\(\int \frac {c+d x^2}{(e x)^{9/2} (a+b x^2)^{9/4}} \, dx\) [475]

Optimal result

Integrand size = 26, antiderivative size = 141 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}+\frac {64 (12 b c-7 a d) \left (a+b x^2\right )^{3/4}}{105 a^4 e^3 (e x)^{3/2}} \] Output:

-2/7*c/a/e/(e*x)^(7/2)/(b*x^2+a)^(5/4)-2/35*(-7*a*d+12*b*c)/a^2/e^3/(e*x)^ 
(3/2)/(b*x^2+a)^(5/4)-16/35*(-7*a*d+12*b*c)/a^3/e^3/(e*x)^(3/2)/(b*x^2+a)^ 
(1/4)+64/105*(-7*a*d+12*b*c)*(b*x^2+a)^(3/4)/a^4/e^3/(e*x)^(3/2)
 

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.62 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 x \left (-384 b^3 c x^6+32 a b^2 x^4 \left (-15 c+7 d x^2\right )+5 a^3 \left (3 c+7 d x^2\right )+a^2 b \left (-60 c x^2+280 d x^4\right )\right )}{105 a^4 (e x)^{9/2} \left (a+b x^2\right )^{5/4}} \] Input:

Integrate[(c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(9/4)),x]
 

Output:

(-2*x*(-384*b^3*c*x^6 + 32*a*b^2*x^4*(-15*c + 7*d*x^2) + 5*a^3*(3*c + 7*d* 
x^2) + a^2*b*(-60*c*x^2 + 280*d*x^4)))/(105*a^4*(e*x)^(9/2)*(a + b*x^2)^(5 
/4))
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {359, 246, 246, 242}

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.

\(\displaystyle \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx\)

\(\Big \downarrow \) 359

\(\displaystyle -\frac {(12 b c-7 a d) \int \frac {1}{(e x)^{5/2} \left (b x^2+a\right )^{9/4}}dx}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 246

\(\displaystyle -\frac {(12 b c-7 a d) \left (\frac {8 \int \frac {1}{(e x)^{5/2} \left (b x^2+a\right )^{5/4}}dx}{5 a}+\frac {2}{5 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 246

\(\displaystyle -\frac {(12 b c-7 a d) \left (\frac {8 \left (\frac {4 \int \frac {1}{(e x)^{5/2} \sqrt [4]{b x^2+a}}dx}{a}+\frac {2}{a e (e x)^{3/2} \sqrt [4]{a+b x^2}}\right )}{5 a}+\frac {2}{5 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}\)

\(\Big \downarrow \) 242

\(\displaystyle -\frac {(12 b c-7 a d) \left (\frac {8 \left (\frac {2}{a e (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac {8 \left (a+b x^2\right )^{3/4}}{3 a^2 e (e x)^{3/2}}\right )}{5 a}+\frac {2}{5 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}\right )}{7 a e^2}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}\)

Input:

Int[(c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(9/4)),x]
 

Output:

(-2*c)/(7*a*e*(e*x)^(7/2)*(a + b*x^2)^(5/4)) - ((12*b*c - 7*a*d)*(2/(5*a*e 
*(e*x)^(3/2)*(a + b*x^2)^(5/4)) + (8*(2/(a*e*(e*x)^(3/2)*(a + b*x^2)^(1/4) 
) - (8*(a + b*x^2)^(3/4))/(3*a^2*e*(e*x)^(3/2))))/(5*a)))/(7*a*e^2)
 

Defintions of rubi rules used

rule 242

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, p}, x 
] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
 

rule 246

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x 
)^(m + 1))*((a + b*x^2)^(p + 1)/(a*c*2*(p + 1))), x] + Simp[(m + 2*p + 3)/( 
a*2*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m 
, p}, x] && ILtQ[Simplify[(m + 1)/2 + p + 1], 0] && NeQ[p, -1]
 

rule 359

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + 
Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1))   Int[(e*x)^(m + 2)* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] 
&& LtQ[m, -1] &&  !ILtQ[p, -1]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61

method	result	size
gosper	\(-\frac {2 x \left (224 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}+280 a^{2} b d \,x^{4}-480 a \,b^{2} c \,x^{4}+35 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+15 a^{3} c \right )}{105 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{4} \left (e x \right )^{\frac {9}{2}}}\)	\(86\)
orering	\(-\frac {2 x \left (224 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}+280 a^{2} b d \,x^{4}-480 a \,b^{2} c \,x^{4}+35 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+15 a^{3} c \right )}{105 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{4} \left (e x \right )^{\frac {9}{2}}}\)	\(86\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{4}} \left (7 a d \,x^{2}-18 x^{2} b c +3 a c \right )}{21 a^{4} x^{3} e^{4} \sqrt {e x}}-\frac {2 b \left (9 a b d \,x^{2}-14 b^{2} c \,x^{2}+10 d \,a^{2}-15 a b c \right ) x}{5 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{4} e^{4} \sqrt {e x}}\)	\(99\)

Input:

int((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(9/4),x,method=_RETURNVERBOSE)
 

Output:

-2/105*x*(224*a*b^2*d*x^6-384*b^3*c*x^6+280*a^2*b*d*x^4-480*a*b^2*c*x^4+35 
*a^3*d*x^2-60*a^2*b*c*x^2+15*a^3*c)/(b*x^2+a)^(5/4)/a^4/(e*x)^(9/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (32 \, {\left (12 \, b^{3} c - 7 \, a b^{2} d\right )} x^{6} + 40 \, {\left (12 \, a b^{2} c - 7 \, a^{2} b d\right )} x^{4} - 15 \, a^{3} c + 5 \, {\left (12 \, a^{2} b c - 7 \, a^{3} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{105 \, {\left (a^{4} b^{2} e^{5} x^{8} + 2 \, a^{5} b e^{5} x^{6} + a^{6} e^{5} x^{4}\right )}} \] Input:

integrate((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")
 

Output:

2/105*(32*(12*b^3*c - 7*a*b^2*d)*x^6 + 40*(12*a*b^2*c - 7*a^2*b*d)*x^4 - 1 
5*a^3*c + 5*(12*a^2*b*c - 7*a^3*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^4*b 
^2*e^5*x^8 + 2*a^5*b*e^5*x^6 + a^6*e^5*x^4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \] Input:

integrate((d*x**2+c)/(e*x)**(9/2)/(b*x**2+a)**(9/4),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(9/2)), x)
 

Giac [F]

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

integrate((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(9/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(9/2)), x)
 

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{7\,a\,b^2\,e^4}+\frac {16\,x^4\,\left (7\,a\,d-12\,b\,c\right )}{21\,a^3\,b\,e^4}+\frac {x^2\,\left (70\,a^3\,d-120\,a^2\,b\,c\right )}{105\,a^4\,b^2\,e^4}-\frac {x^6\,\left (768\,b^3\,c-448\,a\,b^2\,d\right )}{105\,a^4\,b^2\,e^4}\right )}{x^7\,\sqrt {e\,x}+\frac {a^2\,x^3\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^5\,\sqrt {e\,x}}{b}} \] Input:

int((c + d*x^2)/((e*x)^(9/2)*(a + b*x^2)^(9/4)),x)
 

Output:

-((a + b*x^2)^(3/4)*((2*c)/(7*a*b^2*e^4) + (16*x^4*(7*a*d - 12*b*c))/(21*a 
^3*b*e^4) + (x^2*(70*a^3*d - 120*a^2*b*c))/(105*a^4*b^2*e^4) - (x^6*(768*b 
^3*c - 448*a*b^2*d))/(105*a^4*b^2*e^4)))/(x^7*(e*x)^(1/2) + (a^2*x^3*(e*x) 
^(1/2))/b^2 + (2*a*x^5*(e*x)^(1/2))/b)
 

Reduce [B] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.48 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {\sqrt {e}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (35 \left (b \,x^{2}+a \right ) a^{4} d -50 \left (b \,x^{2}+a \right ) a^{3} b c -140 \left (b \,x^{2}+a \right ) a^{3} b d \,x^{2}+200 \left (b \,x^{2}+a \right ) a^{2} b^{2} c \,x^{2}-1120 \left (b \,x^{2}+a \right ) a^{2} b^{2} d \,x^{4}+1600 \left (b \,x^{2}+a \right ) a \,b^{3} c \,x^{4}-896 \left (b \,x^{2}+a \right ) a \,b^{3} d \,x^{6}+1280 \left (b \,x^{2}+a \right ) b^{4} c \,x^{6}-35 a^{5} d -35 a^{4} b d \,x^{2}\right )}{175 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{4} b \,e^{5} x^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((d*x^2+c)/(e*x)^(9/2)/(b*x^2+a)^(9/4),x)
 

Output:

(sqrt(e)*(a + b*x**2)**(1/4)*(35*(a + b*x**2)*a**4*d - 50*(a + b*x**2)*a** 
3*b*c - 140*(a + b*x**2)*a**3*b*d*x**2 + 200*(a + b*x**2)*a**2*b**2*c*x**2 
 - 1120*(a + b*x**2)*a**2*b**2*d*x**4 + 1600*(a + b*x**2)*a*b**3*c*x**4 - 
896*(a + b*x**2)*a*b**3*d*x**6 + 1280*(a + b*x**2)*b**4*c*x**6 - 35*a**5*d 
 - 35*a**4*b*d*x**2))/(175*sqrt(x)*sqrt(a + b*x**2)*a**4*b*e**5*x**3*(a**2 
 + 2*a*b*x**2 + b**2*x**4))