\(\int \frac {c+d x^2}{(e x)^{5/2} (a+b x^2)^{9/4}} \, dx\) [1129]

Optimal result

Integrand size = 26, antiderivative size = 104 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (8 b c-3 a d) \sqrt {e x}}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac {8 (8 b c-3 a d) \sqrt {e x}}{15 a^3 e^3 \sqrt [4]{a+b x^2}} \]

[Out]

-2/3*c/a/e/(e*x)^(3/2)/(b*x^2+a)^(5/4)-2/15*(-3*a*d+8*b*c)*(e*x)^(1/2)/a^2/e^3/(b*x^2+a)^(5/4)-8/15*(-3*a*d+8*
b*c)*(e*x)^(1/2)/a^3/e^3/(b*x^2+a)^(1/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {464, 279, 270} \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {8 \sqrt {e x} (8 b c-3 a d)}{15 a^3 e^3 \sqrt [4]{a+b x^2}}-\frac {2 \sqrt {e x} (8 b c-3 a d)}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}} \]

[In]

Int[(c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(9/4)),x]

[Out]

(-2*c)/(3*a*e*(e*x)^(3/2)*(a + b*x^2)^(5/4)) - (2*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^2*e^3*(a + b*x^2)^(5/4)) -
(8*(8*b*c - 3*a*d)*Sqrt[e*x])/(15*a^3*e^3*(a + b*x^2)^(1/4))

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}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {(8 b c-3 a d) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{9/4}} \, dx}{3 a e^2} \\ & = -\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (8 b c-3 a d) \sqrt {e x}}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac {(4 (8 b c-3 a d)) \int \frac {1}{\sqrt {e x} \left (a+b x^2\right )^{5/4}} \, dx}{15 a^2 e^2} \\ & = -\frac {2 c}{3 a e (e x)^{3/2} \left (a+b x^2\right )^{5/4}}-\frac {2 (8 b c-3 a d) \sqrt {e x}}{15 a^2 e^3 \left (a+b x^2\right )^{5/4}}-\frac {8 (8 b c-3 a d) \sqrt {e x}}{15 a^3 e^3 \sqrt [4]{a+b x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=\frac {2 x \left (-5 a^2 c-40 a b c x^2+15 a^2 d x^2-32 b^2 c x^4+12 a b d x^4\right )}{15 a^3 (e x)^{5/2} \left (a+b x^2\right )^{5/4}} \]

[In]

Integrate[(c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(9/4)),x]

[Out]

(2*x*(-5*a^2*c - 40*a*b*c*x^2 + 15*a^2*d*x^2 - 32*b^2*c*x^4 + 12*a*b*d*x^4))/(15*a^3*(e*x)^(5/2)*(a + b*x^2)^(
5/4))

Maple [A] (verified)

Time = 3.38 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.60

method	result	size
gosper	\(-\frac {2 x \left (-12 a b d \,x^{4}+32 b^{2} c \,x^{4}-15 a^{2} d \,x^{2}+40 a b c \,x^{2}+5 a^{2} c \right )}{15 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{3} \left (e x \right )^{\frac {5}{2}}}\)	\(62\)
risch	\(-\frac {2 c \left (b \,x^{2}+a \right )^{\frac {3}{4}}}{3 a^{3} x \,e^{2} \sqrt {e x}}+\frac {2 \left (4 x^{2} a b d -9 b^{2} c \,x^{2}+5 a^{2} d -10 a b c \right ) x}{5 \left (b \,x^{2}+a \right )^{\frac {5}{4}} a^{3} e^{2} \sqrt {e x}}\)	\(80\)

[In]

int((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(9/4),x,method=_RETURNVERBOSE)

[Out]

-2/15*x*(-12*a*b*d*x^4+32*b^2*c*x^4-15*a^2*d*x^2+40*a*b*c*x^2+5*a^2*c)/(b*x^2+a)^(5/4)/a^3/(e*x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {2 \, {\left (4 \, {\left (8 \, b^{2} c - 3 \, a b d\right )} x^{4} + 5 \, a^{2} c + 5 \, {\left (8 \, a b c - 3 \, a^{2} d\right )} x^{2}\right )} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {e x}}{15 \, {\left (a^{3} b^{2} e^{3} x^{6} + 2 \, a^{4} b e^{3} x^{4} + a^{5} e^{3} x^{2}\right )}} \]

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(9/4),x, algorithm="fricas")

[Out]

-2/15*(4*(8*b^2*c - 3*a*b*d)*x^4 + 5*a^2*c + 5*(8*a*b*c - 3*a^2*d)*x^2)*(b*x^2 + a)^(3/4)*sqrt(e*x)/(a^3*b^2*e
^3*x^6 + 2*a^4*b*e^3*x^4 + a^5*e^3*x^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=\text {Timed out} \]

[In]

integrate((d*x**2+c)/(e*x)**(5/2)/(b*x**2+a)**(9/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(9/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(5/2)), x)

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {9}{4}} \left (e x\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(5/2)/(b*x^2+a)^(9/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(9/4)*(e*x)^(5/2)), x)

Mupad [B] (verification not implemented)

Time = 5.91 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.11 \[ \int \frac {c+d x^2}{(e x)^{5/2} \left (a+b x^2\right )^{9/4}} \, dx=-\frac {{\left (b\,x^2+a\right )}^{3/4}\,\left (\frac {2\,c}{3\,a\,b^2\,e^2}-\frac {x^2\,\left (30\,a^2\,d-80\,a\,b\,c\right )}{15\,a^3\,b^2\,e^2}+\frac {x^4\,\left (64\,b^2\,c-24\,a\,b\,d\right )}{15\,a^3\,b^2\,e^2}\right )}{x^5\,\sqrt {e\,x}+\frac {2\,a\,x^3\,\sqrt {e\,x}}{b}+\frac {a^2\,x\,\sqrt {e\,x}}{b^2}} \]

[In]

int((c + d*x^2)/((e*x)^(5/2)*(a + b*x^2)^(9/4)),x)

[Out]

-((a + b*x^2)^(3/4)*((2*c)/(3*a*b^2*e^2) - (x^2*(30*a^2*d - 80*a*b*c))/(15*a^3*b^2*e^2) + (x^4*(64*b^2*c - 24*
a*b*d))/(15*a^3*b^2*e^2)))/(x^5*(e*x)^(1/2) + (2*a*x^3*(e*x)^(1/2))/b + (a^2*x*(e*x)^(1/2))/b^2)