\(\int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx\) [370]

Optimal result

Integrand size = 19, antiderivative size = 175 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=\frac {2}{3 (b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}+\frac {16 b}{3 (b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}-\frac {32 b^2 \sqrt {c+d x}}{5 (b c-a d)^3 (a+b x)^{5/2}}+\frac {128 b^2 d \sqrt {c+d x}}{15 (b c-a d)^4 (a+b x)^{3/2}}-\frac {256 b^2 d^2 \sqrt {c+d x}}{15 (b c-a d)^5 \sqrt {a+b x}} \] Output:

2/3/(-a*d+b*c)/(b*x+a)^(5/2)/(d*x+c)^(3/2)+16/3*b/(-a*d+b*c)^2/(b*x+a)^(5/ 
2)/(d*x+c)^(1/2)-32/5*b^2*(d*x+c)^(1/2)/(-a*d+b*c)^3/(b*x+a)^(5/2)+128/15* 
b^2*d*(d*x+c)^(1/2)/(-a*d+b*c)^4/(b*x+a)^(3/2)-256/15*b^2*d^2*(d*x+c)^(1/2 
)/(-a*d+b*c)^5/(b*x+a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.97 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=-\frac {2 \left (-5 a^4 d^4+20 a^3 b d^3 (3 c+2 d x)+30 a^2 b^2 d^2 \left (3 c^2+12 c d x+8 d^2 x^2\right )+20 a b^3 d \left (-c^3+6 c^2 d x+24 c d^2 x^2+16 d^3 x^3\right )+b^4 \left (3 c^4-8 c^3 d x+48 c^2 d^2 x^2+192 c d^3 x^3+128 d^4 x^4\right )\right )}{15 (b c-a d)^5 (a+b x)^{5/2} (c+d x)^{3/2}} \] Input:

Integrate[1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x]
 

Output:

(-2*(-5*a^4*d^4 + 20*a^3*b*d^3*(3*c + 2*d*x) + 30*a^2*b^2*d^2*(3*c^2 + 12* 
c*d*x + 8*d^2*x^2) + 20*a*b^3*d*(-c^3 + 6*c^2*d*x + 24*c*d^2*x^2 + 16*d^3* 
x^3) + b^4*(3*c^4 - 8*c^3*d*x + 48*c^2*d^2*x^2 + 192*c*d^3*x^3 + 128*d^4*x 
^4)))/(15*(b*c - a*d)^5*(a + b*x)^(5/2)*(c + d*x)^(3/2))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.17, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {55, 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.

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

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \int \frac {1}{(a+b x)^{5/2} (c+d x)^{5/2}}dx}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \left (-\frac {2 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{5/2}}dx}{b c-a d}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \left (-\frac {2 d \left (-\frac {4 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/2}}dx}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 55

\(\displaystyle -\frac {8 d \left (-\frac {2 d \left (-\frac {4 d \left (\frac {2 b \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 (b c-a d)}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 48

\(\displaystyle -\frac {8 d \left (-\frac {2 d \left (-\frac {4 d \left (\frac {4 b \sqrt {a+b x}}{3 \sqrt {c+d x} (b c-a d)^2}+\frac {2 \sqrt {a+b x}}{3 (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{\sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}\right )}{b c-a d}-\frac {2}{3 (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}\right )}{5 (b c-a d)}-\frac {2}{5 (a+b x)^{5/2} (c+d x)^{3/2} (b c-a d)}\)

Input:

Int[1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x]
 

Output:

-2/(5*(b*c - a*d)*(a + b*x)^(5/2)*(c + d*x)^(3/2)) - (8*d*(-2/(3*(b*c - a* 
d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) - (2*d*(-2/((b*c - a*d)*Sqrt[a + b*x]* 
(c + d*x)^(3/2)) - (4*d*((2*Sqrt[a + b*x])/(3*(b*c - a*d)*(c + d*x)^(3/2)) 
 + (4*b*Sqrt[a + b*x])/(3*(b*c - a*d)^2*Sqrt[c + d*x])))/(b*c - a*d)))/(b* 
c - a*d)))/(5*(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 = 175, normalized size of antiderivative = 1.00

method	result	size
default	\(-\frac {2}{5 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {3}{2}}}-\frac {8 d \left (-\frac {2}{3 \left (-a d +b c \right ) \left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {3}{2}}}-\frac {2 d \left (-\frac {2}{\left (-a d +b c \right ) \sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}-\frac {4 d \left (-\frac {2 \sqrt {b x +a}}{3 \left (a d -b c \right ) \left (x d +c \right )^{\frac {3}{2}}}+\frac {4 b \sqrt {b x +a}}{3 \left (a d -b c \right )^{2} \sqrt {x d +c}}\right )}{-a d +b c}\right )}{-a d +b c}\right )}{5 \left (-a d +b c \right )}\)	\(175\)
gosper	\(-\frac {2 \left (-128 d^{4} x^{4} b^{4}-320 a \,b^{3} d^{4} x^{3}-192 b^{4} c \,d^{3} x^{3}-240 a^{2} b^{2} d^{4} x^{2}-480 a \,b^{3} c \,d^{3} x^{2}-48 b^{4} c^{2} d^{2} x^{2}-40 a^{3} b \,d^{4} x -360 a^{2} b^{2} c \,d^{3} x -120 a \,b^{3} c^{2} d^{2} x +8 b^{4} c^{3} d x +5 d^{4} a^{4}-60 a^{3} b c \,d^{3}-90 a^{2} b^{2} c^{2} d^{2}+20 a \,b^{3} c^{3} d -3 c^{4} b^{4}\right )}{15 \left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -c^{5} b^{5}\right )}\)	\(256\)
orering	\(-\frac {2 \left (-128 d^{4} x^{4} b^{4}-320 a \,b^{3} d^{4} x^{3}-192 b^{4} c \,d^{3} x^{3}-240 a^{2} b^{2} d^{4} x^{2}-480 a \,b^{3} c \,d^{3} x^{2}-48 b^{4} c^{2} d^{2} x^{2}-40 a^{3} b \,d^{4} x -360 a^{2} b^{2} c \,d^{3} x -120 a \,b^{3} c^{2} d^{2} x +8 b^{4} c^{3} d x +5 d^{4} a^{4}-60 a^{3} b c \,d^{3}-90 a^{2} b^{2} c^{2} d^{2}+20 a \,b^{3} c^{3} d -3 c^{4} b^{4}\right )}{15 \left (b x +a \right )^{\frac {5}{2}} \left (x d +c \right )^{\frac {3}{2}} \left (a^{5} d^{5}-5 a^{4} b c \,d^{4}+10 a^{3} b^{2} c^{2} d^{3}-10 a^{2} b^{3} c^{3} d^{2}+5 a \,b^{4} c^{4} d -c^{5} b^{5}\right )}\)	\(256\)

Input:

int(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 715 vs. \(2 (145) = 290\).

Time = 1.95 (sec) , antiderivative size = 715, normalized size of antiderivative = 4.09 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=-\frac {2 \, {\left (128 \, b^{4} d^{4} x^{4} + 3 \, b^{4} c^{4} - 20 \, a b^{3} c^{3} d + 90 \, a^{2} b^{2} c^{2} d^{2} + 60 \, a^{3} b c d^{3} - 5 \, a^{4} d^{4} + 64 \, {\left (3 \, b^{4} c d^{3} + 5 \, a b^{3} d^{4}\right )} x^{3} + 48 \, {\left (b^{4} c^{2} d^{2} + 10 \, a b^{3} c d^{3} + 5 \, a^{2} b^{2} d^{4}\right )} x^{2} - 8 \, {\left (b^{4} c^{3} d - 15 \, a b^{3} c^{2} d^{2} - 45 \, a^{2} b^{2} c d^{3} - 5 \, a^{3} b d^{4}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{15 \, {\left (a^{3} b^{5} c^{7} - 5 \, a^{4} b^{4} c^{6} d + 10 \, a^{5} b^{3} c^{5} d^{2} - 10 \, a^{6} b^{2} c^{4} d^{3} + 5 \, a^{7} b c^{3} d^{4} - a^{8} c^{2} d^{5} + {\left (b^{8} c^{5} d^{2} - 5 \, a b^{7} c^{4} d^{3} + 10 \, a^{2} b^{6} c^{3} d^{4} - 10 \, a^{3} b^{5} c^{2} d^{5} + 5 \, a^{4} b^{4} c d^{6} - a^{5} b^{3} d^{7}\right )} x^{5} + {\left (2 \, b^{8} c^{6} d - 7 \, a b^{7} c^{5} d^{2} + 5 \, a^{2} b^{6} c^{4} d^{3} + 10 \, a^{3} b^{5} c^{3} d^{4} - 20 \, a^{4} b^{4} c^{2} d^{5} + 13 \, a^{5} b^{3} c d^{6} - 3 \, a^{6} b^{2} d^{7}\right )} x^{4} + {\left (b^{8} c^{7} + a b^{7} c^{6} d - 17 \, a^{2} b^{6} c^{5} d^{2} + 35 \, a^{3} b^{5} c^{4} d^{3} - 25 \, a^{4} b^{4} c^{3} d^{4} - a^{5} b^{3} c^{2} d^{5} + 9 \, a^{6} b^{2} c d^{6} - 3 \, a^{7} b d^{7}\right )} x^{3} + {\left (3 \, a b^{7} c^{7} - 9 \, a^{2} b^{6} c^{6} d + a^{3} b^{5} c^{5} d^{2} + 25 \, a^{4} b^{4} c^{4} d^{3} - 35 \, a^{5} b^{3} c^{3} d^{4} + 17 \, a^{6} b^{2} c^{2} d^{5} - a^{7} b c d^{6} - a^{8} d^{7}\right )} x^{2} + {\left (3 \, a^{2} b^{6} c^{7} - 13 \, a^{3} b^{5} c^{6} d + 20 \, a^{4} b^{4} c^{5} d^{2} - 10 \, a^{5} b^{3} c^{4} d^{3} - 5 \, a^{6} b^{2} c^{3} d^{4} + 7 \, a^{7} b c^{2} d^{5} - 2 \, a^{8} c d^{6}\right )} x\right )}} \] Input:

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

-2/15*(128*b^4*d^4*x^4 + 3*b^4*c^4 - 20*a*b^3*c^3*d + 90*a^2*b^2*c^2*d^2 + 
 60*a^3*b*c*d^3 - 5*a^4*d^4 + 64*(3*b^4*c*d^3 + 5*a*b^3*d^4)*x^3 + 48*(b^4 
*c^2*d^2 + 10*a*b^3*c*d^3 + 5*a^2*b^2*d^4)*x^2 - 8*(b^4*c^3*d - 15*a*b^3*c 
^2*d^2 - 45*a^2*b^2*c*d^3 - 5*a^3*b*d^4)*x)*sqrt(b*x + a)*sqrt(d*x + c)/(a 
^3*b^5*c^7 - 5*a^4*b^4*c^6*d + 10*a^5*b^3*c^5*d^2 - 10*a^6*b^2*c^4*d^3 + 5 
*a^7*b*c^3*d^4 - a^8*c^2*d^5 + (b^8*c^5*d^2 - 5*a*b^7*c^4*d^3 + 10*a^2*b^6 
*c^3*d^4 - 10*a^3*b^5*c^2*d^5 + 5*a^4*b^4*c*d^6 - a^5*b^3*d^7)*x^5 + (2*b^ 
8*c^6*d - 7*a*b^7*c^5*d^2 + 5*a^2*b^6*c^4*d^3 + 10*a^3*b^5*c^3*d^4 - 20*a^ 
4*b^4*c^2*d^5 + 13*a^5*b^3*c*d^6 - 3*a^6*b^2*d^7)*x^4 + (b^8*c^7 + a*b^7*c 
^6*d - 17*a^2*b^6*c^5*d^2 + 35*a^3*b^5*c^4*d^3 - 25*a^4*b^4*c^3*d^4 - a^5* 
b^3*c^2*d^5 + 9*a^6*b^2*c*d^6 - 3*a^7*b*d^7)*x^3 + (3*a*b^7*c^7 - 9*a^2*b^ 
6*c^6*d + a^3*b^5*c^5*d^2 + 25*a^4*b^4*c^4*d^3 - 35*a^5*b^3*c^3*d^4 + 17*a 
^6*b^2*c^2*d^5 - a^7*b*c*d^6 - a^8*d^7)*x^2 + (3*a^2*b^6*c^7 - 13*a^3*b^5* 
c^6*d + 20*a^4*b^4*c^5*d^2 - 10*a^5*b^3*c^4*d^3 - 5*a^6*b^2*c^3*d^4 + 7*a^ 
7*b*c^2*d^5 - 2*a^8*c*d^6)*x)
 

Sympy [F]

\[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {7}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(b*x+a)**(7/2)/(d*x+c)**(5/2),x)
                                                                                    
                                                                                    
 

Output:

Integral(1/((a + b*x)**(7/2)*(c + d*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1203 vs. \(2 (145) = 290\).

Time = 0.45 (sec) , antiderivative size = 1203, normalized size of antiderivative = 6.87 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-2/3*sqrt(b*x + a)*(11*(b^8*c^4*d^5*abs(b) - 4*a*b^7*c^3*d^6*abs(b) + 6*a^ 
2*b^6*c^2*d^7*abs(b) - 4*a^3*b^5*c*d^8*abs(b) + a^4*b^4*d^9*abs(b))*(b*x + 
 a)/(b^11*c^9*d - 9*a*b^10*c^8*d^2 + 36*a^2*b^9*c^7*d^3 - 84*a^3*b^8*c^6*d 
^4 + 126*a^4*b^7*c^5*d^5 - 126*a^5*b^6*c^4*d^6 + 84*a^6*b^5*c^3*d^7 - 36*a 
^7*b^4*c^2*d^8 + 9*a^8*b^3*c*d^9 - a^9*b^2*d^10) + 12*(b^9*c^5*d^4*abs(b) 
- 5*a*b^8*c^4*d^5*abs(b) + 10*a^2*b^7*c^3*d^6*abs(b) - 10*a^3*b^6*c^2*d^7* 
abs(b) + 5*a^4*b^5*c*d^8*abs(b) - a^5*b^4*d^9*abs(b))/(b^11*c^9*d - 9*a*b^ 
10*c^8*d^2 + 36*a^2*b^9*c^7*d^3 - 84*a^3*b^8*c^6*d^4 + 126*a^4*b^7*c^5*d^5 
 - 126*a^5*b^6*c^4*d^6 + 84*a^6*b^5*c^3*d^7 - 36*a^7*b^4*c^2*d^8 + 9*a^8*b 
^3*c*d^9 - a^9*b^2*d^10))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 4/15*(73 
*sqrt(b*d)*b^11*c^4*d^2 - 292*sqrt(b*d)*a*b^10*c^3*d^3 + 438*sqrt(b*d)*a^2 
*b^9*c^2*d^4 - 292*sqrt(b*d)*a^3*b^8*c*d^5 + 73*sqrt(b*d)*a^4*b^7*d^6 - 32 
0*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d) 
)^2*b^9*c^3*d^2 + 960*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b 
*x + a)*b*d - a*b*d))^2*a*b^8*c^2*d^3 - 960*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x 
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^7*c*d^4 + 320*sqrt(b*d 
)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^ 
6*d^5 + 490*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b* 
d - a*b*d))^4*b^7*c^2*d^2 - 980*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt( 
b^2*c + (b*x + a)*b*d - a*b*d))^4*a*b^6*c*d^3 + 490*sqrt(b*d)*(sqrt(b*d...
 

Mupad [B] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.98 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx=\frac {\sqrt {c+d\,x}\,\left (\frac {32\,x^2\,\left (5\,a^2\,d^2+10\,a\,b\,c\,d+b^2\,c^2\right )}{5\,{\left (a\,d-b\,c\right )}^5}+\frac {256\,b^2\,d^2\,x^4}{15\,{\left (a\,d-b\,c\right )}^5}+\frac {-10\,a^4\,d^4+120\,a^3\,b\,c\,d^3+180\,a^2\,b^2\,c^2\,d^2-40\,a\,b^3\,c^3\,d+6\,b^4\,c^4}{15\,b^2\,d^2\,{\left (a\,d-b\,c\right )}^5}+\frac {x\,\left (80\,a^3\,b\,d^4+720\,a^2\,b^2\,c\,d^3+240\,a\,b^3\,c^2\,d^2-16\,b^4\,c^3\,d\right )}{15\,b^2\,d^2\,{\left (a\,d-b\,c\right )}^5}+\frac {128\,b\,d\,x^3\,\left (5\,a\,d+3\,b\,c\right )}{15\,{\left (a\,d-b\,c\right )}^5}\right )}{x^4\,\sqrt {a+b\,x}+\frac {x^2\,\sqrt {a+b\,x}\,\left (a^2\,d^2+4\,a\,b\,c\,d+b^2\,c^2\right )}{b^2\,d^2}+\frac {2\,x^3\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b\,d}+\frac {a^2\,c^2\,\sqrt {a+b\,x}}{b^2\,d^2}+\frac {2\,a\,c\,x\,\left (a\,d+b\,c\right )\,\sqrt {a+b\,x}}{b^2\,d^2}} \] Input:

int(1/((a + b*x)^(7/2)*(c + d*x)^(5/2)),x)
 

Output:

((c + d*x)^(1/2)*((32*x^2*(5*a^2*d^2 + b^2*c^2 + 10*a*b*c*d))/(5*(a*d - b* 
c)^5) + (256*b^2*d^2*x^4)/(15*(a*d - b*c)^5) + (6*b^4*c^4 - 10*a^4*d^4 + 1 
80*a^2*b^2*c^2*d^2 - 40*a*b^3*c^3*d + 120*a^3*b*c*d^3)/(15*b^2*d^2*(a*d - 
b*c)^5) + (x*(80*a^3*b*d^4 - 16*b^4*c^3*d + 240*a*b^3*c^2*d^2 + 720*a^2*b^ 
2*c*d^3))/(15*b^2*d^2*(a*d - b*c)^5) + (128*b*d*x^3*(5*a*d + 3*b*c))/(15*( 
a*d - b*c)^5)))/(x^4*(a + b*x)^(1/2) + (x^2*(a + b*x)^(1/2)*(a^2*d^2 + b^2 
*c^2 + 4*a*b*c*d))/(b^2*d^2) + (2*x^3*(a*d + b*c)*(a + b*x)^(1/2))/(b*d) + 
 (a^2*c^2*(a + b*x)^(1/2))/(b^2*d^2) + (2*a*c*x*(a*d + b*c)*(a + b*x)^(1/2 
))/(b^2*d^2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 918, normalized size of antiderivative = 5.25 \[ \int \frac {1}{(a+b x)^{7/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/(b*x+a)^(7/2)/(d*x+c)^(5/2),x)
 

Output:

(2*( - 128*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**2*b*c**2*d**2 - 256*sqrt(d)*sq 
rt(b)*sqrt(a + b*x)*a**2*b*c*d**3*x - 128*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a* 
*2*b*d**4*x**2 - 256*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c**2*d**2*x - 51 
2*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a*b**2*c*d**3*x**2 - 256*sqrt(d)*sqrt(b)*s 
qrt(a + b*x)*a*b**2*d**4*x**3 - 128*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*c** 
2*d**2*x**2 - 256*sqrt(d)*sqrt(b)*sqrt(a + b*x)*b**3*c*d**3*x**3 - 128*sqr 
t(d)*sqrt(b)*sqrt(a + b*x)*b**3*d**4*x**4 - 5*sqrt(c + d*x)*a**4*d**4 + 60 
*sqrt(c + d*x)*a**3*b*c*d**3 + 40*sqrt(c + d*x)*a**3*b*d**4*x + 90*sqrt(c 
+ d*x)*a**2*b**2*c**2*d**2 + 360*sqrt(c + d*x)*a**2*b**2*c*d**3*x + 240*sq 
rt(c + d*x)*a**2*b**2*d**4*x**2 - 20*sqrt(c + d*x)*a*b**3*c**3*d + 120*sqr 
t(c + d*x)*a*b**3*c**2*d**2*x + 480*sqrt(c + d*x)*a*b**3*c*d**3*x**2 + 320 
*sqrt(c + d*x)*a*b**3*d**4*x**3 + 3*sqrt(c + d*x)*b**4*c**4 - 8*sqrt(c + d 
*x)*b**4*c**3*d*x + 48*sqrt(c + d*x)*b**4*c**2*d**2*x**2 + 192*sqrt(c + d* 
x)*b**4*c*d**3*x**3 + 128*sqrt(c + d*x)*b**4*d**4*x**4))/(15*sqrt(a + b*x) 
*(a**7*c**2*d**5 + 2*a**7*c*d**6*x + a**7*d**7*x**2 - 5*a**6*b*c**3*d**4 - 
 8*a**6*b*c**2*d**5*x - a**6*b*c*d**6*x**2 + 2*a**6*b*d**7*x**3 + 10*a**5* 
b**2*c**4*d**3 + 10*a**5*b**2*c**3*d**4*x - 9*a**5*b**2*c**2*d**5*x**2 - 8 
*a**5*b**2*c*d**6*x**3 + a**5*b**2*d**7*x**4 - 10*a**4*b**3*c**5*d**2 + 25 
*a**4*b**3*c**3*d**4*x**2 + 10*a**4*b**3*c**2*d**5*x**3 - 5*a**4*b**3*c*d* 
*6*x**4 + 5*a**3*b**4*c**6*d - 10*a**3*b**4*c**5*d**2*x - 25*a**3*b**4*...