3.9.47 \(\int \frac {(e x)^m (a+b x^4)^2}{(c+d x^4)^{3/2}} \, dx\) [847]

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

[Out]

1/2*(-a*d+b*c)^2*(e*x)^(1+m)/c/d^2/e/(d*x^4+c)^(1/2)+b^2*(e*x)^(1+m)*(d*x^4+c)^(1/2)/d^2/e/(3+m)-1/2*(2*b^2*c^
2*(1+m)-(3+m)*(2*a^2*d^2-(-a*d+b*c)^2*(1+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)/c/d^2/e/(1+m)/(3+m)/(d*x^4+c)^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 198, normalized size of antiderivative = 1.00, 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 = {474, 470, 372, 371} \begin {gather*} -\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)} \end {gather*}

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 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 474

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]

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*}

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Mathematica [A]
time = 10.10, size = 167, normalized size = 0.84 \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 {3}{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 {3}{2},\frac {5+m}{4};\frac {9+m}{4};-\frac {d x^4}{c}\right )+b (5+m) x^4 \, _2F_1\left (\frac {3}{2},\frac {9+m}{4};\frac {13+m}{4};-\frac {d x^4}{c}\right )\right )\right )}{c (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)/(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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )^{2}}{\left (d \,x^{4}+c \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(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 = 0.00, 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)^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x\right )^{m} \left (a + b x^{4}\right )^{2}}{\left (c + d x^{4}\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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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)^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e\,x\right )}^m\,{\left (b\,x^4+a\right )}^2}{{\left (d\,x^4+c\right )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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