\(\int \frac {c+d x^2}{(e x)^{11/2} (a+b x^2)^{5/4}} \, dx\) [1115]

Optimal result

Integrand size = 26, antiderivative size = 182 \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {8 b^{3/2} (10 b c-9 a d) \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {e x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{7/2} e^6 \sqrt [4]{a+b x^2}} \]

[Out]

-2/9*c/a/e/(e*x)^(9/2)/(b*x^2+a)^(1/4)+2/45*(-9*a*d+10*b*c)/a^2/e^3/(e*x)^(5/2)/(b*x^2+a)^(1/4)-4/15*b*(-9*a*d
+10*b*c)/a^3/e^5/(b*x^2+a)^(1/4)/(e*x)^(1/2)+8/15*b^(3/2)*(-9*a*d+10*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))*(e*x)^(1/2)/a^(7/2)/e^6/(b*x^2+a)^(1/4)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {464, 292, 290, 342, 202} \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {8 b^{3/2} \sqrt {e x} \sqrt [4]{\frac {a}{b x^2}+1} (10 b c-9 a d) E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 a^{7/2} e^6 \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}} \]

[In]

Int[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*c)/(9*a*e*(e*x)^(9/2)*(a + b*x^2)^(1/4)) + (2*(10*b*c - 9*a*d))/(45*a^2*e^3*(e*x)^(5/2)*(a + b*x^2)^(1/4))
- (4*b*(10*b*c - 9*a*d))/(15*a^3*e^5*Sqrt[e*x]*(a + b*x^2)^(1/4)) + (8*b^(3/2)*(10*b*c - 9*a*d)*(1 + a/(b*x^2
))^(1/4)*Sqrt[e*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(15*a^(7/2)*e^6*(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}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}-\frac {(10 b c-9 a d) \int \frac {1}{(e x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx}{9 a e^2} \\ & = -\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {(2 b (10 b c-9 a d)) \int \frac {1}{(e x)^{3/2} \left (a+b x^2\right )^{5/4}} \, dx}{15 a^2 e^4} \\ & = -\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {\left (4 b^2 (10 b c-9 a d)\right ) \int \frac {\sqrt {e x}}{\left (a+b x^2\right )^{5/4}} \, dx}{15 a^3 e^6} \\ & = -\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}-\frac {\left (4 b (10 b c-9 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}{15 a^3 e^6 \sqrt [4]{a+b x^2}} \\ & = -\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {\left (4 b (10 b c-9 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 )}{15 a^3 e^6 \sqrt [4]{a+b x^2}} \\ & = -\frac {2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt {e x} \sqrt [4]{a+b x^2}}+\frac {8 b^{3/2} (10 b c-9 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 )}{15 a^{7/2} e^6 \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 = 10.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.45 \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 \sqrt {e x} \left (5 a c+(-10 b c+9 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{4},-\frac {1}{4},-\frac {b x^2}{a}\right )\right )}{45 a^2 e^6 x^5 \sqrt [4]{a+b x^2}} \]

[In]

Integrate[(c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(5/4)),x]

[Out]

(-2*Sqrt[e*x]*(5*a*c + (-10*b*c + 9*a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-5/4, 5/4, -1/4, -((b*x^2
)/a)]))/(45*a^2*e^6*x^5*(a + b*x^2)^(1/4))

Maple [F]

\[\int \frac {d \,x^{2}+c}{\left (e x \right )^{\frac {11}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

[In]

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(5/4),x)

[Out]

int((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(5/4),x)

Fricas [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(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^6*x^10 + 2*a*b*e^6*x^8 + a^2*e^6*x^6), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\text {Timed out} \]

[In]

integrate((d*x**2+c)/(e*x)**(11/2)/(b*x**2+a)**(5/4),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*(e*x)^(11/2)), x)

Giac [F]

\[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (e x\right )^{\frac {11}{2}}} \,d x } \]

[In]

integrate((d*x^2+c)/(e*x)^(11/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((d*x^2 + c)/((b*x^2 + a)^(5/4)*(e*x)^(11/2)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {c+d x^2}{(e x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {d\,x^2+c}{{\left (e\,x\right )}^{11/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

[In]

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(5/4)),x)

[Out]

int((c + d*x^2)/((e*x)^(11/2)*(a + b*x^2)^(5/4)), x)