Integrand size = 17, antiderivative size = 519 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=-\frac {(c+d x)^{3/5}}{(b c-a d) (a+b x)}+\frac {\sqrt {2 \left (5-\sqrt {5}\right )} d \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}-\frac {2 \sqrt {\frac {2}{5+\sqrt {5}}} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{5 b^{3/5} (b c-a d)^{7/5}}+\frac {\sqrt {2 \left (5+\sqrt {5}\right )} d \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}+\frac {\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{5 b^{3/5} (b c-a d)^{7/5}}-\frac {2 d \log \left (\sqrt [5]{b c-a d}-\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{5 b^{3/5} (b c-a d)^{7/5}}+\frac {\left (1+\sqrt {5}\right ) d \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}-\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{10 b^{3/5} (b c-a d)^{7/5}}+\frac {\left (1-\sqrt {5}\right ) d \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{10 b^{3/5} (b c-a d)^{7/5}} \] Output:
-(d*x+c)^(3/5)/(-a*d+b*c)/(b*x+a)-1/5*(10-2*5^(1/2))^(1/2)*d*arctan(-1/5*( 25-10*5^(1/2))^(1/2)+2*2^(1/2)/(5+5^(1/2))^(1/2)*b^(1/5)*(d*x+c)^(1/5)/(-a *d+b*c)^(1/5))/b^(3/5)/(-a*d+b*c)^(7/5)+1/5*(10+2*5^(1/2))^(1/2)*d*arctan( 1/5*(25+10*5^(1/2))^(1/2)+1/5*(50+10*5^(1/2))^(1/2)*b^(1/5)*(d*x+c)^(1/5)/ (-a*d+b*c)^(1/5))/b^(3/5)/(-a*d+b*c)^(7/5)-2/5*d*ln((-a*d+b*c)^(1/5)-b^(1/ 5)*(d*x+c)^(1/5))/b^(3/5)/(-a*d+b*c)^(7/5)+1/10*(5^(1/2)+1)*d*ln(2*(-a*d+b *c)^(2/5)+b^(1/5)*(-a*d+b*c)^(1/5)*(d*x+c)^(1/5)-5^(1/2)*b^(1/5)*(-a*d+b*c )^(1/5)*(d*x+c)^(1/5)+2*b^(2/5)*(d*x+c)^(2/5))/b^(3/5)/(-a*d+b*c)^(7/5)+1/ 10*(-5^(1/2)+1)*d*ln(2*(-a*d+b*c)^(2/5)+b^(1/5)*(-a*d+b*c)^(1/5)*(d*x+c)^( 1/5)+5^(1/2)*b^(1/5)*(-a*d+b*c)^(1/5)*(d*x+c)^(1/5)+2*b^(2/5)*(d*x+c)^(2/5 ))/b^(3/5)/(-a*d+b*c)^(7/5)
Time = 1.78 (sec) , antiderivative size = 458, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=\frac {1}{10} \left (-\frac {10 (c+d x)^{3/5}}{(b c-a d) (a+b x)}-\frac {2 \sqrt {2 \left (5+\sqrt {5}\right )} d \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \left (5+\sqrt {5}-\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{b^{3/5} (-b c+a d)^{7/5}}-\frac {2 \sqrt {10-2 \sqrt {5}} d \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \left (5-\sqrt {5}+\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{b^{3/5} (-b c+a d)^{7/5}}+\frac {4 d \log \left (\sqrt [5]{-b c+a d}+\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{b^{3/5} (-b c+a d)^{7/5}}-\frac {\left (1+\sqrt {5}\right ) d \log \left (2 (-b c+a d)^{2/5}+\left (-1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{b^{3/5} (-b c+a d)^{7/5}}+\frac {\left (-1+\sqrt {5}\right ) d \log \left (2 (-b c+a d)^{2/5}-\left (1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{b^{3/5} (-b c+a d)^{7/5}}\right ) \] Input:
Integrate[1/((a + b*x)^2*(c + d*x)^(2/5)),x]
Output:
((-10*(c + d*x)^(3/5))/((b*c - a*d)*(a + b*x)) - (2*Sqrt[2*(5 + Sqrt[5])]* d*ArcTan[(Sqrt[(5 + Sqrt[5])/2]*(5 + Sqrt[5] - (4*Sqrt[5]*b^(1/5)*(c + d*x )^(1/5))/(-(b*c) + a*d)^(1/5)))/10])/(b^(3/5)*(-(b*c) + a*d)^(7/5)) - (2*S qrt[10 - 2*Sqrt[5]]*d*ArcTan[(Sqrt[(5 - Sqrt[5])/2]*(5 - Sqrt[5] + (4*Sqrt [5]*b^(1/5)*(c + d*x)^(1/5))/(-(b*c) + a*d)^(1/5)))/10])/(b^(3/5)*(-(b*c) + a*d)^(7/5)) + (4*d*Log[(-(b*c) + a*d)^(1/5) + b^(1/5)*(c + d*x)^(1/5)])/ (b^(3/5)*(-(b*c) + a*d)^(7/5)) - ((1 + Sqrt[5])*d*Log[2*(-(b*c) + a*d)^(2/ 5) + (-1 + Sqrt[5])*b^(1/5)*(-(b*c) + a*d)^(1/5)*(c + d*x)^(1/5) + 2*b^(2/ 5)*(c + d*x)^(2/5)])/(b^(3/5)*(-(b*c) + a*d)^(7/5)) + ((-1 + Sqrt[5])*d*Lo g[2*(-(b*c) + a*d)^(2/5) - (1 + Sqrt[5])*b^(1/5)*(-(b*c) + a*d)^(1/5)*(c + d*x)^(1/5) + 2*b^(2/5)*(c + d*x)^(2/5)])/(b^(3/5)*(-(b*c) + a*d)^(7/5)))/ 10
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.16 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {52, 78}
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.
| \(\displaystyle \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {2 d \int \frac {1}{(a+b x) (c+d x)^{2/5}}dx}{5 (b c-a d)}-\frac {(c+d x)^{3/5}}{(a+b x) (b c-a d)}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {2 d (c+d x)^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},1,\frac {8}{5},\frac {b (c+d x)}{b c-a d}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{3/5}}{(a+b x) (b c-a d)}\) |
Input:
Int[1/((a + b*x)^2*(c + d*x)^(2/5)),x]
Output:
-((c + d*x)^(3/5)/((b*c - a*d)*(a + b*x))) + (2*d*(c + d*x)^(3/5)*Hypergeo metric2F1[3/5, 1, 8/5, (b*(c + d*x))/(b*c - a*d)])/(3*(b*c - a*d)^2)
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[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Time = 3.96 (sec) , antiderivative size = 578, normalized size of antiderivative = 1.11
| method | result | size |
| pseudoelliptic | \(-\frac {4 d \left (\frac {d \sqrt {5+\sqrt {5}}\, \sqrt {5-\sqrt {5}}\, \left (\sqrt {5}-1\right ) \left (b x +a \right ) \ln \left (-2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (\sqrt {5}+1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}-2 \left (x d +c \right )^{\frac {2}{5}}\right )}{4}-\frac {d \sqrt {5+\sqrt {5}}\, \sqrt {5-\sqrt {5}}\, \left (\sqrt {5}+1\right ) \left (b x +a \right ) \ln \left (2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (\sqrt {5}-1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+2 \left (x d +c \right )^{\frac {2}{5}}\right )}{4}-d \sqrt {2}\, \sqrt {5}\, \sqrt {5-\sqrt {5}}\, \left (b x +a \right ) \arctan \left (\frac {5 \left (\left (\sqrt {5}-1\right ) \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+4 \left (x d +c \right )^{\frac {1}{5}}\right ) \left (3+\sqrt {5}\right ) \sqrt {2}}{\left (5+\sqrt {5}\right )^{\frac {5}{2}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right )+\left (d \sqrt {5-\sqrt {5}}\, \left (b x +a \right ) \ln \left (\left (x d +c \right )^{\frac {1}{5}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}\right )+\frac {5 b \sqrt {5-\sqrt {5}}\, \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}} \left (x d +c \right )^{\frac {3}{5}}}{2}+\left (b x +a \right ) d \arctan \left (\frac {5 \left (\left (\sqrt {5}+1\right ) \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}-4 \left (x d +c \right )^{\frac {1}{5}}\right ) \left (\sqrt {5}-3\right ) \sqrt {2}}{\left (5-\sqrt {5}\right )^{\frac {5}{2}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right ) \sqrt {5}\, \sqrt {2}\right ) \sqrt {5+\sqrt {5}}\right ) \sqrt {5}}{25 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}} \left (-2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (\sqrt {5}+1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}-2 \left (x d +c \right )^{\frac {2}{5}}\right ) \left (\left (x d +c \right )^{\frac {1}{5}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}\right ) \left (2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (\sqrt {5}-1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+2 \left (x d +c \right )^{\frac {2}{5}}\right ) \left (a d -b c \right ) b^{2}}\) | \(578\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(2830\) |
| default | \(\text {Expression too large to display}\) | \(2830\) |
Input:
int(1/(b*x+a)^2/(d*x+c)^(2/5),x,method=_RETURNVERBOSE)
Output:
-4/25/((a*d-b*c)/b)^(2/5)*d*(1/4*d*(5+5^(1/2))^(1/2)*(5-5^(1/2))^(1/2)*(5^ (1/2)-1)*(b*x+a)*ln(-2*((a*d-b*c)/b)^(2/5)+(5^(1/2)+1)*(d*x+c)^(1/5)*((a*d -b*c)/b)^(1/5)-2*(d*x+c)^(2/5))-1/4*d*(5+5^(1/2))^(1/2)*(5-5^(1/2))^(1/2)* (5^(1/2)+1)*(b*x+a)*ln(2*((a*d-b*c)/b)^(2/5)+(5^(1/2)-1)*(d*x+c)^(1/5)*((a *d-b*c)/b)^(1/5)+2*(d*x+c)^(2/5))-d*2^(1/2)*5^(1/2)*(5-5^(1/2))^(1/2)*(b*x +a)*arctan(5*((5^(1/2)-1)*((a*d-b*c)/b)^(1/5)+4*(d*x+c)^(1/5))/(5+5^(1/2)) ^(5/2)/((a*d-b*c)/b)^(1/5)*(3+5^(1/2))*2^(1/2))+(d*(5-5^(1/2))^(1/2)*(b*x+ a)*ln((d*x+c)^(1/5)+((a*d-b*c)/b)^(1/5))+5/2*b*(5-5^(1/2))^(1/2)*((a*d-b*c )/b)^(2/5)*(d*x+c)^(3/5)+(b*x+a)*d*arctan(5/(5-5^(1/2))^(5/2)*((5^(1/2)+1) *((a*d-b*c)/b)^(1/5)-4*(d*x+c)^(1/5))/((a*d-b*c)/b)^(1/5)*(5^(1/2)-3)*2^(1 /2))*5^(1/2)*2^(1/2))*(5+5^(1/2))^(1/2))*5^(1/2)/(-2*((a*d-b*c)/b)^(2/5)+( 5^(1/2)+1)*(d*x+c)^(1/5)*((a*d-b*c)/b)^(1/5)-2*(d*x+c)^(2/5))/((d*x+c)^(1/ 5)+((a*d-b*c)/b)^(1/5))/(2*((a*d-b*c)/b)^(2/5)+(5^(1/2)-1)*(d*x+c)^(1/5)*( (a*d-b*c)/b)^(1/5)+2*(d*x+c)^(2/5))/(a*d-b*c)/b^2
Timed out. \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)^2/(d*x+c)^(2/5),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=\int \frac {1}{\left (a + b x\right )^{2} \left (c + d x\right )^{\frac {2}{5}}}\, dx \] Input:
integrate(1/(b*x+a)**2/(d*x+c)**(2/5),x)
Output:
Integral(1/((a + b*x)**2*(c + d*x)**(2/5)), x)
Time = 0.13 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx =\text {Too large to display} \] Input:
integrate(1/(b*x+a)^2/(d*x+c)^(2/5),x, algorithm="maxima")
Output:
-1/5*(2*d^2*(log((d*x + c)^(1/5)*b^(1/5) + (-b*c + a*d)^(1/5))/((-b*c + a* d)^(2/5)*b^(3/5)) + sqrt(5)*log(((-b*c + a*d)^(1/5)*b^(1/5)*(sqrt(5) + 1) + (-b*c + a*d)^(1/5)*b^(1/5)*sqrt(2*sqrt(5) - 10) - 4*(d*x + c)^(1/5)*b^(2 /5))/((-b*c + a*d)^(1/5)*b^(1/5)*(sqrt(5) + 1) - (-b*c + a*d)^(1/5)*b^(1/5 )*sqrt(2*sqrt(5) - 10) - 4*(d*x + c)^(1/5)*b^(2/5)))/((-b*c + a*d)^(2/5)*b ^(3/5)*sqrt(2*sqrt(5) - 10)) - sqrt(5)*log(((-b*c + a*d)^(1/5)*b^(1/5)*(sq rt(5) - 1) - (-b*c + a*d)^(1/5)*b^(1/5)*sqrt(-2*sqrt(5) - 10) + 4*(d*x + c )^(1/5)*b^(2/5))/((-b*c + a*d)^(1/5)*b^(1/5)*(sqrt(5) - 1) + (-b*c + a*d)^ (1/5)*b^(1/5)*sqrt(-2*sqrt(5) - 10) + 4*(d*x + c)^(1/5)*b^(2/5)))/((-b*c + a*d)^(2/5)*b^(3/5)*sqrt(-2*sqrt(5) - 10)) + log(-(-b*c + a*d)^(1/5)*(d*x + c)^(1/5)*b^(1/5)*(sqrt(5) + 1) + 2*(d*x + c)^(2/5)*b^(2/5) + 2*(-b*c + a *d)^(2/5))/((-b*c + a*d)^(2/5)*b^(3/5)*(sqrt(5) + 1)) - log((-b*c + a*d)^( 1/5)*(d*x + c)^(1/5)*b^(1/5)*(sqrt(5) - 1) + 2*(d*x + c)^(2/5)*b^(2/5) + 2 *(-b*c + a*d)^(2/5))/((-b*c + a*d)^(2/5)*b^(3/5)*(sqrt(5) - 1)))/(b*c - a* d) - 5*(d*x + c)^(3/5)*d^2/(b^2*c^2 - 2*a*b*c*d + a^2*d^2 - (b^2*c - a*b*d )*(d*x + c)))/d
Time = 0.35 (sec) , antiderivative size = 577, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx =\text {Too large to display} \] Input:
integrate(1/(b*x+a)^2/(d*x+c)^(2/5),x, algorithm="giac")
Output:
1/10*(b^5*c - a*b^4*d)^(3/5)*d*(sqrt(5) - 5)*log(1/2*(d*x + c)^(1/5)*(sqrt (5)*((b*c - a*d)/b)^(1/5) + ((b*c - a*d)/b)^(1/5)) + (d*x + c)^(2/5) + ((b *c - a*d)/b)^(2/5))/(sqrt(5)*b^5*c^2 - 2*sqrt(5)*a*b^4*c*d + sqrt(5)*a^2*b ^3*d^2) + 1/10*(b^5*c - a*b^4*d)^(3/5)*d*(sqrt(5) + 5)*log(-1/2*(d*x + c)^ (1/5)*(sqrt(5)*((b*c - a*d)/b)^(1/5) - ((b*c - a*d)/b)^(1/5)) + (d*x + c)^ (2/5) + ((b*c - a*d)/b)^(2/5))/(sqrt(5)*b^5*c^2 - 2*sqrt(5)*a*b^4*c*d + sq rt(5)*a^2*b^3*d^2) - 1/5*(b^5*c - a*b^4*d)^(3/5)*d*sqrt(-2*sqrt(5) + 10)*a rctan(-((sqrt(5) - 1)*((b*c - a*d)/b)^(1/5) - 4*(d*x + c)^(1/5))/(sqrt(2*s qrt(5) + 10)*((b*c - a*d)/b)^(1/5)))/(b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2) + 1/5*(b^5*c - a*b^4*d)^(3/5)*d*sqrt(2*sqrt(5) + 10)*arctan(((sqrt(5) + 1 )*((b*c - a*d)/b)^(1/5) + 4*(d*x + c)^(1/5))/(sqrt(-2*sqrt(5) + 10)*((b*c - a*d)/b)^(1/5)))/(b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2) - 2/5*d*((b*c - a* d)/b)^(3/5)*log(abs((d*x + c)^(1/5) - ((b*c - a*d)/b)^(1/5)))/(b^2*c^2 - 2 *a*b*c*d + a^2*d^2) - (d*x + c)^(3/5)*d/(((d*x + c)*b - b*c + a*d)*(b*c - a*d))
Time = 0.79 (sec) , antiderivative size = 493, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=\frac {\ln \left (\frac {16\,b^2\,d^4}{{\left (a\,d-b\,c\right )}^3}+\frac {b^{11/5}\,d^4\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}+\sqrt {5}-1\right )}^3}{4\,{\left (a\,d-b\,c\right )}^{16/5}}\right )\,\left (\sqrt {5}\,d-d+d\,\sqrt {-2\,\sqrt {5}-10}\right )}{10\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{7/5}}-\frac {\ln \left (\frac {16\,b^2\,d^4}{{\left (a\,d-b\,c\right )}^3}-\frac {b^{11/5}\,d^4\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}-\sqrt {5}+1\right )}^3}{4\,{\left (a\,d-b\,c\right )}^{16/5}}\right )\,\left (d-\sqrt {5}\,d+d\,\sqrt {-2\,\sqrt {5}-10}\right )}{10\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{7/5}}-\frac {\ln \left (\frac {16\,b^2\,d^4}{{\left (a\,d-b\,c\right )}^3}-\frac {b^{11/5}\,d^4\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {5}+\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{4\,{\left (a\,d-b\,c\right )}^{16/5}}\right )\,\left (d+\sqrt {5}\,d+d\,\sqrt {2\,\sqrt {5}-10}\right )}{10\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{7/5}}-\frac {\ln \left (\frac {16\,b^2\,d^4}{{\left (a\,d-b\,c\right )}^3}-\frac {b^{11/5}\,d^4\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {5}-\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{4\,{\left (a\,d-b\,c\right )}^{16/5}}\right )\,\left (d+\sqrt {5}\,d-d\,\sqrt {2\,\sqrt {5}-10}\right )}{10\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{7/5}}+\frac {2\,d\,\ln \left (\frac {16\,b^2\,d^4}{{\left (a\,d-b\,c\right )}^3}+\frac {16\,b^{11/5}\,d^4\,{\left (c+d\,x\right )}^{1/5}}{{\left (a\,d-b\,c\right )}^{16/5}}\right )}{5\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{7/5}}+\frac {d\,{\left (c+d\,x\right )}^{3/5}}{\left (a\,d-b\,c\right )\,\left (a\,d-b\,c+b\,\left (c+d\,x\right )\right )} \] Input:
int(1/((a + b*x)^2*(c + d*x)^(2/5)),x)
Output:
(log((16*b^2*d^4)/(a*d - b*c)^3 + (b^(11/5)*d^4*(c + d*x)^(1/5)*(2^(1/2)*( - 5^(1/2) - 5)^(1/2) + 5^(1/2) - 1)^3)/(4*(a*d - b*c)^(16/5)))*(5^(1/2)*d - d + d*(- 2*5^(1/2) - 10)^(1/2)))/(10*b^(3/5)*(a*d - b*c)^(7/5)) - (log(( 16*b^2*d^4)/(a*d - b*c)^3 - (b^(11/5)*d^4*(c + d*x)^(1/5)*(2^(1/2)*(- 5^(1 /2) - 5)^(1/2) - 5^(1/2) + 1)^3)/(4*(a*d - b*c)^(16/5)))*(d - 5^(1/2)*d + d*(- 2*5^(1/2) - 10)^(1/2)))/(10*b^(3/5)*(a*d - b*c)^(7/5)) - (log((16*b^2 *d^4)/(a*d - b*c)^3 - (b^(11/5)*d^4*(c + d*x)^(1/5)*(5^(1/2) + 2^(1/2)*(5^ (1/2) - 5)^(1/2) + 1)^3)/(4*(a*d - b*c)^(16/5)))*(d + 5^(1/2)*d + d*(2*5^( 1/2) - 10)^(1/2)))/(10*b^(3/5)*(a*d - b*c)^(7/5)) - (log((16*b^2*d^4)/(a*d - b*c)^3 - (b^(11/5)*d^4*(c + d*x)^(1/5)*(5^(1/2) - 2^(1/2)*(5^(1/2) - 5) ^(1/2) + 1)^3)/(4*(a*d - b*c)^(16/5)))*(d + 5^(1/2)*d - d*(2*5^(1/2) - 10) ^(1/2)))/(10*b^(3/5)*(a*d - b*c)^(7/5)) + (2*d*log((16*b^2*d^4)/(a*d - b*c )^3 + (16*b^(11/5)*d^4*(c + d*x)^(1/5))/(a*d - b*c)^(16/5)))/(5*b^(3/5)*(a *d - b*c)^(7/5)) + (d*(c + d*x)^(3/5))/((a*d - b*c)*(a*d - b*c + b*(c + d* x)))
\[ \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}} \, dx=\text {too large to display} \] Input:
int(1/(b*x+a)^2/(d*x+c)^(2/5),x)
Output:
(2*(c + d*x)**(2/5)*int(x/((c + d*x)**(2/5)*a**4*c*d**2 + (c + d*x)**(2/5) *a**4*d**3*x + 12*(c + d*x)**(2/5)*a**3*b*c**2*d + 14*(c + d*x)**(2/5)*a** 3*b*c*d**2*x + 2*(c + d*x)**(2/5)*a**3*b*d**3*x**2 + 36*(c + d*x)**(2/5)*a **2*b**2*c**3 + 60*(c + d*x)**(2/5)*a**2*b**2*c**2*d*x + 25*(c + d*x)**(2/ 5)*a**2*b**2*c*d**2*x**2 + (c + d*x)**(2/5)*a**2*b**2*d**3*x**3 + 72*(c + d*x)**(2/5)*a*b**3*c**3*x + 84*(c + d*x)**(2/5)*a*b**3*c**2*d*x**2 + 12*(c + d*x)**(2/5)*a*b**3*c*d**2*x**3 + 36*(c + d*x)**(2/5)*b**4*c**3*x**2 + 3 6*(c + d*x)**(2/5)*b**4*c**2*d*x**3),x)*a**5*d**5 + 27*(c + d*x)**(2/5)*in t(x/((c + d*x)**(2/5)*a**4*c*d**2 + (c + d*x)**(2/5)*a**4*d**3*x + 12*(c + d*x)**(2/5)*a**3*b*c**2*d + 14*(c + d*x)**(2/5)*a**3*b*c*d**2*x + 2*(c + d*x)**(2/5)*a**3*b*d**3*x**2 + 36*(c + d*x)**(2/5)*a**2*b**2*c**3 + 60*(c + d*x)**(2/5)*a**2*b**2*c**2*d*x + 25*(c + d*x)**(2/5)*a**2*b**2*c*d**2*x* *2 + (c + d*x)**(2/5)*a**2*b**2*d**3*x**3 + 72*(c + d*x)**(2/5)*a*b**3*c** 3*x + 84*(c + d*x)**(2/5)*a*b**3*c**2*d*x**2 + 12*(c + d*x)**(2/5)*a*b**3* c*d**2*x**3 + 36*(c + d*x)**(2/5)*b**4*c**3*x**2 + 36*(c + d*x)**(2/5)*b** 4*c**2*d*x**3),x)*a**4*b*c*d**4 + 2*(c + d*x)**(2/5)*int(x/((c + d*x)**(2/ 5)*a**4*c*d**2 + (c + d*x)**(2/5)*a**4*d**3*x + 12*(c + d*x)**(2/5)*a**3*b *c**2*d + 14*(c + d*x)**(2/5)*a**3*b*c*d**2*x + 2*(c + d*x)**(2/5)*a**3*b* d**3*x**2 + 36*(c + d*x)**(2/5)*a**2*b**2*c**3 + 60*(c + d*x)**(2/5)*a**2* b**2*c**2*d*x + 25*(c + d*x)**(2/5)*a**2*b**2*c*d**2*x**2 + (c + d*x)**...