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

Optimal. Leaf size=198 \[ -\frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} \left (2 b^2 c^2 (m+1)-(m+3) \left (2 a^2 d^2-(m+1) (b c-a d)^2\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{2 c d^2 e (m+1) (m+3) \sqrt{c+d x^4}}+\frac{(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt{c+d x^4}}+\frac{b^2 \sqrt{c+d x^4} (e x)^{m+1}}{d^2 e (m+3)} \]

[Out]

((b*c - a*d)^2*(e*x)^(1 + m))/(2*c*d^2*e*Sqrt[c + d*x^4]) + (b^2*(e*x)^(1 + m)*Sqrt[c + d*x^4])/(d^2*e*(3 + m)
) - ((2*b^2*c^2*(1 + m) - (3 + m)*(2*a^2*d^2 - (b*c - a*d)^2*(1 + m)))*(e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hyper
geometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(2*c*d^2*e*(1 + m)*(3 + m)*Sqrt[c + d*x^4])

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Rubi [A]  time = 0.17849, antiderivative size = 198, normalized size of antiderivative = 1., 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 = {463, 459, 365, 364} \[ -\frac{\sqrt{\frac{d x^4}{c}+1} (e x)^{m+1} \left (2 b^2 c^2 (m+1)-(m+3) \left (2 a^2 d^2-(m+1) (b c-a d)^2\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )}{2 c d^2 e (m+1) (m+3) \sqrt{c+d x^4}}+\frac{(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt{c+d x^4}}+\frac{b^2 \sqrt{c+d x^4} (e x)^{m+1}}{d^2 e (m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*(e*x)^(1 + m))/(2*c*d^2*e*Sqrt[c + d*x^4]) + (b^2*(e*x)^(1 + m)*Sqrt[c + d*x^4])/(d^2*e*(3 + m)
) - ((2*b^2*c^2*(1 + m) - (3 + m)*(2*a^2*d^2 - (b*c - a*d)^2*(1 + m)))*(e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hyper
geometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(2*c*d^2*e*(1 + m)*(3 + m)*Sqrt[c + d*x^4])

Rule 463

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

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

Mathematica [A]  time = 0.14129, size = 167, normalized size = 0.84 \[ \frac{x \sqrt{\frac{d x^4}{c}+1} (e x)^m \left (a^2 \left (m^2+14 m+45\right ) \, _2F_1\left (\frac{3}{2},\frac{m+1}{4};\frac{m+5}{4};-\frac{d x^4}{c}\right )+b (m+1) x^4 \left (2 a (m+9) \, _2F_1\left (\frac{3}{2},\frac{m+5}{4};\frac{m+9}{4};-\frac{d x^4}{c}\right )+b (m+5) x^4 \, _2F_1\left (\frac{3}{2},\frac{m+9}{4};\frac{m+13}{4};-\frac{d x^4}{c}\right )\right )\right )}{c (m+1) (m+5) (m+9) \sqrt{c+d x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(e*x)^m*Sqrt[1 + (d*x^4)/c]*(a^2*(45 + 14*m + m^2)*Hypergeometric2F1[3/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c
)] + b*(1 + m)*x^4*(2*a*(9 + m)*Hypergeometric2F1[3/2, (5 + m)/4, (9 + m)/4, -((d*x^4)/c)] + b*(5 + m)*x^4*Hyp
ergeometric2F1[3/2, (9 + m)/4, (13 + m)/4, -((d*x^4)/c)])))/(c*(1 + m)*(5 + m)*(9 + m)*Sqrt[c + d*x^4])

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex \right ) ^{m} \left ( b{x}^{4}+a \right ) ^{2} \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)^2/(d*x^4+c)^(3/2),x)

[Out]

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

[Out]

integrate((b*x^4 + a)^2*(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^{2} x^{8} + 2 \, a b x^{4} + a^{2}\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)^2/(d*x^4+c)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*x^8 + 2*a*b*x^4 + a^2)*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 [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)**m*(b*x**4+a)**2/(d*x**4+c)**(3/2),x)

[Out]

Timed out

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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}}{{\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)^2/(d*x^4+c)^(3/2),x, algorithm="giac")

[Out]

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