\(\int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx\) [617]

Optimal result

Integrand size = 19, antiderivative size = 98 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {4}{(b c-a d) (a+b x)^{7/4} \sqrt [4]{c+d x}}-\frac {32 b (c+d x)^{3/4}}{7 (b c-a d)^2 (a+b x)^{7/4}}+\frac {128 b d (c+d x)^{3/4}}{21 (b c-a d)^3 (a+b x)^{3/4}} \] Output:

4/(-a*d+b*c)/(b*x+a)^(7/4)/(d*x+c)^(1/4)-32/7*b*(d*x+c)^(3/4)/(-a*d+b*c)^2 
/(b*x+a)^(7/4)+128/21*b*d*(d*x+c)^(3/4)/(-a*d+b*c)^3/(b*x+a)^(3/4)
 

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {84 a^2 d^2+56 a b d (c+4 d x)+4 b^2 \left (-3 c^2+8 c d x+32 d^2 x^2\right )}{21 (b c-a d)^3 (a+b x)^{7/4} \sqrt [4]{c+d x}} \] Input:

Integrate[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]
 

Output:

(84*a^2*d^2 + 56*a*b*d*(c + 4*d*x) + 4*b^2*(-3*c^2 + 8*c*d*x + 32*d^2*x^2) 
)/(21*(b*c - a*d)^3*(a + b*x)^(7/4)*(c + d*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {55, 55, 48}

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.

\(\displaystyle \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{7/4} (c+d x)^{5/4}}dx}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{3/4} (c+d x)^{5/4}}dx}{3 (b c-a d)}-\frac {4}{3 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {8 d \left (-\frac {16 d \sqrt [4]{a+b x}}{3 \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{3 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\)

Input:

Int[1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x]
 

Output:

-4/(7*(b*c - a*d)*(a + b*x)^(7/4)*(c + d*x)^(1/4)) - (8*d*(-4/(3*(b*c - a* 
d)*(a + b*x)^(3/4)*(c + d*x)^(1/4)) - (16*d*(a + b*x)^(1/4))/(3*(b*c - a*d 
)^2*(c + d*x)^(1/4))))/(7*(b*c - a*d))
 

Defintions of rubi rules used

rule 48

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] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55

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*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07

method	result	size
gosper	\(-\frac {4 \left (32 d^{2} x^{2} b^{2}+56 x a b \,d^{2}+8 x \,b^{2} c d +21 a^{2} d^{2}+14 a b c d -3 b^{2} c^{2}\right )}{21 \left (b x +a \right )^{\frac {7}{4}} \left (x d +c \right )^{\frac {1}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)
orering	\(-\frac {4 \left (32 d^{2} x^{2} b^{2}+56 x a b \,d^{2}+8 x \,b^{2} c d +21 a^{2} d^{2}+14 a b c d -3 b^{2} c^{2}\right )}{21 \left (b x +a \right )^{\frac {7}{4}} \left (x d +c \right )^{\frac {1}{4}} \left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right )}\)	\(105\)

Input:

int(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

-4/21*(32*b^2*d^2*x^2+56*a*b*d^2*x+8*b^2*c*d*x+21*a^2*d^2+14*a*b*c*d-3*b^2 
*c^2)/(b*x+a)^(7/4)/(d*x+c)^(1/4)/(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3 
*c^3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (82) = 164\).

Time = 0.18 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.79 \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\frac {4 \, {\left (32 \, b^{2} d^{2} x^{2} - 3 \, b^{2} c^{2} + 14 \, a b c d + 21 \, a^{2} d^{2} + 8 \, {\left (b^{2} c d + 7 \, a b d^{2}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{21 \, {\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} + {\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} + {\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} + {\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="fricas")
 

Output:

4/21*(32*b^2*d^2*x^2 - 3*b^2*c^2 + 14*a*b*c*d + 21*a^2*d^2 + 8*(b^2*c*d + 
7*a*b*d^2)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3 
*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^ 
3*c*d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 
5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^ 
3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)
 

Sympy [F]

\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {11}{4}} \left (c + d x\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/(b*x+a)**(11/4)/(d*x+c)**(5/4),x)
 

Output:

Integral(1/((a + b*x)**(11/4)*(c + d*x)**(5/4)), x)
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(5/4)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {11}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(11/4)*(d*x + c)^(5/4)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{11/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \] Input:

int(1/((a + b*x)^(11/4)*(c + d*x)^(5/4)),x)
 

Output:

int(1/((a + b*x)^(11/4)*(c + d*x)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{2} c +\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{2} d x +2 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a b c x +2 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a b d \,x^{2}+\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} b^{2} c \,x^{2}+\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} b^{2} d \,x^{3}}d x \] Input:

int(1/(b*x+a)^(11/4)/(d*x+c)^(5/4),x)
 

Output:

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