Integrand size = 17, antiderivative size = 566 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=-\frac {(c+d x)^{3/5}}{2 (b c-a d) (a+b x)^2}+\frac {7 d (c+d x)^{3/5}}{10 (b c-a d)^2 (a+b x)}-\frac {7 \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}-\frac {2 \sqrt {\frac {2}{5+\sqrt {5}}} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 b^{3/5} (b c-a d)^{12/5}}-\frac {7 \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}+\frac {\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 b^{3/5} (b c-a d)^{12/5}}+\frac {7 d^2 \log \left (\sqrt [5]{b c-a d}-\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{25 b^{3/5} (b c-a d)^{12/5}}-\frac {7 \left (1+\sqrt {5}\right ) d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}-\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{100 b^{3/5} (b c-a d)^{12/5}}-\frac {7 \left (1-\sqrt {5}\right ) d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{100 b^{3/5} (b c-a d)^{12/5}} \] Output:
-1/2*(d*x+c)^(3/5)/(-a*d+b*c)/(b*x+a)^2+7/10*d*(d*x+c)^(3/5)/(-a*d+b*c)^2/ (b*x+a)+7/50*(10-2*5^(1/2))^(1/2)*d^2*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)^(12/5)-7/50*(10+2*5^(1/2))^(1/2)*d^2*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)^(12/5)+7/25*d^2*ln((-a*d+b*c)^(1/5)-b^(1/5)*(d*x+c)^(1/5 ))/b^(3/5)/(-a*d+b*c)^(12/5)-7/100*(5^(1/2)+1)*d^2*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)^(12/5)-7/100*(-5^(1 /2)+1)*d^2*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)^(12/5)
Time = 2.67 (sec) , antiderivative size = 474, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=\frac {1}{100} d^2 \left (\frac {10 (c+d x)^{3/5} (-5 b c+12 a d+7 b d x)}{d^2 (b c-a d)^2 (a+b x)^2}-\frac {14 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \left (5+\sqrt {5}-\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{b^{3/5} (-b c+a d)^{12/5}}-\frac {14 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \left (5-\sqrt {5}+\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{b^{3/5} (-b c+a d)^{12/5}}+\frac {28 \log \left (\sqrt [5]{-b c+a d}+\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{b^{3/5} (-b c+a d)^{12/5}}-\frac {7 \left (1+\sqrt {5}\right ) \log \left (2 (-b c+a d)^{2/5}+\left (-1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{b^{3/5} (-b c+a d)^{12/5}}+\frac {7 \left (-1+\sqrt {5}\right ) \log \left (2 (-b c+a d)^{2/5}-\left (1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{b^{3/5} (-b c+a d)^{12/5}}\right ) \] Input:
Integrate[1/((a + b*x)^3*(c + d*x)^(2/5)),x]
Output:
(d^2*((10*(c + d*x)^(3/5)*(-5*b*c + 12*a*d + 7*b*d*x))/(d^2*(b*c - a*d)^2* (a + b*x)^2) - (14*Sqrt[2*(5 + Sqrt[5])]*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)^(12/5)) - (14*Sqrt[10 - 2*Sqrt[5]]*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)^(12/5)) + (28*Log[(-(b*c) + a*d)^(1/5) + b^(1/5)*(c + d*x)^(1/5)])/(b^(3/5)*(-(b*c) + a*d)^(12/5)) - ( 7*(1 + Sqrt[5])*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)^(12/5)) + (7*(-1 + Sqrt[5])*Log[2*(-(b*c) + a*d)^(2/5) - (1 + S qrt[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)^(12/5))))/100
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.22, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {52, 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.
| \(\displaystyle \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {7 d \int \frac {1}{(a+b x)^2 (c+d x)^{2/5}}dx}{10 (b c-a d)}-\frac {(c+d x)^{3/5}}{2 (a+b x)^2 (b c-a d)}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {7 d \left (-\frac {2 d \int \frac {1}{(a+b x) (c+d x)^{2/5}}dx}{5 (b c-a d)}-\frac {(c+d x)^{3/5}}{(a+b x) (b c-a d)}\right )}{10 (b c-a d)}-\frac {(c+d x)^{3/5}}{2 (a+b x)^2 (b c-a d)}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle -\frac {7 d \left (\frac {2 d (c+d x)^{3/5} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},1,\frac {8}{5},\frac {b (c+d x)}{b c-a d}\right )}{3 (b c-a d)^2}-\frac {(c+d x)^{3/5}}{(a+b x) (b c-a d)}\right )}{10 (b c-a d)}-\frac {(c+d x)^{3/5}}{2 (a+b x)^2 (b c-a d)}\) |
Input:
Int[1/((a + b*x)^3*(c + d*x)^(2/5)),x]
Output:
-1/2*(c + d*x)^(3/5)/((b*c - a*d)*(a + b*x)^2) - (7*d*(-((c + d*x)^(3/5)/( (b*c - a*d)*(a + b*x))) + (2*d*(c + d*x)^(3/5)*Hypergeometric2F1[3/5, 1, 8 /5, (b*(c + d*x))/(b*c - a*d)])/(3*(b*c - a*d)^2)))/(10*(b*c - a*d))
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 = 6.48 (sec) , antiderivative size = 601, normalized size of antiderivative = 1.06
| method | result | size |
| pseudoelliptic | \(\frac {56 \left (-d^{2} \left (b x +a \right )^{2} \arctan \left (\frac {300 \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}+\frac {7}{3}\right ) \sqrt {2}}{\left (5+\sqrt {5}\right )^{\frac {9}{2}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right ) \sqrt {5}\, \sqrt {2}\, \sqrt {5-\sqrt {5}}+\left (\frac {d^{2} \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \left (\sqrt {5}-1\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^{2} \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \left (\sqrt {5}+1\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^{2} \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \ln \left (\left (x d +c \right )^{\frac {1}{5}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}\right )+\frac {30 \left (x d +c \right )^{\frac {3}{5}} b \sqrt {5-\sqrt {5}}\, \left (\frac {\left (7 x d -5 c \right ) b}{12}+a d \right ) \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}}{7}+\left (b x +a \right )^{2} \arctan \left (\frac {300 \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}-\frac {7}{3}\right ) \sqrt {2}}{\left (5-\sqrt {5}\right )^{\frac {9}{2}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right ) d^{2} \sqrt {5}\, \sqrt {2}\right ) \sqrt {5+\sqrt {5}}\right ) d^{2} \sqrt {5}}{125 \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 )^{2} \left (\left (x d +c \right )^{\frac {1}{5}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}\right )^{2} \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 )^{2} \left (a d -b c \right )^{2} b^{3}}\) | \(601\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(5709\) |
| default | \(\text {Expression too large to display}\) | \(5709\) |
Input:
int(1/(b*x+a)^3/(d*x+c)^(2/5),x,method=_RETURNVERBOSE)
Output:
56/125*(-d^2*(b*x+a)^2*arctan(300*((5^(1/2)-1)*((a*d-b*c)/b)^(1/5)+4*(d*x+ c)^(1/5))/(5+5^(1/2))^(9/2)/((a*d-b*c)/b)^(1/5)*(5^(1/2)+7/3)*2^(1/2))*5^( 1/2)*2^(1/2)*(5-5^(1/2))^(1/2)+(1/4*d^2*(5-5^(1/2))^(1/2)*(b*x+a)^2*(5^(1/ 2)-1)*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^2*(5-5^(1/2))^(1/2)*(b*x+a)^2*(5^(1/2)+1)*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*(5-5^(1/2))^(1/2)*(b*x+a)^2*ln((d*x+c)^(1/5)+((a*d-b*c)/b)^(1 /5))+30/7*(d*x+c)^(3/5)*b*(5-5^(1/2))^(1/2)*(1/12*(7*d*x-5*c)*b+a*d)*((a*d -b*c)/b)^(2/5)+(b*x+a)^2*arctan(300/(5-5^(1/2))^(9/2)*((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)-7/3)*2^(1/2))*d ^2*5^(1/2)*2^(1/2))*(5+5^(1/2))^(1/2))/((a*d-b*c)/b)^(2/5)*d^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))^2/((d*x+c)^(1/5)+((a*d-b*c)/b)^(1/5))^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))^2/(a*d-b*c)^ 2/b^3
Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=\text {Timed out} \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(2/5),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=\int \frac {1}{\left (a + b x\right )^{3} \left (c + d x\right )^{\frac {2}{5}}}\, dx \] Input:
integrate(1/(b*x+a)**3/(d*x+c)**(2/5),x)
Output:
Integral(1/((a + b*x)**3*(c + d*x)**(2/5)), x)
Time = 0.14 (sec) , antiderivative size = 672, normalized size of antiderivative = 1.19 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx =\text {Too large to display} \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(2/5),x, algorithm="maxima")
Output:
1/50*(14*d^3*(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)*(s qrt(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^2*c^2 - 2*a*b*c*d + a^2*d^2) + 5*(7*(d*x + c)^(8/5)*b*d^3 - 12*(b*c*d^3 - a*d^4 )*(d*x + c)^(3/5))/(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 + (b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*(d*x + c)^2 - 2*(b ^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3)*(d*x + c)))/d
Time = 0.40 (sec) , antiderivative size = 714, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx =\text {Too large to display} \] Input:
integrate(1/(b*x+a)^3/(d*x+c)^(2/5),x, algorithm="giac")
Output:
-7/100*(b^5*c - a*b^4*d)^(3/5)*d^2*(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^6*c^3 - 3*sqrt(5)*a*b^5*c^2*d + 3*sqrt( 5)*a^2*b^4*c*d^2 - sqrt(5)*a^3*b^3*d^3) - 7/100*(b^5*c - a*b^4*d)^(3/5)*d^ 2*(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^6*c^3 - 3*sqrt(5)*a*b^5*c^2*d + 3*sqrt(5)*a^2*b^4*c*d^2 - sqrt(5)*a^3*b ^3*d^3) + 7/50*(b^5*c - a*b^4*d)^(3/5)*d^2*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^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^2 - a^3*b^3*d^3) - 7/50*(b^5*c - a*b^4*d)^(3/5)*d^2*sqrt(2*sqrt(5) + 10)*arcta n(((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^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^ 2 - a^3*b^3*d^3) + 7/25*d^2*((b*c - a*d)/b)^(3/5)*log(abs((d*x + c)^(1/5) - ((b*c - a*d)/b)^(1/5)))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d ^3) + 1/10*(7*(d*x + c)^(8/5)*b*d^2 - 12*(d*x + c)^(3/5)*b*c*d^2 + 12*(d*x + c)^(3/5)*a*d^3)/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((d*x + c)*b - b*c + a *d)^2)
Time = 1.19 (sec) , antiderivative size = 567, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=\frac {\frac {6\,d^2\,{\left (c+d\,x\right )}^{3/5}}{5\,\left (a\,d-b\,c\right )}+\frac {7\,b\,d^2\,{\left (c+d\,x\right )}^{8/5}}{10\,{\left (a\,d-b\,c\right )}^2}}{b^2\,{\left (c+d\,x\right )}^2-\left (2\,b^2\,c-2\,a\,b\,d\right )\,\left (c+d\,x\right )+a^2\,d^2+b^2\,c^2-2\,a\,b\,c\,d}+\frac {7\,d^2\,\ln \left (\frac {2401\,b^2\,d^8}{625\,{\left (a\,d-b\,c\right )}^7}+\frac {2401\,b^{11/5}\,d^8\,{\left (c+d\,x\right )}^{1/5}}{625\,{\left (a\,d-b\,c\right )}^{36/5}}\right )}{25\,b^{3/5}\,{\left (a\,d-b\,c\right )}^{12/5}}+\frac {d^2\,\ln \left (\frac {2401\,b^2\,d^8}{625\,{\left (a\,d-b\,c\right )}^7}+\frac {2401\,b^{11/5}\,d^8\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}+\sqrt {5}-1\right )}^3}{40000\,{\left (a\,d-b\,c\right )}^{36/5}}\right )\,\left (\frac {7\,\sqrt {5}}{100}+\frac {7\,\sqrt {-2\,\sqrt {5}-10}}{100}-\frac {7}{100}\right )}{b^{3/5}\,{\left (a\,d-b\,c\right )}^{12/5}}-\frac {d^2\,\ln \left (\frac {2401\,b^2\,d^8}{625\,{\left (a\,d-b\,c\right )}^7}-\frac {2401\,b^{11/5}\,d^8\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}-\sqrt {5}+1\right )}^3}{40000\,{\left (a\,d-b\,c\right )}^{36/5}}\right )\,\left (\frac {7\,\sqrt {-2\,\sqrt {5}-10}}{100}-\frac {7\,\sqrt {5}}{100}+\frac {7}{100}\right )}{b^{3/5}\,{\left (a\,d-b\,c\right )}^{12/5}}-\frac {d^2\,\ln \left (\frac {2401\,b^2\,d^8}{625\,{\left (a\,d-b\,c\right )}^7}-\frac {2401\,b^{11/5}\,d^8\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {5}+\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{40000\,{\left (a\,d-b\,c\right )}^{36/5}}\right )\,\left (\frac {7\,\sqrt {5}}{100}+\frac {7\,\sqrt {2\,\sqrt {5}-10}}{100}+\frac {7}{100}\right )}{b^{3/5}\,{\left (a\,d-b\,c\right )}^{12/5}}-\frac {d^2\,\ln \left (\frac {2401\,b^2\,d^8}{625\,{\left (a\,d-b\,c\right )}^7}-\frac {2401\,b^{11/5}\,d^8\,{\left (c+d\,x\right )}^{1/5}\,{\left (\sqrt {5}-\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}^3}{40000\,{\left (a\,d-b\,c\right )}^{36/5}}\right )\,\left (\frac {7\,\sqrt {5}}{100}-\frac {7\,\sqrt {2\,\sqrt {5}-10}}{100}+\frac {7}{100}\right )}{b^{3/5}\,{\left (a\,d-b\,c\right )}^{12/5}} \] Input:
int(1/((a + b*x)^3*(c + d*x)^(2/5)),x)
Output:
((6*d^2*(c + d*x)^(3/5))/(5*(a*d - b*c)) + (7*b*d^2*(c + d*x)^(8/5))/(10*( a*d - b*c)^2))/(b^2*(c + d*x)^2 - (2*b^2*c - 2*a*b*d)*(c + d*x) + a^2*d^2 + b^2*c^2 - 2*a*b*c*d) + (7*d^2*log((2401*b^2*d^8)/(625*(a*d - b*c)^7) + ( 2401*b^(11/5)*d^8*(c + d*x)^(1/5))/(625*(a*d - b*c)^(36/5))))/(25*b^(3/5)* (a*d - b*c)^(12/5)) + (d^2*log((2401*b^2*d^8)/(625*(a*d - b*c)^7) + (2401* b^(11/5)*d^8*(c + d*x)^(1/5)*(2^(1/2)*(- 5^(1/2) - 5)^(1/2) + 5^(1/2) - 1) ^3)/(40000*(a*d - b*c)^(36/5)))*((7*5^(1/2))/100 + (7*(- 2*5^(1/2) - 10)^( 1/2))/100 - 7/100))/(b^(3/5)*(a*d - b*c)^(12/5)) - (d^2*log((2401*b^2*d^8) /(625*(a*d - b*c)^7) - (2401*b^(11/5)*d^8*(c + d*x)^(1/5)*(2^(1/2)*(- 5^(1 /2) - 5)^(1/2) - 5^(1/2) + 1)^3)/(40000*(a*d - b*c)^(36/5)))*((7*(- 2*5^(1 /2) - 10)^(1/2))/100 - (7*5^(1/2))/100 + 7/100))/(b^(3/5)*(a*d - b*c)^(12/ 5)) - (d^2*log((2401*b^2*d^8)/(625*(a*d - b*c)^7) - (2401*b^(11/5)*d^8*(c + d*x)^(1/5)*(5^(1/2) + 2^(1/2)*(5^(1/2) - 5)^(1/2) + 1)^3)/(40000*(a*d - b*c)^(36/5)))*((7*5^(1/2))/100 + (7*(2*5^(1/2) - 10)^(1/2))/100 + 7/100))/ (b^(3/5)*(a*d - b*c)^(12/5)) - (d^2*log((2401*b^2*d^8)/(625*(a*d - b*c)^7) - (2401*b^(11/5)*d^8*(c + d*x)^(1/5)*(5^(1/2) - 2^(1/2)*(5^(1/2) - 5)^(1/ 2) + 1)^3)/(40000*(a*d - b*c)^(36/5)))*((7*5^(1/2))/100 - (7*(2*5^(1/2) - 10)^(1/2))/100 + 7/100))/(b^(3/5)*(a*d - b*c)^(12/5))
\[ \int \frac {1}{(a+b x)^3 (c+d x)^{2/5}} \, dx=\text {too large to display} \] Input:
int(1/(b*x+a)^3/(d*x+c)^(2/5),x)
Output:
(2*(c + d*x)**(2/5)*int(x/((c + d*x)**(2/5)*a**5*c*d**2 + (c + d*x)**(2/5) *a**5*d**3*x + 22*(c + d*x)**(2/5)*a**4*b*c**2*d + 25*(c + d*x)**(2/5)*a** 4*b*c*d**2*x + 3*(c + d*x)**(2/5)*a**4*b*d**3*x**2 + 121*(c + d*x)**(2/5)* a**3*b**2*c**3 + 187*(c + d*x)**(2/5)*a**3*b**2*c**2*d*x + 69*(c + d*x)**( 2/5)*a**3*b**2*c*d**2*x**2 + 3*(c + d*x)**(2/5)*a**3*b**2*d**3*x**3 + 363* (c + d*x)**(2/5)*a**2*b**3*c**3*x + 429*(c + d*x)**(2/5)*a**2*b**3*c**2*d* x**2 + 67*(c + d*x)**(2/5)*a**2*b**3*c*d**2*x**3 + (c + d*x)**(2/5)*a**2*b **3*d**3*x**4 + 363*(c + d*x)**(2/5)*a*b**4*c**3*x**2 + 385*(c + d*x)**(2/ 5)*a*b**4*c**2*d*x**3 + 22*(c + d*x)**(2/5)*a*b**4*c*d**2*x**4 + 121*(c + d*x)**(2/5)*b**5*c**3*x**3 + 121*(c + d*x)**(2/5)*b**5*c**2*d*x**4),x)*a** 6*d**5 + 52*(c + d*x)**(2/5)*int(x/((c + d*x)**(2/5)*a**5*c*d**2 + (c + d* x)**(2/5)*a**5*d**3*x + 22*(c + d*x)**(2/5)*a**4*b*c**2*d + 25*(c + d*x)** (2/5)*a**4*b*c*d**2*x + 3*(c + d*x)**(2/5)*a**4*b*d**3*x**2 + 121*(c + d*x )**(2/5)*a**3*b**2*c**3 + 187*(c + d*x)**(2/5)*a**3*b**2*c**2*d*x + 69*(c + d*x)**(2/5)*a**3*b**2*c*d**2*x**2 + 3*(c + d*x)**(2/5)*a**3*b**2*d**3*x* *3 + 363*(c + d*x)**(2/5)*a**2*b**3*c**3*x + 429*(c + d*x)**(2/5)*a**2*b** 3*c**2*d*x**2 + 67*(c + d*x)**(2/5)*a**2*b**3*c*d**2*x**3 + (c + d*x)**(2/ 5)*a**2*b**3*d**3*x**4 + 363*(c + d*x)**(2/5)*a*b**4*c**3*x**2 + 385*(c + d*x)**(2/5)*a*b**4*c**2*d*x**3 + 22*(c + d*x)**(2/5)*a*b**4*c*d**2*x**4 + 121*(c + d*x)**(2/5)*b**5*c**3*x**3 + 121*(c + d*x)**(2/5)*b**5*c**2*d*...