3.1117 \(\int \frac{\sqrt{e x} (c+d x^2)}{(a+b x^2)^{7/4}} \, dx\)

Optimal. Leaf size=125 \[ -\frac{d \sqrt{e} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac{d \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac{2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0764689, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {452, 329, 331, 298, 205, 208} \[ -\frac{d \sqrt{e} \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac{d \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{e x}}{\sqrt{e} \sqrt [4]{a+b x^2}}\right )}{b^{7/4}}+\frac{2 (e x)^{3/2} (b c-a d)}{3 a b e \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 452

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

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

Mathematica [C]  time = 0.059424, size = 69, normalized size = 0.55 \[ \frac{2 \sqrt{e x} \left (3 d x^3 \left (\frac{b x^2}{a}+1\right )^{3/4} \, _2F_1\left (\frac{7}{4},\frac{7}{4};\frac{11}{4};-\frac{b x^2}{a}\right )+7 c x\right )}{21 a \left (a+b x^2\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e*x]*(7*c*x + 3*d*x^3*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[7/4, 7/4, 11/4, -((b*x^2)/a)]))/(21*a*(a
+ b*x^2)^(3/4))

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Maple [F]  time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{(d{x}^{2}+c)\sqrt{ex} \left ( b{x}^{2}+a \right ) ^{-{\frac{7}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

[Out]

int((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((d*x^2 + c)*sqrt(e*x)/(b*x^2 + a)^(7/4), 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)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

Timed out

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Sympy [C]  time = 75.2104, size = 87, normalized size = 0.7 \begin{align*} \frac{c \sqrt{e} x^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right )}{2 a^{\frac{7}{4}} \left (1 + \frac{b x^{2}}{a}\right )^{\frac{3}{4}} \Gamma \left (\frac{7}{4}\right )} + \frac{d \sqrt{e} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{7}{4}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{7}{4}} \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)**(1/2)*(d*x**2+c)/(b*x**2+a)**(7/4),x)

[Out]

c*sqrt(e)*x**(3/2)*gamma(3/4)/(2*a**(7/4)*(1 + b*x**2/a)**(3/4)*gamma(7/4)) + d*sqrt(e)*x**(7/2)*gamma(7/4)*hy
per((7/4, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(11/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x^{2} + c\right )} \sqrt{e x}}{{\left (b x^{2} + a\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x)^(1/2)*(d*x^2+c)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

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