Integrand size = 26, antiderivative size = 141 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}+\frac {2 (12 b c-13 a d) \sqrt [4]{a+b x^2}}{13 a^2 e^3 (e x)^{9/2}}-\frac {16 (12 b c-13 a d) \left (a+b x^2\right )^{5/4}}{65 a^3 e^3 (e x)^{9/2}}+\frac {64 (12 b c-13 a d) \left (a+b x^2\right )^{9/4}}{585 a^4 e^3 (e x)^{9/2}} \]
-2/13*c*(b*x^2+a)^(1/4)/a/e/(e*x)^(13/2)+2/13*(-13*a*d+12*b*c)*(b*x^2+a)^( 1/4)/a^2/e^3/(e*x)^(9/2)-16/65*(-13*a*d+12*b*c)*(b*x^2+a)^(5/4)/a^3/e^3/(e *x)^(9/2)+64/585*(-13*a*d+12*b*c)*(b*x^2+a)^(9/4)/a^4/e^3/(e*x)^(9/2)
Time = 0.59 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 x \sqrt [4]{a+b x^2} \left (45 a^3 c-60 a^2 b c x^2+65 a^3 d x^2+96 a b^2 c x^4-104 a^2 b d x^4-384 b^3 c x^6+416 a b^2 d x^6\right )}{585 a^4 (e x)^{15/2}} \]
(-2*x*(a + b*x^2)^(1/4)*(45*a^3*c - 60*a^2*b*c*x^2 + 65*a^3*d*x^2 + 96*a*b ^2*c*x^4 - 104*a^2*b*d*x^4 - 384*b^3*c*x^6 + 416*a*b^2*d*x^6))/(585*a^4*(e *x)^(15/2))
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 359 |
\(\displaystyle -\frac {(12 b c-13 a d) \int \frac {1}{(e x)^{11/2} \left (b x^2+a\right )^{3/4}}dx}{13 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {(12 b c-13 a d) \left (-\frac {8 \int \frac {\sqrt [4]{b x^2+a}}{(e x)^{11/2}}dx}{a}-\frac {2 \sqrt [4]{a+b x^2}}{a e (e x)^{9/2}}\right )}{13 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}\) |
\(\Big \downarrow \) 246 |
\(\displaystyle -\frac {(12 b c-13 a d) \left (-\frac {8 \left (-\frac {4 \int \frac {\left (b x^2+a\right )^{5/4}}{(e x)^{11/2}}dx}{5 a}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a e (e x)^{9/2}}\right )}{a}-\frac {2 \sqrt [4]{a+b x^2}}{a e (e x)^{9/2}}\right )}{13 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle -\frac {(12 b c-13 a d) \left (-\frac {8 \left (\frac {8 \left (a+b x^2\right )^{9/4}}{45 a^2 e (e x)^{9/2}}-\frac {2 \left (a+b x^2\right )^{5/4}}{5 a e (e x)^{9/2}}\right )}{a}-\frac {2 \sqrt [4]{a+b x^2}}{a e (e x)^{9/2}}\right )}{13 a e^2}-\frac {2 c \sqrt [4]{a+b x^2}}{13 a e (e x)^{13/2}}\) |
(-2*c*(a + b*x^2)^(1/4))/(13*a*e*(e*x)^(13/2)) - ((12*b*c - 13*a*d)*((-2*( a + b*x^2)^(1/4))/(a*e*(e*x)^(9/2)) - (8*((-2*(a + b*x^2)^(5/4))/(5*a*e*(e *x)^(9/2)) + (8*(a + b*x^2)^(9/4))/(45*a^2*e*(e*x)^(9/2))))/a))/(13*a*e^2)
3.11.98.3.1 Defintions of rubi rules used
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]
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]
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]
Time = 3.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61
method | result | size |
gosper | \(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (416 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}-104 a^{2} b d \,x^{4}+96 a \,b^{2} c \,x^{4}+65 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+45 c \,a^{3}\right )}{585 a^{4} \left (e x \right )^{\frac {15}{2}}}\) | \(86\) |
risch | \(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (416 a \,b^{2} d \,x^{6}-384 b^{3} c \,x^{6}-104 a^{2} b d \,x^{4}+96 a \,b^{2} c \,x^{4}+65 a^{3} d \,x^{2}-60 a^{2} b c \,x^{2}+45 c \,a^{3}\right )}{585 e^{7} \sqrt {e x}\, a^{4} x^{6}}\) | \(91\) |
-2/585*x*(b*x^2+a)^(1/4)*(416*a*b^2*d*x^6-384*b^3*c*x^6-104*a^2*b*d*x^4+96 *a*b^2*c*x^4+65*a^3*d*x^2-60*a^2*b*c*x^2+45*a^3*c)/a^4/(e*x)^(15/2)
Time = 0.31 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \, {\left (32 \, {\left (12 \, b^{3} c - 13 \, a b^{2} d\right )} x^{6} - 8 \, {\left (12 \, a b^{2} c - 13 \, a^{2} b d\right )} x^{4} - 45 \, a^{3} c + 5 \, {\left (12 \, a^{2} b c - 13 \, a^{3} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{585 \, a^{4} e^{8} x^{7}} \]
2/585*(32*(12*b^3*c - 13*a*b^2*d)*x^6 - 8*(12*a*b^2*c - 13*a^2*b*d)*x^4 - 45*a^3*c + 5*(12*a^2*b*c - 13*a^3*d)*x^2)*(b*x^2 + a)^(1/4)*sqrt(e*x)/(a^4 *e^8*x^7)
Timed out. \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \]
\[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {15}{2}}} \,d x } \]
\[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {15}{2}}} \,d x } \]
Time = 5.67 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.71 \[ \int \frac {c+d x^2}{(e x)^{15/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{1/4}\,\left (\frac {2\,c}{13\,a\,e^7}+\frac {x^2\,\left (130\,a^3\,d-120\,a^2\,b\,c\right )}{585\,a^4\,e^7}-\frac {x^6\,\left (768\,b^3\,c-832\,a\,b^2\,d\right )}{585\,a^4\,e^7}-\frac {16\,b\,x^4\,\left (13\,a\,d-12\,b\,c\right )}{585\,a^3\,e^7}\right )}{x^6\,\sqrt {e\,x}} \]