3.1115 \(\int \frac{c+d x^2}{(e x)^{11/2} (a+b x^2)^{5/4}} \, dx\)

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

[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))

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Rubi [A]  time = 0.0996455, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {453, 286, 284, 335, 196} \[ \frac{8 b^{3/2} \sqrt{e x} \sqrt [4]{\frac{a}{b x^2}+1} (10 b c-9 a d) E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{15 a^{7/2} e^6 \sqrt [4]{a+b x^2}}-\frac{4 b (10 b c-9 a d)}{15 a^3 e^5 \sqrt{e x} \sqrt [4]{a+b x^2}}+\frac{2 (10 b c-9 a d)}{45 a^2 e^3 (e x)^{5/2} \sqrt [4]{a+b x^2}}-\frac{2 c}{9 a e (e x)^{9/2} \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[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 453

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]

Rule 286

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 284

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 335

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 196

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

Rubi steps

\begin{align*} \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{(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 ) \operatorname{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]  time = 0.0686539, size = 82, normalized size = 0.45 \[ -\frac{2 \sqrt{e x} \left (x^2 \sqrt [4]{\frac{b x^2}{a}+1} (9 a d-10 b c) \, _2F_1\left (-\frac{5}{4},\frac{5}{4};-\frac{1}{4};-\frac{b x^2}{a}\right )+5 a c\right )}{45 a^2 e^6 x^5 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c) \left ( ex \right ) ^{-{\frac{11}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac{5}{4}} \left (e x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{4}}{\left (d x^{2} + c\right )} \sqrt{e x}}{b^{2} e^{6} x^{10} + 2 \, a b e^{6} x^{8} + a^{2} e^{6} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x^{2} + c}{{\left (b x^{2} + a\right )}^{\frac{5}{4}} \left (e x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)