\(\int \frac {c+d x^2}{(e x)^{7/2} (a+b x^2)^{5/4}} \, dx\) [1114]

Optimal result

Integrand size = 26, antiderivative size = 144 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {4 \sqrt {b} (6 b c-5 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} e^4 \sqrt [4]{a+b x^2}} \]

[Out]

-2/5*c/a/e/(e*x)^(5/2)/(b*x^2+a)^(1/4)+2/5*(-5*a*d+6*b*c)/a^2/e^3/(b*x^2+a)^(1/4)/(e*x)^(1/2)-4/5*(-5*a*d+6*b*
c)*(1+a/b/x^2)^(1/4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticE
(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))*b^(1/2)*(e*x)^(1/2)/a^(5/2)/e^4/(b*x^2+a)^(1/4)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {464, 292, 290, 342, 202} \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {4 \sqrt {b} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (6 b c-5 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} e^4 \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}} \]

[In]

Int[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*c)/(5*a*e*(e*x)^(5/2)*(a + b*x^2)^(1/4)) + (2*(6*b*c - 5*a*d))/(5*a^2*e^3*Sqrt[e*x]*(a + b*x^2)^(1/4)) - (
4*Sqrt[b]*(6*b*c - 5*a*d)*(1 + a/(b*x^2))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*a^(5
/2)*e^4*(a + b*x^2)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 290

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[Sqrt[c*x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b
*x^2)^(1/4))), Int[1/(x^2*(1 + a/(b*x^2))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]

Rule 292

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^(m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1
/4)), x] - Dist[b*((2*m + 1)/(2*a*c^2*(m + 1))), Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c
}, x] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

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}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {(6 b c-5 a d) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx}{5 a e^2} \\ & = -\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {(2 b (6 b c-5 a d)) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{5 a^2 e^4} \\ & = -\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {\left (2 (6 b c-5 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{5 a^2 e^4 \sqrt [4]{a+b x^2}} \\ & = -\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {\left (2 (6 b c-5 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{5 a^2 e^4 \sqrt [4]{a+b x^2}} \\ & = -\frac {2 c}{5 a e (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {2 (6 b c-5 a d)}{5 a^2 e^3 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {4 \sqrt {b} (6 b c-5 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} e^4 \sqrt [4]{a+b x^2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.54 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (-2 a c+2 (6 b c-5 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{4},\frac {3}{4},-\frac {b x^2}{a}\right )\right )}{5 a^2 (e x)^{7/2} \sqrt [4]{a+b x^2}} \]

[In]

Integrate[(c + d*x^2)/((e*x)^(7/2)*(a + b*x^2)^(5/4)),x]

[Out]

(x*(-2*a*c + 2*(6*b*c - 5*a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-1/4, 5/4, 3/4, -((b*x^2)/a)]))/(5*
a^2*(e*x)^(7/2)*(a + b*x^2)^(1/4))

Maple [F]

\[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*(d*x^2 + c)*sqrt(e*x)/(b^2*e^4*x^8 + 2*a*b*e^4*x^6 + a^2*e^4*x^4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 77.57 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.59 \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=- \frac {c {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{5 b^{\frac {5}{4}} e^{\frac {7}{2}} x^{5}} + \frac {d \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} e^{\frac {7}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

[In]

integrate((d*x**2+c)/(e*x)**(7/2)/(b*x**2+a)**(5/4),x)

[Out]

-c*hyper((5/4, 5/2), (7/2,), a*exp_polar(I*pi)/(b*x**2))/(5*b**(5/4)*e**(7/2)*x**5) + d*gamma(-1/4)*hyper((-1/
4, 5/4), (3/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*e**(7/2)*sqrt(x)*gamma(3/4))

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

[In]

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

[Out]

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