\(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{3/4}} \, dx\) [1096]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 c \sqrt [4]{a+b x^2}}{5 a e (e x)^{5/2}}+\frac {2 (4 b c-5 a d) \sqrt [4]{a+b x^2}}{5 a^2 e^3 \sqrt {e x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {464, 270} \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \sqrt [4]{a+b x^2} (4 b c-5 a d)}{5 a^2 e^3 \sqrt {e x}}-\frac {2 c \sqrt [4]{a+b x^2}}{5 a e (e x)^{5/2}} \]

[In]

Int[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/4)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 x \sqrt [4]{a+b x^2} \left (a c-4 b c x^2+5 a d x^2\right )}{5 a^2 (e x)^{7/2}} \]

[In]

Integrate[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/4)),x]

[Out]

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

Maple [A] (verified)

Time = 3.14 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.58

method	result	size
gosper	\(-\frac {2 x \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (5 a d \,x^{2}-4 c b \,x^{2}+a c \right )}{5 a^{2} \left (e x \right )^{\frac {7}{2}}}\)	\(39\)
risch	\(-\frac {2 \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (5 a d \,x^{2}-4 c b \,x^{2}+a c \right )}{5 e^{3} \sqrt {e x}\, a^{2} x^{2}}\)	\(44\)

[In]

int((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(3/4),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.64 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 \, {\left ({\left (4 \, b c - 5 \, a d\right )} x^{2} - a c\right )} {\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {e x}}{5 \, a^{2} e^{4} x^{3}} \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (60) = 120\).

Time = 32.11 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.81 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=- \frac {\sqrt [4]{b} c \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{8 a e^{\frac {7}{2}} x^{2} \Gamma \left (\frac {3}{4}\right )} + \frac {\sqrt [4]{b} d \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {1}{4}\right )}{2 a e^{\frac {7}{2}} \Gamma \left (\frac {3}{4}\right )} + \frac {b^{\frac {5}{4}} c \sqrt [4]{\frac {a}{b x^{2}} + 1} \Gamma \left (- \frac {5}{4}\right )}{2 a^{2} e^{\frac {7}{2}} \Gamma \left (\frac {3}{4}\right )} \]

[In]

integrate((d*x**2+c)/(e*x)**(7/2)/(b*x**2+a)**(3/4),x)

[Out]

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

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(7/2)), x)

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(3/4)*(e*x)^(7/2)), x)

Mupad [B] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {\left (\frac {2\,c}{5\,a\,e^3}+\frac {x^2\,\left (10\,a\,d-8\,b\,c\right )}{5\,a^2\,e^3}\right )\,{\left (b\,x^2+a\right )}^{1/4}}{x^2\,\sqrt {e\,x}} \]

[In]

int((c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(3/4)),x)

[Out]

-(((2*c)/(5*a*e^3) + (x^2*(10*a*d - 8*b*c))/(5*a^2*e^3))*(a + b*x^2)^(1/4))/(x^2*(e*x)^(1/2))