\(\int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx\) [675]

Optimal result

Integrand size = 17, antiderivative size = 592 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\frac {63 d^2}{20 (b c-a d)^3 (c+d x)^{4/5}}-\frac {1}{2 (b c-a d) (a+b x)^2 (c+d x)^{4/5}}+\frac {7 d}{5 (b c-a d)^2 (a+b x) (c+d x)^{4/5}}+\frac {63 \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} b^{4/5} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5-2 \sqrt {5}\right )}-\frac {2 \sqrt {\frac {2}{5+\sqrt {5}}} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 (b c-a d)^{19/5}}-\frac {63 \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} b^{4/5} d^2 \arctan \left (\sqrt {\frac {1}{5} \left (5+2 \sqrt {5}\right )}+\frac {\sqrt {\frac {2}{5} \left (5+\sqrt {5}\right )} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{b c-a d}}\right )}{25 (b c-a d)^{19/5}}+\frac {63 b^{4/5} d^2 \log \left (\sqrt [5]{b c-a d}-\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{25 (b c-a d)^{19/5}}-\frac {63 \left (1-\sqrt {5}\right ) b^{4/5} d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}-\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{100 (b c-a d)^{19/5}}-\frac {63 \left (1+\sqrt {5}\right ) b^{4/5} d^2 \log \left (2 (b c-a d)^{2/5}+\sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+\sqrt {5} \sqrt [5]{b} \sqrt [5]{b c-a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{100 (b c-a d)^{19/5}} \] Output:

63/20*d^2/(-a*d+b*c)^3/(d*x+c)^(4/5)-1/2/(-a*d+b*c)/(b*x+a)^2/(d*x+c)^(4/5 
)+7/5*d/(-a*d+b*c)^2/(b*x+a)/(d*x+c)^(4/5)-63/50*(10+2*5^(1/2))^(1/2)*b^(4 
/5)*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))/(-a*d+b*c)^(19/5)-63/50*(10-2*5^(1/2)) 
^(1/2)*b^(4/5)*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))/(-a*d+b*c)^(19/5)+63/25*b^(4/5 
)*d^2*ln((-a*d+b*c)^(1/5)-b^(1/5)*(d*x+c)^(1/5))/(-a*d+b*c)^(19/5)-63/100* 
(-5^(1/2)+1)*b^(4/5)*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))/(-a*d+b*c)^(19/5)-63/100*(5^(1/2)+1)*b^(4/5)*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))/(-a*d+b*c)^(19/5)
 

Mathematica [A] (verified)

Time = 4.81 (sec) , antiderivative size = 505, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\frac {1}{100} d^2 \left (-\frac {5 \left (25 a^2 d^2+2 a b d (24 c+49 d x)+b^2 \left (-10 c^2+28 c d x+63 d^2 x^2\right )\right )}{d^2 (-b c+a d)^3 (a+b x)^2 (c+d x)^{4/5}}+\frac {126 \sqrt {10-2 \sqrt {5}} b^{4/5} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \left (5+\sqrt {5}-\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{(-b c+a d)^{19/5}}-\frac {126 \sqrt {2 \left (5+\sqrt {5}\right )} b^{4/5} \arctan \left (\frac {1}{10} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \left (5-\sqrt {5}+\frac {4 \sqrt {5} \sqrt [5]{b} \sqrt [5]{c+d x}}{\sqrt [5]{-b c+a d}}\right )\right )}{(-b c+a d)^{19/5}}-\frac {252 b^{4/5} \log \left (\sqrt [5]{-b c+a d}+\sqrt [5]{b} \sqrt [5]{c+d x}\right )}{(-b c+a d)^{19/5}}-\frac {63 \left (-1+\sqrt {5}\right ) b^{4/5} \log \left (2 (-b c+a d)^{2/5}+\left (-1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{(-b c+a d)^{19/5}}+\frac {63 \left (1+\sqrt {5}\right ) b^{4/5} \log \left (2 (-b c+a d)^{2/5}-\left (1+\sqrt {5}\right ) \sqrt [5]{b} \sqrt [5]{-b c+a d} \sqrt [5]{c+d x}+2 b^{2/5} (c+d x)^{2/5}\right )}{(-b c+a d)^{19/5}}\right ) \] Input:

Integrate[1/((a + b*x)^3*(c + d*x)^(9/5)),x]
 

Output:

(d^2*((-5*(25*a^2*d^2 + 2*a*b*d*(24*c + 49*d*x) + b^2*(-10*c^2 + 28*c*d*x 
+ 63*d^2*x^2)))/(d^2*(-(b*c) + a*d)^3*(a + b*x)^2*(c + d*x)^(4/5)) + (126* 
Sqrt[10 - 2*Sqrt[5]]*b^(4/5)*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)^(19/5) - (126*Sqrt[2*(5 + Sqrt[5])]*b^(4/5)*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)^(19/5) - (252*b^(4/5)*Log[(-(b*c) + a*d)^(1/5) + 
b^(1/5)*(c + d*x)^(1/5)])/(-(b*c) + a*d)^(19/5) - (63*(-1 + Sqrt[5])*b^(4/ 
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)^(19/5) + (6 
3*(1 + Sqrt[5])*b^(4/5)*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)^(19/5)))/100
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.18 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.21, 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.

\(\displaystyle \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 d \int \frac {1}{(a+b x)^2 (c+d x)^{9/5}}dx}{5 (b c-a d)}-\frac {1}{2 (a+b x)^2 (c+d x)^{4/5} (b c-a d)}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 d \left (-\frac {9 d \int \frac {1}{(a+b x) (c+d x)^{9/5}}dx}{5 (b c-a d)}-\frac {1}{(a+b x) (c+d x)^{4/5} (b c-a d)}\right )}{5 (b c-a d)}-\frac {1}{2 (a+b x)^2 (c+d x)^{4/5} (b c-a d)}\)

\(\Big \downarrow \) 78

\(\displaystyle -\frac {7 d \left (-\frac {9 d \operatorname {Hypergeometric2F1}\left (-\frac {4}{5},1,\frac {1}{5},\frac {b (c+d x)}{b c-a d}\right )}{4 (c+d x)^{4/5} (b c-a d)^2}-\frac {1}{(a+b x) (c+d x)^{4/5} (b c-a d)}\right )}{5 (b c-a d)}-\frac {1}{2 (a+b x)^2 (c+d x)^{4/5} (b c-a d)}\)

Input:

Int[1/((a + b*x)^3*(c + d*x)^(9/5)),x]
 

Output:

-1/2*1/((b*c - a*d)*(a + b*x)^2*(c + d*x)^(4/5)) - (7*d*(-(1/((b*c - a*d)* 
(a + b*x)*(c + d*x)^(4/5))) - (9*d*Hypergeometric2F1[-4/5, 1, 1/5, (b*(c + 
 d*x))/(b*c - a*d)])/(4*(b*c - a*d)^2*(c + d*x)^(4/5))))/(5*(b*c - a*d))
 

Defintions of rubi rules used

rule 52

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]
 

rule 78

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]
 

Maple [A] (verified)

Time = 1.62 (sec) , antiderivative size = 497, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(-\frac {63 \left (-\left (x d +c \right )^{\frac {4}{5}} d^{2} \left (b x +a \right )^{2} \left (5+\sqrt {5}\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 )-\left (x d +c \right )^{\frac {4}{5}} d^{2} \left (b x +a \right )^{2} \left (\sqrt {5}-5\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 )+\left (x d +c \right )^{\frac {4}{5}} d^{2} \sqrt {2}\, \sqrt {5-\sqrt {5}}\, \left (b x +a \right )^{2} \left (5+\sqrt {5}\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 )+4 \left (x d +c \right )^{\frac {4}{5}} d^{2} \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 {125 \left (\left (\frac {63}{25} b^{2} x^{2}+\frac {98}{25} a b x +a^{2}\right ) d^{2}+\frac {48 \left (\frac {7 b x}{12}+a \right ) c b d}{25}-\frac {2 b^{2} c^{2}}{5}\right ) \sqrt {5}\, \left (\frac {a d -b c}{b}\right )^{\frac {4}{5}}}{63}+\left (b x +a \right )^{2} \left (x d +c \right )^{\frac {4}{5}} d^{2} \sqrt {5+\sqrt {5}}\, \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 ) \left (\sqrt {5}-5\right ) \sqrt {2}\right ) \sqrt {5}}{500 \left (x d +c \right )^{\frac {4}{5}} \left (\frac {a d -b c}{b}\right )^{\frac {4}{5}} \left (b x +a \right )^{2} \left (a d -b c \right )^{3}}\)	\(497\)
derivativedivides	\(\text {Expression too large to display}\)	\(1408\)
default	\(\text {Expression too large to display}\)	\(1408\)

Input:

int(1/(b*x+a)^3/(d*x+c)^(9/5),x,method=_RETURNVERBOSE)
 

Output:

-63/500/(d*x+c)^(4/5)*(-(d*x+c)^(4/5)*d^2*(b*x+a)^2*(5+5^(1/2))*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*x+c)^(4/5)*d^2*(b*x+a)^2*(5^(1/2)-5)*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*x+c)^(4/5)*d 
^2*2^(1/2)*(5-5^(1/2))^(1/2)*(b*x+a)^2*(5+5^(1/2))*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))+4*(d*x+c)^(4/5)*d^2*5^(1/2)*(b*x+a)^2*ln((d*x+c)^(1/5 
)+((a*d-b*c)/b)^(1/5))+125/63*((63/25*b^2*x^2+98/25*a*b*x+a^2)*d^2+48/25*( 
7/12*b*x+a)*c*b*d-2/5*b^2*c^2)*5^(1/2)*((a*d-b*c)/b)^(4/5)+(b*x+a)^2*(d*x+ 
c)^(4/5)*d^2*(5+5^(1/2))^(1/2)*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)))*(5^(1/2)- 
5)*2^(1/2))*5^(1/2)/((a*d-b*c)/b)^(4/5)/(b*x+a)^2/(a*d-b*c)^3
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(9/5),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)**3/(d*x+c)**(9/5),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 805, normalized size of antiderivative = 1.36 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx =\text {Too large to display} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(9/5),x, algorithm="maxima")
 

Output:

1/100*(126*(sqrt(5)*d^3*(sqrt(5) - 1)*log(((-b*c + a*d)^(1/5)*b^(1/5)*(sqr 
t(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)^(4/5)*b^(1/5)*sqrt(2*sqrt(5) - 10)) + sqrt(5)*d^3*(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)^(4/5)*b^(1/5)*sqrt(-2*sqrt(5) - 10)) - d 
^3*(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)^(4/ 
5)*b^(1/5)*(sqrt(5) + 1)) - d^3*(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)^(4/5)*b^(1/5)*(sqrt(5) - 1)) + 2*d^3*log((d*x + c 
)^(1/5)*b^(1/5) + (-b*c + a*d)^(1/5))/((-b*c + a*d)^(4/5)*b^(1/5)))*b/(b^3 
*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) + 5*(63*(d*x + c)^2*b^2*d^ 
3 + 25*b^2*c^2*d^3 - 50*a*b*c*d^4 + 25*a^2*d^5 - 98*(b^2*c*d^3 - a*b*d^4)* 
(d*x + c))/((b^5*c^3 - 3*a*b^4*c^2*d + 3*a^2*b^3*c*d^2 - a^3*b^2*d^3)*(d*x 
 + c)^(14/5) - 2*(b^5*c^4 - 4*a*b^4*c^3*d + 6*a^2*b^3*c^2*d^2 - 4*a^3*b^2* 
c*d^3 + a^4*b*d^4)*(d*x + c)^(9/5) + (b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*...
 

Giac [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 846, normalized size of antiderivative = 1.43 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx =\text {Too large to display} \] Input:

integrate(1/(b*x+a)^3/(d*x+c)^(9/5),x, algorithm="giac")
 

Output:

63/25*b*d^2*((b*c - a*d)/b)^(1/5)*log(abs((d*x + c)^(1/5) - ((b*c - a*d)/b 
)^(1/5)))/(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) - 63/50*(b^5*c - a*b^4*d)^(1/5)*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^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) - 63/50*(b^5*c - a*b^4*d)^(1/5)*d^2*sqrt(-2*sqr 
t(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^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) - 63/25*(b^5*c - a*b^4*d)^( 
1/5)*d^2*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^4*c^4*(sqrt(5 
) - 1) - 4*a*b^3*c^3*d*(sqrt(5) - 1) + 6*a^2*b^2*c^2*d^2*(sqrt(5) - 1) - 4 
*a^3*b*c*d^3*(sqrt(5) - 1) + a^4*d^4*(sqrt(5) - 1)) + 63/25*(b^5*c - a*b^4 
*d)^(1/5)*d^2*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^4*c^4*( 
sqrt(5) + 1) - 4*a*b^3*c^3*d*(sqrt(5) + 1) + 6*a^2*b^2*c^2*d^2*(sqrt(5) + 
1) - 4*a^3*b*c*d^3*(sqrt(5) + 1) + a^4*d^4*(sqrt(5) + 1)) + 5/4*d^2/((b^3* 
c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(d*x + c)^(4/5)) + 1/10*(19 
*(d*x + c)^(6/5)*b^2*d^2 - 24*(d*x + c)^(1/5)*b^2*c*d^2 + 24*(d*x + c)^(1/ 
5)*a*b*d^3)/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((d*x ...
 

Mupad [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 2510, normalized size of antiderivative = 4.24 \[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*x)^3*(c + d*x)^(9/5)),x)
 

Output:

(63*b^(4/5)*d^2*log((c + d*x)^(1/5)*(781396875*a^12*b^7*d^18 + 781396875*b 
^19*c^12*d^6 - 9376762500*a*b^18*c^11*d^7 - 9376762500*a^11*b^8*c*d^17 + 5 
1572193750*a^2*b^17*c^10*d^8 - 171907312500*a^3*b^16*c^9*d^9 + 38679145312 
5*a^4*b^15*c^8*d^10 - 618866325000*a^5*b^14*c^7*d^11 + 722010712500*a^6*b^ 
13*c^6*d^12 - 618866325000*a^7*b^12*c^5*d^13 + 386791453125*a^8*b^11*c^4*d 
^14 - 171907312500*a^9*b^10*c^3*d^15 + 51572193750*a^10*b^9*c^2*d^16) - (6 
3*b^(4/5)*d^2*(310078125*a^16*b^6*d^20 + 310078125*b^22*c^16*d^4 - 4961250 
000*a*b^21*c^15*d^5 - 4961250000*a^15*b^7*c*d^19 + 37209375000*a^2*b^20*c^ 
14*d^6 - 173643750000*a^3*b^19*c^13*d^7 + 564342187500*a^4*b^18*c^12*d^8 - 
 1354421250000*a^5*b^17*c^11*d^9 + 2483105625000*a^6*b^16*c^10*d^10 - 3547 
293750000*a^7*b^15*c^9*d^11 + 3990705468750*a^8*b^14*c^8*d^12 - 3547293750 
000*a^9*b^13*c^7*d^13 + 2483105625000*a^10*b^12*c^6*d^14 - 1354421250000*a 
^11*b^11*c^5*d^15 + 564342187500*a^12*b^10*c^4*d^16 - 173643750000*a^13*b^ 
9*c^3*d^17 + 37209375000*a^14*b^8*c^2*d^18))/(25*(b*c - a*d)^(19/5))))/(25 
*(b*c - a*d)^(19/5)) - ((5*d^2)/(4*(a*d - b*c)) + (63*b^2*d^2*(c + d*x)^2) 
/(20*(a*d - b*c)^3) + (49*b*d^2*(c + d*x))/(10*(a*d - b*c)^2))/(b^2*(c + d 
*x)^(14/5) - (2*b^2*c - 2*a*b*d)*(c + d*x)^(9/5) + (c + d*x)^(4/5)*(a^2*d^ 
2 + b^2*c^2 - 2*a*b*c*d)) - (b^(4/5)*d^2*log((c + d*x)^(1/5)*(781396875*a^ 
12*b^7*d^18 + 781396875*b^19*c^12*d^6 - 9376762500*a*b^18*c^11*d^7 - 93767 
62500*a^11*b^8*c*d^17 + 51572193750*a^2*b^17*c^10*d^8 - 171907312500*a^...
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^3 (c+d x)^{9/5}} \, dx=\int \frac {\left (d x +c \right )^{\frac {1}{5}}}{b^{3} d^{2} x^{5}+3 a \,b^{2} d^{2} x^{4}+2 b^{3} c d \,x^{4}+3 a^{2} b \,d^{2} x^{3}+6 a \,b^{2} c d \,x^{3}+b^{3} c^{2} x^{3}+a^{3} d^{2} x^{2}+6 a^{2} b c d \,x^{2}+3 a \,b^{2} c^{2} x^{2}+2 a^{3} c d x +3 a^{2} b \,c^{2} x +a^{3} c^{2}}d x \] Input:

int(1/(b*x+a)^3/(d*x+c)^(9/5),x)
 

Output:

int((c + d*x)**(1/5)/(a**3*c**2 + 2*a**3*c*d*x + a**3*d**2*x**2 + 3*a**2*b 
*c**2*x + 6*a**2*b*c*d*x**2 + 3*a**2*b*d**2*x**3 + 3*a*b**2*c**2*x**2 + 6* 
a*b**2*c*d*x**3 + 3*a*b**2*d**2*x**4 + b**3*c**2*x**3 + 2*b**3*c*d*x**4 + 
b**3*d**2*x**5),x)