Integrand size = 17, antiderivative size = 558 \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=-\frac {(c+d x)^{4/5}}{2 b (a+b x)^2}-\frac {2 d (c+d x)^{4/5}}{5 b (b c-a d) (a+b x)}+\frac {\sqrt {2 \left (5+\sqrt {5}\right )} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}-\frac {2 \sqrt {\frac {2}{5+\sqrt {5}}} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 b^{9/5} (b c-a d)^{6/5}}-\frac {\sqrt {2 \left (5-\sqrt {5}\right )} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}+\frac {\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 b^{9/5} (b c-a d)^{6/5}}-\frac {2 d^2 \log \left (\sqrt [5]{b c-a d}-\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{25 b^{9/5} (b c-a d)^{6/5}}+\frac {\left (1-\sqrt {5}\right ) d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}-\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{50 b^{9/5} (b c-a d)^{6/5}}+\frac {\left (1+\sqrt {5}\right ) d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{50 b^{9/5} (b c-a d)^{6/5}} \] Output:
-1/2*(d*x+c)^(4/5)/b/(b*x+a)^2-2/5*d*(d*x+c)^(4/5)/b/(-a*d+b*c)/(b*x+a)-1/ 25*(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^(9/5)/(-a*d+b*c) ^(6/5)-1/25*(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^(9/5)/(-a* d+b*c)^(6/5)-2/25*d^2*ln((-a*d+b*c)^(1/5)-b^(1/5)*(d*x+c)^(1/5))/b^(9/5)/( -a*d+b*c)^(6/5)+1/50*(-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^(9/5)/(-a*d+b*c)^(6/5)+1/50*(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^(9/5)/(-a*d+b*c)^( 6/5)
Time = 3.84 (sec) , antiderivative size = 458, normalized size of antiderivative = 0.82 \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=\frac {d^2 \left (\frac {5 b^{4/5} (c+d x)^{4/5} (5 b c-a d+4 b d x)}{d^2 (-b c+a d) (a+b x)^2}-\frac {2 \sqrt {10-2 \sqrt {5}} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \left (5+\sqrt {5}-\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{(-b c+a d)^{6/5}}+\frac {2 \sqrt {2 \left (5+\sqrt {5}\right )} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \left (5-\sqrt {5}+\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{(-b c+a d)^{6/5}}-\frac {4 \log \left (\sqrt [5]{-b c+a d}+\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{(-b c+a d)^{6/5}}-\frac {\left (-1+\sqrt {5}\right ) \log \left (2 (-b c+a d)^{2/5}+\left (-1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{(-b c+a d)^{6/5}}+\frac {\left (1+\sqrt {5}\right ) \log \left (2 (-b c+a d)^{2/5}-\left (1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{(-b c+a d)^{6/5}}\right )}{50 b^{9/5}} \] Input:
Integrate[(c + d*x)^(4/5)/(a + b*x)^3,x]
Output:
(d^2*((5*b^(4/5)*(c + d*x)^(4/5)*(5*b*c - a*d + 4*b*d*x))/(d^2*(-(b*c) + a *d)*(a + b*x)^2) - (2*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)))/1 0])/(-(b*c) + a*d)^(6/5) + (2*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*c) + a*d)^(6/5) - (4*Log[(-(b*c) + a*d)^(1/5) + b^(1/5)* (c + d*x)^(1/5)])/(-(b*c) + a*d)^(6/5) - ((-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*c) + a*d)^(6/5) + ((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*c) + a*d)^(6/5)))/(50*b^(9/5))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.20, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {51, 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.
| \(\displaystyle \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {2 d \int \frac {1}{(a+b x)^2 \sqrt [5]{c+d x}}dx}{5 b}-\frac {(c+d x)^{4/5}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {2 d \left (-\frac {d \int \frac {1}{(a+b x) \sqrt [5]{c+d x}}dx}{5 (b c-a d)}-\frac {(c+d x)^{4/5}}{(a+b x) (b c-a d)}\right )}{5 b}-\frac {(c+d x)^{4/5}}{2 b (a+b x)^2}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {2 d \left (\frac {d (c+d x)^{4/5} \operatorname {Hypergeometric2F1}\left (\frac {4}{5},1,\frac {9}{5},\frac {b (c+d x)}{b c-a d}\right )}{4 (b c-a d)^2}-\frac {(c+d x)^{4/5}}{(a+b x) (b c-a d)}\right )}{5 b}-\frac {(c+d x)^{4/5}}{2 b (a+b x)^2}\) |
Input:
Int[(c + d*x)^(4/5)/(a + b*x)^3,x]
Output:
-1/2*(c + d*x)^(4/5)/(b*(a + b*x)^2) + (2*d*(-((c + d*x)^(4/5)/((b*c - a*d )*(a + b*x))) + (d*(c + d*x)^(4/5)*Hypergeometric2F1[4/5, 1, 9/5, (b*(c + d*x))/(b*c - a*d)])/(4*(b*c - a*d)^2)))/(5*b)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[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] && 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 = 1.44 (sec) , antiderivative size = 471, normalized size of antiderivative = 0.84
| method | result | size |
| pseudoelliptic | \(-\frac {\left (-\frac {d^{2} \sqrt {5+\sqrt {5}}\, \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \left (\sqrt {5}+1\right ) \ln \left (2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (-\sqrt {5}-1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+2 \left (x d +c \right )^{\frac {2}{5}}\right )}{4}+\frac {d^{2} \sqrt {5+\sqrt {5}}\, \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \left (\sqrt {5}-1\right ) \ln \left (2 \left (\frac {a d -b c}{b}\right )^{\frac {2}{5}}+\left (\sqrt {5}-1\right ) \left (x d +c \right )^{\frac {1}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+2 \left (x d +c \right )^{\frac {2}{5}}\right )}{4}-\frac {\sqrt {2}\, d^{2} \sqrt {5+\sqrt {5}}\, \left (b x +a \right )^{2} \left (\sqrt {5}-5\right ) \arctan \left (\frac {\sqrt {2}\, \left (\left (\sqrt {5}+1\right ) \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}-4 \left (x d +c \right )^{\frac {1}{5}}\right )}{2 \sqrt {5-\sqrt {5}}\, \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right )}{2}+\left (-\frac {\sqrt {2}\, d^{2} \left (b x +a \right )^{2} \left (5+\sqrt {5}\right ) \arctan \left (\frac {\left (\left (\sqrt {5}-1\right ) \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}+4 \left (x d +c \right )^{\frac {1}{5}}\right ) \sqrt {2}}{2 \sqrt {5+\sqrt {5}}\, \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}\right )}{2}+\sqrt {5+\sqrt {5}}\, \left (d^{2} \left (b x +a \right )^{2} \ln \left (\left (x d +c \right )^{\frac {1}{5}}+\left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}\right )+\frac {5 \left (\left (-4 b x +a \right ) d -5 b c \right ) \left (x d +c \right )^{\frac {4}{5}} b \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}}}{4}\right )\right ) \sqrt {5-\sqrt {5}}\right ) \sqrt {5}}{125 \left (\frac {a d -b c}{b}\right )^{\frac {1}{5}} \left (a d -b c \right ) \left (b x +a \right )^{2} b^{2}}\) | \(471\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1346\) |
| default | \(\text {Expression too large to display}\) | \(1346\) |
Input:
int((d*x+c)^(4/5)/(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
-1/125/((a*d-b*c)/b)^(1/5)*(-1/4*d^2*(5+5^(1/2))^(1/2)*(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)*(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))-1/2*2^(1/2)*d^2*(5+5^(1/2))^(1/2 )*(b*x+a)^2*(5^(1/2)-5)*arctan(1/2/(5-5^(1/2))^(1/2)*2^(1/2)/((a*d-b*c)/b) ^(1/5)*((5^(1/2)+1)*((a*d-b*c)/b)^(1/5)-4*(d*x+c)^(1/5)))+(-1/2*2^(1/2)*d^ 2*(b*x+a)^2*(5+5^(1/2))*arctan(1/2*((5^(1/2)-1)*((a*d-b*c)/b)^(1/5)+4*(d*x +c)^(1/5))/(5+5^(1/2))^(1/2)/((a*d-b*c)/b)^(1/5)*2^(1/2))+(5+5^(1/2))^(1/2 )*(d^2*(b*x+a)^2*ln((d*x+c)^(1/5)+((a*d-b*c)/b)^(1/5))+5/4*((-4*b*x+a)*d-5 *b*c)*(d*x+c)^(4/5)*b*((a*d-b*c)/b)^(1/5)))*(5-5^(1/2))^(1/2))*5^(1/2)/(a* d-b*c)/(b*x+a)^2/b^2
Timed out. \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=\text {Timed out} \] Input:
integrate((d*x+c)^(4/5)/(b*x+a)^3,x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=\int \frac {\left (c + d x\right )^{\frac {4}{5}}}{\left (a + b x\right )^{3}}\, dx \] Input:
integrate((d*x+c)**(4/5)/(b*x+a)**3,x)
Output:
Integral((c + d*x)**(4/5)/(a + b*x)**3, x)
Time = 0.15 (sec) , antiderivative size = 644, normalized size of antiderivative = 1.15 \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)^(4/5)/(b*x+a)^3,x, algorithm="maxima")
Output:
-1/50*(2*d^3*(sqrt(5)*(sqrt(5) - 1)*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) ^(1/5)*b^(4/5)*sqrt(2*sqrt(5) - 10)) + sqrt(5)*(sqrt(5) + 1)*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)^(1/5)*b^(4/5)*sqrt(-2*sqrt(5) - 10)) + (sqrt(5 ) + 3)*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)^(1/5)*b^(4/5) *(sqrt(5) + 1)) + (sqrt(5) - 3)*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)^(1/5)*b^(4/5)*(sqrt(5) - 1)) - 2*log((d*x + c)^(1/5)*b^(1/5) + (-b*c + a*d)^(1/5))/((-b*c + a*d)^(1/5)*b^(4/5)))/(b^2*c - a*b*d) + 5*(4*( d*x + c)^(9/5)*b*d^3 + (b*c*d^3 - a*d^4)*(d*x + c)^(4/5))/(b^4*c^3 - 3*a*b ^3*c^2*d + 3*a^2*b^2*c*d^2 - a^3*b*d^3 + (b^4*c - a*b^3*d)*(d*x + c)^2 - 2 *(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*(d*x + c)))/d
Time = 0.45 (sec) , antiderivative size = 637, normalized size of antiderivative = 1.14 \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx =\text {Too large to display} \] Input:
integrate((d*x+c)^(4/5)/(b*x+a)^3,x, algorithm="giac")
Output:
1/50*(b^5*c - a*b^4*d)^(4/5)*d^2*(sqrt(5) + 1)*log(1/2*(d*x + c)^(1/5)*(sq rt(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))/(b^7*c^2 - 2*a*b^6*c*d + a^2*b^5*d^2) - 1/50*(b^5*c - a*b^4*d)^(4/5)*d^2*(sqrt(5) - 1)*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))/(b^7*c^2 - 2*a*b^6*c*d + a^2*b^5*d^2) - 2/25*(b^5*c - a*b^4*d)^ (4/5)*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^7 *c^2*(sqrt(5) - 1) - 2*a*b^6*c*d*(sqrt(5) - 1) + a^2*b^5*d^2*(sqrt(5) - 1) ) - 2/25*(b^5*c - a*b^4*d)^(4/5)*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^7*c^2*(sqrt(5) + 1) - 2*a*b^6*c*d*(sqrt(5) + 1) + a^2*b^5*d^2*(sqrt(5) + 1)) - 2/25*d^2*((b*c - a*d)/b)^(4/5)*log(abs((d*x + c)^(1/5) - ((b*c - a*d)/b)^(1/5)))/(b^3*c^2 - 2*a*b^2*c*d + a^2*b*d^2) - 1/10*(4*(d*x + c)^(9/5)*b*d^2 + (d*x + c)^(4/5)*b*c*d^2 - (d*x + c)^(4/5) *a*d^3)/((b^2*c - a*b*d)*((d*x + c)*b - b*c + a*d)^2)
Time = 0.65 (sec) , antiderivative size = 591, normalized size of antiderivative = 1.06 \[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=\frac {2\,d^2\,\ln \left (\frac {8\,d^6}{25\,\left (b^2\,c-a\,b\,d\right )}-\frac {8\,d^6\,{\left (c+d\,x\right )}^{1/5}}{25\,{\left (-b\right )}^{4/5}\,{\left (a\,d-b\,c\right )}^{6/5}}\right )}{25\,{\left (-b\right )}^{9/5}\,{\left (a\,d-b\,c\right )}^{6/5}}-\frac {\ln \left (\frac {8\,d^6}{25\,\left (b^2\,c-a\,b\,d\right )}+\frac {2\,d^6\,{\left (c+d\,x\right )}^{1/5}\,\left (\sqrt {2}\,\sqrt {-\sqrt {5}-5}-\sqrt {5}+1\right )}{25\,{\left (-b\right )}^{4/5}\,{\left (a\,d-b\,c\right )}^{6/5}}\right )\,\left (d^2\,\sqrt {-2\,\sqrt {5}-10}-\sqrt {5}\,d^2+d^2\right )}{50\,{\left (-b\right )}^{9/5}\,{\left (a\,d-b\,c\right )}^{6/5}}-\frac {\frac {d^2\,{\left (c+d\,x\right )}^{4/5}}{10\,b}-\frac {2\,d^2\,{\left (c+d\,x\right )}^{9/5}}{5\,\left (a\,d-b\,c\right )}}{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 {\ln \left (\frac {8\,d^6}{25\,\left (b^2\,c-a\,b\,d\right )}+\frac {2\,d^6\,{\left (c+d\,x\right )}^{1/5}\,\left (\sqrt {5}+\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}{25\,{\left (-b\right )}^{4/5}\,{\left (a\,d-b\,c\right )}^{6/5}}\right )\,\left (\sqrt {5}\,d^2+d^2\,\sqrt {2\,\sqrt {5}-10}+d^2\right )}{50\,{\left (-b\right )}^{9/5}\,{\left (a\,d-b\,c\right )}^{6/5}}-\frac {\ln \left (\frac {8\,d^6}{25\,\left (b^2\,c-a\,b\,d\right )}+\frac {2\,d^6\,{\left (c+d\,x\right )}^{1/5}\,\left (\sqrt {5}-\sqrt {2}\,\sqrt {\sqrt {5}-5}+1\right )}{25\,{\left (-b\right )}^{4/5}\,{\left (a\,d-b\,c\right )}^{6/5}}\right )\,\left (\sqrt {5}\,d^2-d^2\,\sqrt {2\,\sqrt {5}-10}+d^2\right )}{50\,{\left (-b\right )}^{9/5}\,{\left (a\,d-b\,c\right )}^{6/5}}+\frac {d^2\,\ln \left (\frac {8\,d^6}{25\,\left (b^2\,c-a\,b\,d\right )}-\frac {4\,d^6\,{\left (c+d\,x\right )}^{1/5}\,\left (\frac {\sqrt {2}\,\sqrt {-\sqrt {5}-5}}{50}+\frac {\sqrt {5}}{50}-\frac {1}{50}\right )}{{\left (-b\right )}^{4/5}\,{\left (a\,d-b\,c\right )}^{6/5}}\right )\,\left (\frac {\sqrt {5}}{50}+\frac {\sqrt {-2\,\sqrt {5}-10}}{50}-\frac {1}{50}\right )}{{\left (-b\right )}^{9/5}\,{\left (a\,d-b\,c\right )}^{6/5}} \] Input:
int((c + d*x)^(4/5)/(a + b*x)^3,x)
Output:
(2*d^2*log((8*d^6)/(25*(b^2*c - a*b*d)) - (8*d^6*(c + d*x)^(1/5))/(25*(-b) ^(4/5)*(a*d - b*c)^(6/5))))/(25*(-b)^(9/5)*(a*d - b*c)^(6/5)) - (log((8*d^ 6)/(25*(b^2*c - a*b*d)) + (2*d^6*(c + d*x)^(1/5)*(2^(1/2)*(- 5^(1/2) - 5)^ (1/2) - 5^(1/2) + 1))/(25*(-b)^(4/5)*(a*d - b*c)^(6/5)))*(d^2*(- 2*5^(1/2) - 10)^(1/2) - 5^(1/2)*d^2 + d^2))/(50*(-b)^(9/5)*(a*d - b*c)^(6/5)) - ((d ^2*(c + d*x)^(4/5))/(10*b) - (2*d^2*(c + d*x)^(9/5))/(5*(a*d - b*c)))/(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) - (log((8*d^6)/(25*(b^2*c - a*b*d)) + (2*d^6*(c + d*x)^(1/5)*(5^(1/2) + 2^(1/2)*(5^(1/2) - 5)^(1/2) + 1))/(25*(-b)^(4/5)*(a*d - b*c)^(6/5)))*(5^ (1/2)*d^2 + d^2*(2*5^(1/2) - 10)^(1/2) + d^2))/(50*(-b)^(9/5)*(a*d - b*c)^ (6/5)) - (log((8*d^6)/(25*(b^2*c - a*b*d)) + (2*d^6*(c + d*x)^(1/5)*(5^(1/ 2) - 2^(1/2)*(5^(1/2) - 5)^(1/2) + 1))/(25*(-b)^(4/5)*(a*d - b*c)^(6/5)))* (5^(1/2)*d^2 - d^2*(2*5^(1/2) - 10)^(1/2) + d^2))/(50*(-b)^(9/5)*(a*d - b* c)^(6/5)) + (d^2*log((8*d^6)/(25*(b^2*c - a*b*d)) - (4*d^6*(c + d*x)^(1/5) *((2^(1/2)*(- 5^(1/2) - 5)^(1/2))/50 + 5^(1/2)/50 - 1/50))/((-b)^(4/5)*(a* d - b*c)^(6/5)))*(5^(1/2)/50 + (- 2*5^(1/2) - 10)^(1/2)/50 - 1/50))/((-b)^ (9/5)*(a*d - b*c)^(6/5))
\[ \int \frac {(c+d x)^{4/5}}{(a+b x)^3} \, dx=\text {too large to display} \] Input:
int((d*x+c)^(4/5)/(b*x+a)^3,x)
Output:
(4*(c + d*x)**(1/5)*int(x/((c + d*x)**(1/5)*a**5*c*d**2 + (c + d*x)**(1/5) *a**5*d**3*x + 42*(c + d*x)**(1/5)*a**4*b*c**2*d + 45*(c + d*x)**(1/5)*a** 4*b*c*d**2*x + 3*(c + d*x)**(1/5)*a**4*b*d**3*x**2 + 441*(c + d*x)**(1/5)* a**3*b**2*c**3 + 567*(c + d*x)**(1/5)*a**3*b**2*c**2*d*x + 129*(c + d*x)** (1/5)*a**3*b**2*c*d**2*x**2 + 3*(c + d*x)**(1/5)*a**3*b**2*d**3*x**3 + 132 3*(c + d*x)**(1/5)*a**2*b**3*c**3*x + 1449*(c + d*x)**(1/5)*a**2*b**3*c**2 *d*x**2 + 127*(c + d*x)**(1/5)*a**2*b**3*c*d**2*x**3 + (c + d*x)**(1/5)*a* *2*b**3*d**3*x**4 + 1323*(c + d*x)**(1/5)*a*b**4*c**3*x**2 + 1365*(c + d*x )**(1/5)*a*b**4*c**2*d*x**3 + 42*(c + d*x)**(1/5)*a*b**4*c*d**2*x**4 + 441 *(c + d*x)**(1/5)*b**5*c**3*x**3 + 441*(c + d*x)**(1/5)*b**5*c**2*d*x**4), x)*a**7*d**6 + 200*(c + d*x)**(1/5)*int(x/((c + d*x)**(1/5)*a**5*c*d**2 + (c + d*x)**(1/5)*a**5*d**3*x + 42*(c + d*x)**(1/5)*a**4*b*c**2*d + 45*(c + d*x)**(1/5)*a**4*b*c*d**2*x + 3*(c + d*x)**(1/5)*a**4*b*d**3*x**2 + 441*( c + d*x)**(1/5)*a**3*b**2*c**3 + 567*(c + d*x)**(1/5)*a**3*b**2*c**2*d*x + 129*(c + d*x)**(1/5)*a**3*b**2*c*d**2*x**2 + 3*(c + d*x)**(1/5)*a**3*b**2 *d**3*x**3 + 1323*(c + d*x)**(1/5)*a**2*b**3*c**3*x + 1449*(c + d*x)**(1/5 )*a**2*b**3*c**2*d*x**2 + 127*(c + d*x)**(1/5)*a**2*b**3*c*d**2*x**3 + (c + d*x)**(1/5)*a**2*b**3*d**3*x**4 + 1323*(c + d*x)**(1/5)*a*b**4*c**3*x**2 + 1365*(c + d*x)**(1/5)*a*b**4*c**2*d*x**3 + 42*(c + d*x)**(1/5)*a*b**4*c *d**2*x**4 + 441*(c + d*x)**(1/5)*b**5*c**3*x**3 + 441*(c + d*x)**(1/5)...