3.133 \(\int (a+b \sec ^2(e+f x))^p \sin ^5(e+f x) \, dx\)

Optimal. Leaf size=182 \[ -\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )}{15 a^2 f}+\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a f} \]

[Out]

1/15*(10*a+b*(3-2*p))*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(1+p)/a^2/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1+p)/a/
f-1/15*(15*a^2+10*a*b*(1-2*p)+b^2*(4*p^2-8*p+3))*cos(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*sec(f*x+e)^2/a)*(a+b
*sec(f*x+e)^2)^p/a^2/f/((1+b*sec(f*x+e)^2/a)^p)

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Rubi [A]  time = 0.19, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4134, 462, 453, 365, 364} \[ -\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )}{15 a^2 f}+\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^5,x]

[Out]

((10*a + b*(3 - 2*p))*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(1 + p))/(15*a^2*f) - (Cos[e + f*x]^5*(a + b*Sec[e
+ f*x]^2)^(1 + p))/(5*a*f) - ((15*a^2 + 10*a*b*(1 - 2*p) + b^2*(3 - 8*p + 4*p^2))*Cos[e + f*x]*Hypergeometric
2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)]*(a + b*Sec[e + f*x]^2)^p)/(15*a^2*f*(1 + (b*Sec[e + f*x]^2)/a)^p)

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 453

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

Rule 462

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

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

\begin {align*} \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\operatorname {Subst}\left (\int \frac {\left (-10 a-b (3-2 p)+5 a x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\left (\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \cos (e+f x) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f}\\ \end {align*}

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Mathematica [A]  time = 7.77, size = 253, normalized size = 1.39 \[ \frac {2 \sin ^4(e+f x) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (4 \left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a}\right )+(a \cos (2 (e+f x))+a+2 b) (3 a \cos (2 (e+f x))-17 a+4 b p-6 b) \left (\frac {a+b \tan ^2(e+f x)+b}{a}\right )^p\right )}{15 a^2 f \left (4 \cos (2 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)+b}{a}\right )^p-2^{-p} \left (2^p \cos (4 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)+b}{a}\right )^p+3 \left (\frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)}{a}\right )^p\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^5,x]

[Out]

(2*Cos[e + f*x]*(a + b*Sec[e + f*x]^2)^p*Sin[e + f*x]^4*(4*(15*a^2 + 10*a*b*(1 - 2*p) + b^2*(3 - 8*p + 4*p^2))
*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/a)] + (a + 2*b + a*Cos[2*(e + f*x)])*(-17*a - 6*b + 4*b
*p + 3*a*Cos[2*(e + f*x)])*((a + b + b*Tan[e + f*x]^2)/a)^p))/(15*a^2*f*(4*Cos[2*(e + f*x)]*((a + b + b*Tan[e
+ f*x]^2)/a)^p - (3*(((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2)/a)^p + 2^p*Cos[4*(e + f*x)]*((a + b + b*T
an[e + f*x]^2)/a)^p)/2^p))

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )} {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="fricas")

[Out]

integral((cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*(b*sec(f*x + e)^2 + a)^p*sin(f*x + e), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)

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maple [F]  time = 4.77, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p} \left (\sin ^{5}\left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x)

[Out]

int((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)^2)^p*sin(f*x+e)^5,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sin(f*x + e)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^p,x)

[Out]

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^p, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(f*x+e)**2)**p*sin(f*x+e)**5,x)

[Out]

Timed out

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