3.9.41 \(\int \frac {(e x)^m (a+b x^4)^2}{\sqrt {c+d x^4}} \, dx\) [841]

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

[Out]

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

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Rubi [A]
time = 0.16, 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 = {475, 470, 372, 371} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

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 470

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 475

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]

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*}

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Mathematica [A]
time = 7.75, size = 164, normalized size = 0.82 \begin {gather*} \frac {x (e x)^m \sqrt {1+\frac {d x^4}{c}} \left (a^2 \left (45+14 m+m^2\right ) \, _2F_1\left (\frac {1}{2},\frac {1+m}{4};\frac {5+m}{4};-\frac {d x^4}{c}\right )+b (1+m) x^4 \left (2 a (9+m) \, _2F_1\left (\frac {1}{2},\frac {5+m}{4};\frac {9+m}{4};-\frac {d x^4}{c}\right )+b (5+m) x^4 \, _2F_1\left (\frac {1}{2},\frac {9+m}{4};\frac {13+m}{4};-\frac {d x^4}{c}\right )\right )\right )}{(1+m) (5+m) (9+m) \sqrt {c+d x^4}} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )^{2}}{\sqrt {d \,x^{4}+c}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*(x*e)^m/sqrt(d*x^4 + c), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*(x*e)^m/sqrt(d*x^4 + c), x)

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Sympy [C] Result contains complex when optimal does not.
time = 7.60, size = 185, normalized size = 0.92 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*(x*e)^m/sqrt(d*x^4 + c), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,{\left (b\,x^4+a\right )}^2}{\sqrt {d\,x^4+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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