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

Optimal. Leaf size=132 \[ \frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} (a d (1-m)+b c (m+1)) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{2 c d e (m+1) \sqrt{c+d x^4}}-\frac{(e x)^{m+1} (b c-a d)}{2 c d e \sqrt{c+d x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0599819, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {457, 365, 364} \[ \frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} (a d (1-m)+b c (m+1)) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{2 c d e (m+1) \sqrt{c+d x^4}}-\frac{(e x)^{m+1} (b c-a d)}{2 c d e \sqrt{c+d x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 457

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*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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

Mathematica [A]  time = 0.0914583, size = 113, normalized size = 0.86 \[ \frac{x \sqrt{\frac{d x^4}{c}+1} (e x)^m \left (a (m+5) \, _2F_1\left (\frac{3}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )+b (m+1) x^4 \, _2F_1\left (\frac{3}{2},\frac{m+5}{4};\frac{m+9}{4};-\frac{d x^4}{c}\right )\right )}{c (m+1) (m+5) \sqrt{c+d x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.02, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( b{x}^{4}+a \right ) \left ( d{x}^{4}+c \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 163.386, size = 119, normalized size = 0.9 \begin{align*} \frac{a e^{m} x x^{m} \Gamma \left (\frac{m}{4} + \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{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 c^{\frac{3}{2}} \Gamma \left (\frac{m}{4} + \frac{5}{4}\right )} + \frac{b e^{m} x^{5} x^{m} \Gamma \left (\frac{m}{4} + \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{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 )}}{4 c^{\frac{3}{2}} \Gamma \left (\frac{m}{4} + \frac{9}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e**m*x*x**m*gamma(m/4 + 1/4)*hyper((3/2, m/4 + 1/4), (m/4 + 5/4,), d*x**4*exp_polar(I*pi)/c)/(4*c**(3/2)*gam
ma(m/4 + 5/4)) + b*e**m*x**5*x**m*gamma(m/4 + 5/4)*hyper((3/2, m/4 + 5/4), (m/4 + 9/4,), d*x**4*exp_polar(I*pi
)/c)/(4*c**(3/2)*gamma(m/4 + 9/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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