3.1093 \(\int \frac{(e x)^{5/2} (c+d x^2)}{(a+b x^2)^{3/4}} \, dx\)

Optimal. Leaf size=173 \[ \frac{3 a e^{5/2} (8 b c-7 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac{3 a e^{5/2} (8 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}+\frac{e (e x)^{3/2} \sqrt [4]{a+b x^2} (8 b c-7 a d)}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e} \]

[Out]

((8*b*c - 7*a*d)*e*(e*x)^(3/2)*(a + b*x^2)^(1/4))/(16*b^2) + (d*(e*x)^(7/2)*(a + b*x^2)^(1/4))/(4*b*e) + (3*a*
(8*b*c - 7*a*d)*e^(5/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4)) - (3*a*(8*b*c -
7*a*d)*e^(5/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4))

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Rubi [A]  time = 0.126232, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {459, 321, 329, 331, 298, 205, 208} \[ \frac{3 a e^{5/2} (8 b c-7 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac{3 a e^{5/2} (8 b c-7 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}+\frac{e (e x)^{3/2} \sqrt [4]{a+b x^2} (8 b c-7 a d)}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e} \]

Antiderivative was successfully verified.

[In]

Int[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(3/4),x]

[Out]

((8*b*c - 7*a*d)*e*(e*x)^(3/2)*(a + b*x^2)^(1/4))/(16*b^2) + (d*(e*x)^(7/2)*(a + b*x^2)^(1/4))/(4*b*e) + (3*a*
(8*b*c - 7*a*d)*e^(5/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4)) - (3*a*(8*b*c -
7*a*d)*e^(5/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(32*b^(11/4))

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 205

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

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{(e x)^{5/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{3/4}} \, dx &=\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac{\left (-4 b c+\frac{7 a d}{2}\right ) \int \frac{(e x)^{5/2}}{\left (a+b x^2\right )^{3/4}} \, dx}{4 b}\\ &=\frac{(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac{\left (3 a (8 b c-7 a d) e^2\right ) \int \frac{\sqrt{e x}}{\left (a+b x^2\right )^{3/4}} \, dx}{32 b^2}\\ &=\frac{(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac{(3 a (8 b c-7 a d) e) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+\frac{b x^4}{e^2}\right )^{3/4}} \, dx,x,\sqrt{e x}\right )}{16 b^2}\\ &=\frac{(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac{(3 a (8 b c-7 a d) e) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{b x^4}{e^2}} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{16 b^2}\\ &=\frac{(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}-\frac{\left (3 a (8 b c-7 a d) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e-\sqrt{b} x^2} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}}+\frac{\left (3 a (8 b c-7 a d) e^3\right ) \operatorname{Subst}\left (\int \frac{1}{e+\sqrt{b} x^2} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{32 b^{5/2}}\\ &=\frac{(8 b c-7 a d) e (e x)^{3/2} \sqrt [4]{a+b x^2}}{16 b^2}+\frac{d (e x)^{7/2} \sqrt [4]{a+b x^2}}{4 b e}+\frac{3 a (8 b c-7 a d) e^{5/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}-\frac{3 a (8 b c-7 a d) e^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{32 b^{11/4}}\\ \end{align*}

Mathematica [A]  time = 0.165653, size = 131, normalized size = 0.76 \[ \frac{(e x)^{5/2} \left (2 b^{3/4} x^{3/2} \sqrt [4]{a+b x^2} \left (-7 a d+8 b c+4 b d x^2\right )-3 a (7 a d-8 b c) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )+3 a (7 a d-8 b c) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a+b x^2}}\right )\right )}{32 b^{11/4} x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((e*x)^(5/2)*(c + d*x^2))/(a + b*x^2)^(3/4),x]

[Out]

((e*x)^(5/2)*(2*b^(3/4)*x^(3/2)*(a + b*x^2)^(1/4)*(8*b*c - 7*a*d + 4*b*d*x^2) - 3*a*(-8*b*c + 7*a*d)*ArcTan[(b
^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + 3*a*(-8*b*c + 7*a*d)*ArcTanh[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)]))/(32*b
^(11/4)*x^(5/2))

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Maple [F]  time = 0.031, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c) \left ( ex \right ) ^{{\frac{5}{2}}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \left (e x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="maxima")

[Out]

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

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Fricas [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((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C]  time = 142.568, size = 94, normalized size = 0.54 \begin{align*} \frac{c e^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{11}{4}\right )} + \frac{d e^{\frac{5}{2}} x^{\frac{11}{2}} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{3}{4}} \Gamma \left (\frac{15}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(5/2)*(d*x**2+c)/(b*x**2+a)**(3/4),x)

[Out]

c*e**(5/2)*x**(7/2)*gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(11/4)) +
d*e**(5/2)*x**(11/2)*gamma(11/4)*hyper((3/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/4)*gamma(15/4
))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \left (e x\right )^{\frac{5}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(5/2)*(d*x^2+c)/(b*x^2+a)^(3/4),x, algorithm="giac")

[Out]

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