\(\int \frac {c+d x^2}{(e x)^{9/2} (a+b x^2)^{9/4}} \, dx\) [1130]

Optimal result

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

[Out]

-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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {464, 279, 270} \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {64 \left (a+b x^2\right )^{3/4} (12 b c-7 a d)}{105 a^4 e^3 (e x)^{3/2}}-\frac {16 (12 b c-7 a d)}{35 a^3 e^3 (e x)^{3/2} \sqrt [4]{a+b x^2}}-\frac {2 (12 b c-7 a d)}{35 a^2 e^3 (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}} \]

[In]

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

[Out]

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

Rule 270

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

Rule 279

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 c}{7 a e (e x)^{7/2} \left (a+b x^2\right )^{5/4}}-\frac {(12 b c-7 a d) \int \frac {1}{(e x)^{5/2} \left (a+b x^2\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}}-\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 {(8 (12 b c-7 a d)) \int \frac {1}{(e x)^{5/2} \left (a+b x^2\right )^{5/4}} \, dx}{35 a^2 e^2} \\ & = -\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 {(32 (12 b c-7 a d)) \int \frac {1}{(e x)^{5/2} \sqrt [4]{a+b x^2}} \, dx}{35 a^3 e^2} \\ & = -\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}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(-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))

Maple [A] (verified)

Time = 3.25 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61

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

[In]

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

[Out]

-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)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {2 \, {\left (32 \, {\left (12 \, b^{3} c - 7 \, a b^{2} d\right )} x^{6} + 40 \, {\left (12 \, a b^{2} c - 7 \, a^{2} b d\right )} x^{4} - 15 \, a^{3} c + 5 \, {\left (12 \, a^{2} b c - 7 \, a^{3} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{105 \, {\left (a^{4} b^{2} e^{5} x^{8} + 2 \, a^{5} b e^{5} x^{6} + a^{6} e^{5} x^{4}\right )}} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.02 \[ \int \frac {c+d x^2}{(e x)^{9/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{7\,a\,b^2\,e^4}+\frac {16\,x^4\,\left (7\,a\,d-12\,b\,c\right )}{21\,a^3\,b\,e^4}+\frac {x^2\,\left (70\,a^3\,d-120\,a^2\,b\,c\right )}{105\,a^4\,b^2\,e^4}-\frac {x^6\,\left (768\,b^3\,c-448\,a\,b^2\,d\right )}{105\,a^4\,b^2\,e^4}\right )}{x^7\,\sqrt {e\,x}+\frac {a^2\,x^3\,\sqrt {e\,x}}{b^2}+\frac {2\,a\,x^5\,\sqrt {e\,x}}{b}} \]

[In]

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

[Out]

-((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)