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

Optimal. Leaf size=200 \[ \frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} \left (a^2 d^2 (m+3) (m+7)+b c (m+1) (b c (m+5)-2 a d (m+7))\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{d^2 e (m+1) (m+3) (m+7) \sqrt{c+d x^4}}-\frac{b \sqrt{c+d x^4} (e x)^{m+1} (b c (m+5)-2 a d (m+7))}{d^2 e (m+3) (m+7)}+\frac{b^2 \sqrt{c+d x^4} (e x)^{m+5}}{d e^5 (m+7)} \]

[Out]

-((b*(b*c*(5 + m) - 2*a*d*(7 + m))*(e*x)^(1 + m)*Sqrt[c + d*x^4])/(d^2*e*(3 + m)*(7 + m))) + (b^2*(e*x)^(5 + m
)*Sqrt[c + d*x^4])/(d*e^5*(7 + m)) + ((a^2*d^2*(3 + m)*(7 + m) + b*c*(1 + m)*(b*c*(5 + m) - 2*a*d*(7 + m)))*(e
*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(d^2*e*(1 + m)*(3
+ m)*(7 + m)*Sqrt[c + d*x^4])

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Rubi [A]  time = 0.212091, antiderivative size = 194, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {464, 459, 365, 364} \[ \frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} \left (\frac{a^2 d^2 (m+7)}{m+1}+\frac{b c (b c (m+5)-2 a d (m+7))}{m+3}\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{d^2 e (m+7) \sqrt{c+d x^4}}-\frac{b \sqrt{c+d x^4} (e x)^{m+1} (b c (m+5)-2 a d (m+7))}{d^2 e (m+3) (m+7)}+\frac{b^2 \sqrt{c+d x^4} (e x)^{m+5}}{d e^5 (m+7)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b*c*(5 + m) - 2*a*d*(7 + m))*(e*x)^(1 + m)*Sqrt[c + d*x^4])/(d^2*e*(3 + m)*(7 + m))) + (b^2*(e*x)^(5 + m
)*Sqrt[c + d*x^4])/(d*e^5*(7 + m)) + (((a^2*d^2*(7 + m))/(1 + m) + (b*c*(b*c*(5 + m) - 2*a*d*(7 + m)))/(3 + m)
)*(e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(d^2*e*(7 + m)
*Sqrt[c + d*x^4])

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(d^2*(e*x)^
(m + n + 1)*(a + b*x^n)^(p + 1))/(b*e^(n + 1)*(m + n*(p + 2) + 1)), x] + Dist[1/(b*(m + n*(p + 2) + 1)), Int[(
e*x)^m*(a + b*x^n)^p*Simp[b*c^2*(m + n*(p + 2) + 1) + d*((2*b*c - a*d)*(m + n + 1) + 2*b*c*n*(p + 1))*x^n, x],
x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && NeQ[m + n*(p + 2) + 1, 0]

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 365

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

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(e x)^m \left (a+b x^4\right )^2}{\sqrt{c+d x^4}} \, dx &=\frac{b^2 (e x)^{5+m} \sqrt{c+d x^4}}{d e^5 (7+m)}+\frac{\int \frac{(e x)^m \left (a^2 d (7+m)-b (b c (5+m)-2 a d (7+m)) x^4\right )}{\sqrt{c+d x^4}} \, dx}{d (7+m)}\\ &=-\frac{b (b c (5+m)-2 a d (7+m)) (e x)^{1+m} \sqrt{c+d x^4}}{d^2 e (3+m) (7+m)}+\frac{b^2 (e x)^{5+m} \sqrt{c+d x^4}}{d e^5 (7+m)}-\left (-a^2-\frac{b c (1+m) (b c (5+m)-2 a d (7+m))}{d^2 (3+m) (7+m)}\right ) \int \frac{(e x)^m}{\sqrt{c+d x^4}} \, dx\\ &=-\frac{b (b c (5+m)-2 a d (7+m)) (e x)^{1+m} \sqrt{c+d x^4}}{d^2 e (3+m) (7+m)}+\frac{b^2 (e x)^{5+m} \sqrt{c+d x^4}}{d e^5 (7+m)}-\frac{\left (\left (-a^2-\frac{b c (1+m) (b c (5+m)-2 a d (7+m))}{d^2 (3+m) (7+m)}\right ) \sqrt{1+\frac{d x^4}{c}}\right ) \int \frac{(e x)^m}{\sqrt{1+\frac{d x^4}{c}}} \, dx}{\sqrt{c+d x^4}}\\ &=-\frac{b (b c (5+m)-2 a d (7+m)) (e x)^{1+m} \sqrt{c+d x^4}}{d^2 e (3+m) (7+m)}+\frac{b^2 (e x)^{5+m} \sqrt{c+d x^4}}{d e^5 (7+m)}+\frac{\left (a^2+\frac{b c (1+m) (b c (5+m)-2 a d (7+m))}{d^2 (3+m) (7+m)}\right ) (e x)^{1+m} \sqrt{1+\frac{d x^4}{c}} \, _2F_1\left (\frac{1}{2},\frac{1+m}{4};\frac{5+m}{4};-\frac{d x^4}{c}\right )}{e (1+m) \sqrt{c+d x^4}}\\ \end{align*}

Mathematica [A]  time = 0.137103, size = 164, normalized size = 0.82 \[ \frac{x \sqrt{\frac{d x^4}{c}+1} (e x)^m \left (a^2 \left (m^2+14 m+45\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )+b (m+1) x^4 \left (2 a (m+9) \, _2F_1\left (\frac{1}{2},\frac{m+5}{4};\frac{m+9}{4};-\frac{d x^4}{c}\right )+b (m+5) x^4 \, _2F_1\left (\frac{1}{2},\frac{m+9}{4};\frac{m+13}{4};-\frac{d x^4}{c}\right )\right )\right )}{(m+1) (m+5) (m+9) \sqrt{c+d x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[1 + (d*x^4)/c]*(a^2*(45 + 14*m + m^2)*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c
)] + b*(1 + m)*x^4*(2*a*(9 + m)*Hypergeometric2F1[1/2, (5 + m)/4, (9 + m)/4, -((d*x^4)/c)] + b*(5 + m)*x^4*Hyp
ergeometric2F1[1/2, (9 + m)/4, (13 + m)/4, -((d*x^4)/c)])))/((1 + m)*(5 + m)*(9 + m)*Sqrt[c + d*x^4])

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( b{x}^{4}+a \right ) ^{2}{\frac{1}{\sqrt{d{x}^{4}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^8 + 2*a*b*x^4 + a^2)*(e*x)^m/sqrt(d*x^4 + c), x)

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Sympy [C]  time = 55.6202, size = 185, normalized size = 0.92 \begin{align*} \frac{a^{2} e^{m} x x^{m} \Gamma \left (\frac{m}{4} + \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{4} + \frac{1}{4} \\ \frac{m}{4} + \frac{5}{4} \end{matrix}\middle |{\frac{d x^{4} e^{i \pi }}{c}} \right )}}{4 \sqrt{c} \Gamma \left (\frac{m}{4} + \frac{5}{4}\right )} + \frac{a b e^{m} x^{5} x^{m} \Gamma \left (\frac{m}{4} + \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{4} + \frac{5}{4} \\ \frac{m}{4} + \frac{9}{4} \end{matrix}\middle |{\frac{d x^{4} e^{i \pi }}{c}} \right )}}{2 \sqrt{c} \Gamma \left (\frac{m}{4} + \frac{9}{4}\right )} + \frac{b^{2} e^{m} x^{9} x^{m} \Gamma \left (\frac{m}{4} + \frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, \frac{m}{4} + \frac{9}{4} \\ \frac{m}{4} + \frac{13}{4} \end{matrix}\middle |{\frac{d x^{4} e^{i \pi }}{c}} \right )}}{4 \sqrt{c} \Gamma \left (\frac{m}{4} + \frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*e**m*x*x**m*gamma(m/4 + 1/4)*hyper((1/2, m/4 + 1/4), (m/4 + 5/4,), d*x**4*exp_polar(I*pi)/c)/(4*sqrt(c)*g
amma(m/4 + 5/4)) + a*b*e**m*x**5*x**m*gamma(m/4 + 5/4)*hyper((1/2, m/4 + 5/4), (m/4 + 9/4,), d*x**4*exp_polar(
I*pi)/c)/(2*sqrt(c)*gamma(m/4 + 9/4)) + b**2*e**m*x**9*x**m*gamma(m/4 + 9/4)*hyper((1/2, m/4 + 9/4), (m/4 + 13
/4,), d*x**4*exp_polar(I*pi)/c)/(4*sqrt(c)*gamma(m/4 + 13/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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