3.176 \(\int (c+e x^2)^3 (a+b x^4)^p \, dx\)

Optimal. Leaf size=204 \[ c^3 x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )-\frac {e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (a e^2-b c^2 (4 p+7)\right ) \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )}{b (4 p+7)}+\frac {3}{5} c e^2 x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+\frac {e^3 x^3 \left (a+b x^4\right )^{p+1}}{b (4 p+7)} \]

[Out]

e^3*x^3*(b*x^4+a)^(1+p)/b/(7+4*p)+c^3*x*(b*x^4+a)^p*hypergeom([1/4, -p],[5/4],-b*x^4/a)/((1+b*x^4/a)^p)-e*(a*e
^2-b*c^2*(7+4*p))*x^3*(b*x^4+a)^p*hypergeom([3/4, -p],[7/4],-b*x^4/a)/b/(7+4*p)/((1+b*x^4/a)^p)+3/5*c*e^2*x^5*
(b*x^4+a)^p*hypergeom([5/4, -p],[9/4],-b*x^4/a)/((1+b*x^4/a)^p)

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Rubi [A]  time = 0.23, antiderivative size = 196, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {1207, 1893, 246, 245, 365, 364} \[ e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (c^2-\frac {a e^2}{4 b p+7 b}\right ) \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )+c^3 x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+\frac {3}{5} c e^2 x^5 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+\frac {e^3 x^3 \left (a+b x^4\right )^{p+1}}{b (4 p+7)} \]

Antiderivative was successfully verified.

[In]

Int[(c + e*x^2)^3*(a + b*x^4)^p,x]

[Out]

(e^3*x^3*(a + b*x^4)^(1 + p))/(b*(7 + 4*p)) + (c^3*x*(a + b*x^4)^p*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a
)])/(1 + (b*x^4)/a)^p + (e*(c^2 - (a*e^2)/(7*b + 4*b*p))*x^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p, 7/4, -((
b*x^4)/a)])/(1 + (b*x^4)/a)^p + (3*c*e^2*x^5*(a + b*x^4)^p*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)])/(5*(
1 + (b*x^4)/a)^p)

Rule 245

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || 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])

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 1207

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e^q*x^(2*q - 3)*(a + c*x^4)^(p +
1))/(c*(4*p + 2*q + 1)), x] + Dist[1/(c*(4*p + 2*q + 1)), Int[(a + c*x^4)^p*ExpandToSum[c*(4*p + 2*q + 1)*(d
+ e*x^2)^q - a*(2*q - 3)*e^q*x^(2*q - 4) - c*(4*p + 2*q + 1)*e^q*x^(2*q), x], x], x] /; FreeQ[{a, c, d, e, p},
x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[q, 1]

Rule 1893

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n])

Rubi steps

\begin {align*} \int \left (c+e x^2\right )^3 \left (a+b x^4\right )^p \, dx &=\frac {e^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (a+b x^4\right )^p \left (b c^3 (7+4 p)-3 e \left (a e^2-b c^2 (7+4 p)\right ) x^2+3 b c e^2 (7+4 p) x^4\right ) \, dx}{b (7+4 p)}\\ &=\frac {e^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+\frac {\int \left (b c^3 (7+4 p) \left (a+b x^4\right )^p+3 e \left (-a e^2+b c^2 (7+4 p)\right ) x^2 \left (a+b x^4\right )^p+3 b c e^2 (7+4 p) x^4 \left (a+b x^4\right )^p\right ) \, dx}{b (7+4 p)}\\ &=\frac {e^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+c^3 \int \left (a+b x^4\right )^p \, dx+\left (3 c e^2\right ) \int x^4 \left (a+b x^4\right )^p \, dx+\left (3 e \left (c^2-\frac {a e^2}{7 b+4 b p}\right )\right ) \int x^2 \left (a+b x^4\right )^p \, dx\\ &=\frac {e^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+\left (c^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^4}{a}\right )^p \, dx+\left (3 c e^2 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int x^4 \left (1+\frac {b x^4}{a}\right )^p \, dx+\left (3 e \left (c^2-\frac {a e^2}{7 b+4 b p}\right ) \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac {b x^4}{a}\right )^p \, dx\\ &=\frac {e^3 x^3 \left (a+b x^4\right )^{1+p}}{b (7+4 p)}+c^3 x \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+e \left (c^2-\frac {a e^2}{7 b+4 b p}\right ) x^3 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )+\frac {3}{5} c e^2 x^5 \left (a+b x^4\right )^p \left (1+\frac {b x^4}{a}\right )^{-p} \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 136, normalized size = 0.67 \[ \frac {1}{35} x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \left (35 c^3 \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-\frac {b x^4}{a}\right )+e x^2 \left (35 c^2 \, _2F_1\left (\frac {3}{4},-p;\frac {7}{4};-\frac {b x^4}{a}\right )+e x^2 \left (21 c \, _2F_1\left (\frac {5}{4},-p;\frac {9}{4};-\frac {b x^4}{a}\right )+5 e x^2 \, _2F_1\left (\frac {7}{4},-p;\frac {11}{4};-\frac {b x^4}{a}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a + b*x^4)^p*(35*c^3*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + e*x^2*(35*c^2*Hypergeometric2F1[3/4,
-p, 7/4, -((b*x^4)/a)] + e*x^2*(21*c*Hypergeometric2F1[5/4, -p, 9/4, -((b*x^4)/a)] + 5*e*x^2*Hypergeometric2F1
[7/4, -p, 11/4, -((b*x^4)/a)]))))/(35*(1 + (b*x^4)/a)^p)

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{6} + 3 \, c e^{2} x^{4} + 3 \, c^{2} e x^{2} + c^{3}\right )} {\left (b x^{4} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^3*x^6 + 3*c*e^2*x^4 + 3*c^2*e*x^2 + c^3)*(b*x^4 + a)^p, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + c\right )}^{3} {\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + c)^3*(b*x^4 + a)^p, x)

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (e \,x^{2}+c \right )^{3} \left (b \,x^{4}+a \right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+c)^3*(b*x^4+a)^p,x)

[Out]

int((e*x^2+c)^3*(b*x^4+a)^p,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x^{2} + c\right )}^{3} {\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + c)^3*(b*x^4 + a)^p, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (b\,x^4+a\right )}^p\,{\left (e\,x^2+c\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^p*(c + e*x^2)^3,x)

[Out]

int((a + b*x^4)^p*(c + e*x^2)^3, x)

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sympy [C]  time = 139.10, size = 167, normalized size = 0.82 \[ \frac {a^{p} c^{3} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 a^{p} c^{2} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 a^{p} c e^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, - p \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {a^{p} e^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{4}, - p \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+c)**3*(b*x**4+a)**p,x)

[Out]

a**p*c**3*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4)) + 3*a**p*c**2*e*x**3*
gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/4)) + 3*a**p*c*e**2*x**5*gamma(5/4)*h
yper((5/4, -p), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(9/4)) + a**p*e**3*x**7*gamma(7/4)*hyper((7/4, -p),
(11/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(11/4))

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