3.2.54 \(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [154]

Optimal. Leaf size=208 \[ \frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a-b}\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{15 (a-b)^2 f} \]

[Out]

1/15*(-2*b*p+10*a-7*b)*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(1+p)/(a-b)^2/f-1/5*cos(f*x+e)^5*(a-b+b*sec(f*x+e)^2)
^(1+p)/(a-b)/f-1/15*(15*a^2-20*a*b*(1+p)+4*b^2*(p^2+3*p+2))*cos(f*x+e)*hypergeom([-1/2, -p],[1/2],-b*sec(f*x+e
)^2/(a-b))*(a-b+b*sec(f*x+e)^2)^p/(a-b)^2/f/((1+b*sec(f*x+e)^2/(a-b))^p)

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Rubi [A]
time = 0.15, antiderivative size = 208, 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 = {3745, 473, 464, 372, 371} \begin {gather*} -\frac {\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac {(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((10*a - 7*b - 2*b*p)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(1 + p))/(15*(a - b)^2*f) - (Cos[e + f*x]^5*(a
- b + b*Sec[e + f*x]^2)^(1 + p))/(5*(a - b)*f) - ((15*a^2 - 20*a*b*(1 + p) + 4*b^2*(2 + 3*p + p^2))*Cos[e + f
*x]*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))]*(a - b + b*Sec[e + f*x]^2)^p)/(15*(a - b)^
2*f*(1 + (b*Sec[e + f*x]^2)/(a - b))^p)

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 464

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 473

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 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\text {Subst}\left (\int \frac {\left (-10 a+b (7+2 p)+5 (a-b) x^2\right ) \left (a-b+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \frac {\left (a-b+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=\frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a-b}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=\frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+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-b}\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{15 (a-b)^2 f}\\ \end {align*}

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Mathematica [A]
time = 5.02, size = 283, normalized size = 1.36 \begin {gather*} -\frac {2^{3+p} \cos (e+f x) \sin ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \, _2F_1\left (-\frac {1}{2},-p;\frac {1}{2};-\frac {b \sec ^2(e+f x)}{a-b}\right )+\frac {1}{4} (a+b+(a-b) \cos (2 (e+f x))) (-17 a+b (11+4 p)+3 (a-b) \cos (2 (e+f x))) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p\right )}{15 (a-b)^2 f \left (3 \left (\frac {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}{a-b}\right )^p-2^{2+p} \cos (2 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p+2^p \cos (4 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/15*(2^(3 + p)*Cos[e + f*x]*Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^p*((15*a^2 - 20*a*b*(1 + p) + 4*b^2*(2 + 3
*p + p^2))*Hypergeometric2F1[-1/2, -p, 1/2, -((b*Sec[e + f*x]^2)/(a - b))] + ((a + b + (a - b)*Cos[2*(e + f*x)
])*(-17*a + b*(11 + 4*p) + 3*(a - b)*Cos[2*(e + f*x)])*((a + b*Tan[e + f*x]^2)/(a - b))^p)/4))/((a - b)^2*f*(3
*(((a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2)/(a - b))^p - 2^(2 + p)*Cos[2*(e + f*x)]*((a + b*Tan[e +
f*x]^2)/(a - b))^p + 2^p*Cos[4*(e + f*x)]*((a + b*Tan[e + f*x]^2)/(a - b))^p))

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Maple [F]
time = 0.80, size = 0, normalized size = 0.00 \[\int \left (\sin ^{5}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integrate((b*tan(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 = -0.00 \begin {gather*} \int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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