\(\int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx\) [618]

Optimal result

Integrand size = 19, antiderivative size = 134 \[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\frac {4}{(b c-a d) (a+b x)^{11/4} \sqrt [4]{c+d x}}-\frac {48 b (c+d x)^{3/4}}{11 (b c-a d)^2 (a+b x)^{11/4}}+\frac {384 b d (c+d x)^{3/4}}{77 (b c-a d)^3 (a+b x)^{7/4}}-\frac {512 b d^2 (c+d x)^{3/4}}{77 (b c-a d)^4 (a+b x)^{3/4}} \] Output:

4/(-a*d+b*c)/(b*x+a)^(11/4)/(d*x+c)^(1/4)-48/11*b*(d*x+c)^(3/4)/(-a*d+b*c) 
^2/(b*x+a)^(11/4)+384/77*b*d*(d*x+c)^(3/4)/(-a*d+b*c)^3/(b*x+a)^(7/4)-512/ 
77*b*d^2*(d*x+c)^(3/4)/(-a*d+b*c)^4/(b*x+a)^(3/4)
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=-\frac {4 \left (77 a^3 d^3+77 a^2 b d^2 (c+4 d x)+11 a b^2 d \left (-3 c^2+8 c d x+32 d^2 x^2\right )+b^3 \left (7 c^3-12 c^2 d x+32 c d^2 x^2+128 d^3 x^3\right )\right )}{77 (b c-a d)^4 (a+b x)^{11/4} \sqrt [4]{c+d x}} \] Input:

Integrate[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]
 

Output:

(-4*(77*a^3*d^3 + 77*a^2*b*d^2*(c + 4*d*x) + 11*a*b^2*d*(-3*c^2 + 8*c*d*x 
+ 32*d^2*x^2) + b^3*(7*c^3 - 12*c^2*d*x + 32*c*d^2*x^2 + 128*d^3*x^3)))/(7 
7*(b*c - a*d)^4*(a + b*x)^(11/4)*(c + d*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.21, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {55, 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.

\(\displaystyle \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {12 d \int \frac {1}{(a+b x)^{11/4} (c+d x)^{5/4}}dx}{11 (b c-a d)}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {12 d \left (-\frac {8 d \int \frac {1}{(a+b x)^{7/4} (c+d x)^{5/4}}dx}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{11 (b c-a d)}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {12 d \left (-\frac {8 d \left (-\frac {4 d \int \frac {1}{(a+b x)^{3/4} (c+d x)^{5/4}}dx}{3 (b c-a d)}-\frac {4}{3 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{11 (b c-a d)}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {12 d \left (-\frac {8 d \left (-\frac {16 d \sqrt [4]{a+b x}}{3 \sqrt [4]{c+d x} (b c-a d)^2}-\frac {4}{3 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{7 (b c-a d)}-\frac {4}{7 (a+b x)^{7/4} \sqrt [4]{c+d x} (b c-a d)}\right )}{11 (b c-a d)}-\frac {4}{11 (a+b x)^{11/4} \sqrt [4]{c+d x} (b c-a d)}\)

Input:

Int[1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x]
 

Output:

-4/(11*(b*c - a*d)*(a + b*x)^(11/4)*(c + d*x)^(1/4)) - (12*d*(-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))))/(11*(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 = 0.16 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28

method	result	size
gosper	\(-\frac {4 \left (128 d^{3} x^{3} b^{3}+352 x^{2} a \,b^{2} d^{3}+32 x^{2} b^{3} c \,d^{2}+308 x \,a^{2} b \,d^{3}+88 x a \,b^{2} c \,d^{2}-12 x \,b^{3} c^{2} d +77 a^{3} d^{3}+77 a^{2} b c \,d^{2}-33 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right )}{77 \left (b x +a \right )^{\frac {11}{4}} \left (x d +c \right )^{\frac {1}{4}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)
orering	\(-\frac {4 \left (128 d^{3} x^{3} b^{3}+352 x^{2} a \,b^{2} d^{3}+32 x^{2} b^{3} c \,d^{2}+308 x \,a^{2} b \,d^{3}+88 x a \,b^{2} c \,d^{2}-12 x \,b^{3} c^{2} d +77 a^{3} d^{3}+77 a^{2} b c \,d^{2}-33 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right )}{77 \left (b x +a \right )^{\frac {11}{4}} \left (x d +c \right )^{\frac {1}{4}} \left (d^{4} a^{4}-4 a^{3} b c \,d^{3}+6 a^{2} b^{2} c^{2} d^{2}-4 a \,b^{3} c^{3} d +c^{4} b^{4}\right )}\)	\(171\)

Input:

int(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

-4/77*(128*b^3*d^3*x^3+352*a*b^2*d^3*x^2+32*b^3*c*d^2*x^2+308*a^2*b*d^3*x+ 
88*a*b^2*c*d^2*x-12*b^3*c^2*d*x+77*a^3*d^3+77*a^2*b*c*d^2-33*a*b^2*c^2*d+7 
*b^3*c^3)/(b*x+a)^(11/4)/(d*x+c)^(1/4)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^ 
2*d^2-4*a*b^3*c^3*d+b^4*c^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (112) = 224\).

Time = 0.39 (sec) , antiderivative size = 457, normalized size of antiderivative = 3.41 \[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=-\frac {4 \, {\left (128 \, b^{3} d^{3} x^{3} + 7 \, b^{3} c^{3} - 33 \, a b^{2} c^{2} d + 77 \, a^{2} b c d^{2} + 77 \, a^{3} d^{3} + 32 \, {\left (b^{3} c d^{2} + 11 \, a b^{2} d^{3}\right )} x^{2} - 4 \, {\left (3 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} - 77 \, a^{2} b d^{3}\right )} x\right )} {\left (b x + a\right )}^{\frac {1}{4}} {\left (d x + c\right )}^{\frac {3}{4}}}{77 \, {\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} + {\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} + {\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \, {\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="fricas")
 

Output:

-4/77*(128*b^3*d^3*x^3 + 7*b^3*c^3 - 33*a*b^2*c^2*d + 77*a^2*b*c*d^2 + 77* 
a^3*d^3 + 32*(b^3*c*d^2 + 11*a*b^2*d^3)*x^2 - 4*(3*b^3*c^2*d - 22*a*b^2*c* 
d^2 - 77*a^2*b*d^3)*x)*(b*x + a)^(1/4)*(d*x + c)^(3/4)/(a^3*b^4*c^5 - 4*a^ 
4*b^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d 
 - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3 - 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^ 
4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a 
^4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3 
*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a 
^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 - a 
^6*b*c*d^4 + a^7*d^5)*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*x+a)**(15/4)/(d*x+c)**(5/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x + a)^(15/4)*(d*x + c)^(5/4)), x)
 

Giac [F]

\[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {15}{4}} {\left (d x + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x + a)^(15/4)*(d*x + c)^(5/4)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{{\left (a+b\,x\right )}^{15/4}\,{\left (c+d\,x\right )}^{5/4}} \,d x \] Input:

int(1/((a + b*x)^(15/4)*(c + d*x)^(5/4)),x)
 

Output:

int(1/((a + b*x)^(15/4)*(c + d*x)^(5/4)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+b x)^{15/4} (c+d x)^{5/4}} \, dx=\int \frac {1}{\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{3} c +\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{3} d x +3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{2} b c x +3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a^{2} b d \,x^{2}+3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a \,b^{2} c \,x^{2}+3 \left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} a \,b^{2} d \,x^{3}+\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} b^{3} c \,x^{3}+\left (d x +c \right )^{\frac {1}{4}} \left (b x +a \right )^{\frac {3}{4}} b^{3} d \,x^{4}}d x \] Input:

int(1/(b*x+a)^(15/4)/(d*x+c)^(5/4),x)
 

Output:

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