d-b\,c\right )}^{9/2}} \]
- ((2*d^3)/(a*d - b*c) + (35*b^2*d^3*(c + d*x)^2)/(3*(a*d - b*c)^3) + (35* b^3*d^3*(c + d*x)^3)/(8*(a*d - b*c)^4) + (77*b*d^3*(c + d*x))/(8*(a*d - b* c)^2))/((c + d*x)^(1/2)*(a^3*d^3 - b^3*c^3 + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 ) + b^3*(c + d*x)^(7/2) - (3*b^3*c - 3*a*b^2*d)*(c + d*x)^(5/2) + (c + d*x )^(3/2)*(3*b^3*c^2 + 3*a^2*b*d^2 - 6*a*b^2*c*d)) - (35*b^(1/2)*d^3*atan((b ^(1/2)*(c + d*x)^(1/2)*(a^4*d^4 + b^4*c^4 + 6*a^2*b^2*c^2*d^2 - 4*a*b^3*c^ 3*d - 4*a^3*b*c*d^3))/(a*d - b*c)^(9/2)))/(8*(a*d - b*c)^(9/2))
Time = 0.00 (sec) , antiderivative size = 702, normalized size of antiderivative = 4.06 \[ \int \frac {1}{(a+b x)^4 (c+d x)^{3/2}} \, dx=\frac {-105 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{3} d^{3}-315 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a^{2} b \,d^{3} x -315 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) a \,b^{2} d^{3} x^{2}-105 \sqrt {b}\, \sqrt {d x +c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {d x +c}\, b}{\sqrt {b}\, \sqrt {a d -b c}}\right ) b^{3} d^{3} x^{3}-48 a^{4} d^{4}-39 a^{3} b c \,d^{3}-231 a^{3} b \,d^{4} x +125 a^{2} b^{2} c^{2} d^{2}+133 a^{2} b^{2} c \,d^{3} x -280 a^{2} b^{2} d^{4} x^{2}-46 a \,b^{3} c^{3} d +112 a \,b^{3} c^{2} d^{2} x +245 a \,b^{3} c \,d^{3} x^{2}-105 a \,b^{3} d^{4} x^{3}+8 b^{4} c^{4}-14 b^{4} c^{3} d x +35 b^{4} c^{2} d^{2} x^{2}+105 b^{4} c \,d^{3} x^{3}}{24 \sqrt {d x +c}\, \left (a^{5} b^{3} d^{5} x^{3}-5 a^{4} b^{4} c \,d^{4} x^{3}+10 a^{3} b^{5} c^{2} d^{3} x^{3}-10 a^{2} b^{6} c^{3} d^{2} x^{3}+5 a \,b^{7} c^{4} d \,x^{3}-b^{8} c^{5} x^{3}+3 a^{6} b^{2} d^{5} x^{2}-15 a^{5} b^{3} c \,d^{4} x^{2}+30 a^{4} b^{4} c^{2} d^{3} x^{2}-30 a^{3} b^{5} c^{3} d^{2} x^{2}+15 a^{2} b^{6} c^{4} d \,x^{2}-3 a \,b^{7} c^{5} x^{2}+3 a^{7} b \,d^{5} x -15 a^{6} b^{2} c \,d^{4} x +30 a^{5} b^{3} c^{2} d^{3} x -30 a^{4} b^{4} c^{3} d^{2} x +15 a^{3} b^{5} c^{4} d x -3 a^{2} b^{6} c^{5} x +a^{8} d^{5}-5 a^{7} b c \,d^{4}+10 a^{6} b^{2} c^{2} d^{3}-10 a^{5} b^{3} c^{3} d^{2}+5 a^{4} b^{4} c^{4} d -a^{3} b^{5} c^{5}\right )} \]
int(1/(sqrt(c + d*x)*(a**4*c + a**4*d*x + 4*a**3*b*c*x + 4*a**3*b*d*x**2 + 6*a**2*b**2*c*x**2 + 6*a**2*b**2*d*x**3 + 4*a*b**3*c*x**3 + 4*a*b**3*d*x* *4 + b**4*c*x**4 + b**4*d*x**5)),x)
( - 105*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt (b)*sqrt(a*d - b*c)))*a**3*d**3 - 315*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c )*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*a**2*b*d**3*x - 315*sq rt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*atan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a *d - b*c)))*a*b**2*d**3*x**2 - 105*sqrt(b)*sqrt(c + d*x)*sqrt(a*d - b*c)*a tan((sqrt(c + d*x)*b)/(sqrt(b)*sqrt(a*d - b*c)))*b**3*d**3*x**3 - 48*a**4* d**4 - 39*a**3*b*c*d**3 - 231*a**3*b*d**4*x + 125*a**2*b**2*c**2*d**2 + 13 3*a**2*b**2*c*d**3*x - 280*a**2*b**2*d**4*x**2 - 46*a*b**3*c**3*d + 112*a* b**3*c**2*d**2*x + 245*a*b**3*c*d**3*x**2 - 105*a*b**3*d**4*x**3 + 8*b**4* c**4 - 14*b**4*c**3*d*x + 35*b**4*c**2*d**2*x**2 + 105*b**4*c*d**3*x**3)/( 24*sqrt(c + d*x)*(a**8*d**5 - 5*a**7*b*c*d**4 + 3*a**7*b*d**5*x + 10*a**6* b**2*c**2*d**3 - 15*a**6*b**2*c*d**4*x + 3*a**6*b**2*d**5*x**2 - 10*a**5*b **3*c**3*d**2 + 30*a**5*b**3*c**2*d**3*x - 15*a**5*b**3*c*d**4*x**2 + a**5 *b**3*d**5*x**3 + 5*a**4*b**4*c**4*d - 30*a**4*b**4*c**3*d**2*x + 30*a**4* b**4*c**2*d**3*x**2 - 5*a**4*b**4*c*d**4*x**3 - a**3*b**5*c**5 + 15*a**3*b **5*c**4*d*x - 30*a**3*b**5*c**3*d**2*x**2 + 10*a**3*b**5*c**2*d**3*x**3 - 3*a**2*b**6*c**5*x + 15*a**2*b**6*c**4*d*x**2 - 10*a**2*b**6*c**3*d**2*x* *3 - 3*a*b**7*c**5*x**2 + 5*a*b**7*c**4*d*x**3 - b**8*c**5*x**3))