3.1105 \(\int \frac{(e x)^{3/2} (c+d x^2)}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=171 \[ \frac{e^{3/2} (4 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac{e^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac{e \sqrt{e x} (4 b c-5 a d)}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}} \]

[Out]

-((4*b*c - 5*a*d)*e*Sqrt[e*x])/(2*b^2*(a + b*x^2)^(1/4)) + (d*(e*x)^(5/2))/(2*b*e*(a + b*x^2)^(1/4)) + ((4*b*c
- 5*a*d)*e^(3/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4)) + ((4*b*c - 5*a*d)*e^(3
/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4))

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Rubi [A]  time = 0.107683, antiderivative size = 171, 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, 288, 329, 240, 212, 208, 205} \[ \frac{e^{3/2} (4 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac{e^{3/2} (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}-\frac{e \sqrt{e x} (4 b c-5 a d)}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*b*c - 5*a*d)*e*Sqrt[e*x])/(2*b^2*(a + b*x^2)^(1/4)) + (d*(e*x)^(5/2))/(2*b*e*(a + b*x^2)^(1/4)) + ((4*b*c
- 5*a*d)*e^(3/2)*ArcTan[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4)) + ((4*b*c - 5*a*d)*e^(3
/2)*ArcTanh[(b^(1/4)*Sqrt[e*x])/(Sqrt[e]*(a + b*x^2)^(1/4))])/(4*b^(9/4))

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 212

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

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]

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]

Rubi steps

\begin{align*} \int \frac{(e x)^{3/2} \left (c+d x^2\right )}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}-\frac{\left (-2 b c+\frac{5 a d}{2}\right ) \int \frac{(e x)^{3/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{2 b}\\ &=-\frac{(4 b c-5 a d) e \sqrt{e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-5 a d) e^2\right ) \int \frac{1}{\sqrt{e x} \sqrt [4]{a+b x^2}} \, dx}{4 b^2}\\ &=-\frac{(4 b c-5 a d) e \sqrt{e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac{((4 b c-5 a d) e) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+\frac{b x^4}{e^2}}} \, dx,x,\sqrt{e x}\right )}{2 b^2}\\ &=-\frac{(4 b c-5 a d) e \sqrt{e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac{((4 b c-5 a d) e) \operatorname{Subst}\left (\int \frac{1}{1-\frac{b x^4}{e^2}} \, dx,x,\frac{\sqrt{e x}}{\sqrt [4]{a+b x^2}}\right )}{2 b^2}\\ &=-\frac{(4 b c-5 a d) e \sqrt{e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac{\left ((4 b c-5 a d) e^2\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 )}{4 b^2}+\frac{\left ((4 b c-5 a d) e^2\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 )}{4 b^2}\\ &=-\frac{(4 b c-5 a d) e \sqrt{e x}}{2 b^2 \sqrt [4]{a+b x^2}}+\frac{d (e x)^{5/2}}{2 b e \sqrt [4]{a+b x^2}}+\frac{(4 b c-5 a d) e^{3/2} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}+\frac{(4 b c-5 a d) e^{3/2} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{4 b^{9/4}}\\ \end{align*}

Mathematica [C]  time = 0.114269, size = 77, normalized size = 0.45 \[ \frac{x (e x)^{3/2} \left (\sqrt [4]{\frac{b x^2}{a}+1} (4 b c-5 a d) \, _2F_1\left (\frac{5}{4},\frac{5}{4};\frac{9}{4};-\frac{b x^2}{a}\right )+5 a d\right )}{10 a b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^(3/2)*(5*a*d + (4*b*c - 5*a*d)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[5/4, 5/4, 9/4, -((b*x^2)/a)]))
/(10*a*b*(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{3}{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((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),x)

[Out]

int((e*x)^(3/2)*(d*x^2+c)/(b*x^2+a)^(5/4),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{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06611, size = 2084, normalized size = 12.19 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(4*(b*d*e*x^2 - (4*b*c - 5*a*d)*e)*(b*x^2 + a)^(3/4)*sqrt(e*x) + 4*(b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*
a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4)*arctan(((4*b^8*c - 5*a*b^7
*d)*(b*x^2 + a)^(3/4)*sqrt(e*x)*e*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 +
625*a^4*d^4)*e^6/b^9)^(3/4) + (b^8*x^2 + a*b^7)*sqrt(((16*b^2*c^2 - 40*a*b*c*d + 25*a^2*d^2)*sqrt(b*x^2 + a)*
e^3*x + (b^5*x^2 + a*b^4)*sqrt((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625
*a^4*d^4)*e^6/b^9))/(b*x^2 + a))*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 +
625*a^4*d^4)*e^6/b^9)^(3/4))/((256*b^5*c^4 - 1280*a*b^4*c^3*d + 2400*a^2*b^3*c^2*d^2 - 2000*a^3*b^2*c*d^3 + 62
5*a^4*b*d^4)*e^6*x^2 + (256*a*b^4*c^4 - 1280*a^2*b^3*c^3*d + 2400*a^3*b^2*c^2*d^2 - 2000*a^4*b*c*d^3 + 625*a^5
*d^4)*e^6)) + (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 6
25*a^4*d^4)*e^6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e + (b^3*x^2 + a*b^2)*((256*b^4*c
^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/(b*x^2 + a)) -
(b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^
6/b^9)^(1/4)*log(-((b*x^2 + a)^(3/4)*(4*b*c - 5*a*d)*sqrt(e*x)*e - (b^3*x^2 + a*b^2)*((256*b^4*c^4 - 1280*a*b^
3*c^3*d + 2400*a^2*b^2*c^2*d^2 - 2000*a^3*b*c*d^3 + 625*a^4*d^4)*e^6/b^9)^(1/4))/(b*x^2 + a)))/(b^3*x^2 + a*b^
2)

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Sympy [C]  time = 60.2405, size = 94, normalized size = 0.55 \begin{align*} \frac{c e^{\frac{3}{2}} x^{\frac{5}{2}} \Gamma \left (\frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{5}{4} \\ \frac{9}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{4}} \Gamma \left (\frac{9}{4}\right )} + \frac{d e^{\frac{3}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{9}{4} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{4}} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*e**(3/2)*x**(5/2)*gamma(5/4)*hyper((5/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(9/4)) + d
*e**(3/2)*x**(9/2)*gamma(9/4)*hyper((5/4, 9/4), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(13/4))

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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{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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