3.1.59 \(\int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx\) [59]

Optimal. Leaf size=55 \[ \frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\sin ^4(c+d x)}{4 a d} \]

[Out]

1/3*cos(d*x+c)^3/a/d-1/5*cos(d*x+c)^5/a/d+1/4*sin(d*x+c)^4/a/d

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Rubi [A]
time = 0.11, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3957, 2914, 2644, 30, 2645, 14} \begin {gather*} \frac {\sin ^4(c+d x)}{4 a d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\cos ^3(c+d x)}{3 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^5/(a + a*Sec[c + d*x]),x]

[Out]

Cos[c + d*x]^3/(3*a*d) - Cos[c + d*x]^5/(5*a*d) + Sin[c + d*x]^4/(4*a*d)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2914

Int[(cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]
), x_Symbol] :> Dist[1/a, Int[Cos[e + f*x]^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[1/(b*d), Int[Cos[e + f*x]
^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (LtQ[0, n, (p + 1)/2] || (LeQ[p, -n] && LtQ[-n, 2*p - 3]) || (GtQ[n, 0] && LeQ[n,
-p]))

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^5(c+d x)}{a+a \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^5(c+d x)}{-a-a \cos (c+d x)} \, dx\\ &=\frac {\int \cos (c+d x) \sin ^3(c+d x) \, dx}{a}-\frac {\int \cos ^2(c+d x) \sin ^3(c+d x) \, dx}{a}\\ &=\frac {\text {Subst}\left (\int x^3 \, dx,x,\sin (c+d x)\right )}{a d}+\frac {\text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\sin ^4(c+d x)}{4 a d}+\frac {\text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {\cos ^3(c+d x)}{3 a d}-\frac {\cos ^5(c+d x)}{5 a d}+\frac {\sin ^4(c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 42, normalized size = 0.76 \begin {gather*} \frac {2 (13+21 \cos (c+d x)+6 \cos (2 (c+d x))) \sin ^6\left (\frac {1}{2} (c+d x)\right )}{15 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^5/(a + a*Sec[c + d*x]),x]

[Out]

(2*(13 + 21*Cos[c + d*x] + 6*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]^6)/(15*a*d)

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Maple [A]
time = 0.10, size = 49, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {1}{5 \sec \left (d x +c \right )^{5}}+\frac {1}{4 \sec \left (d x +c \right )^{4}}+\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{2}}}{d a}\)	\(49\)
default	\(\frac {-\frac {1}{5 \sec \left (d x +c \right )^{5}}+\frac {1}{4 \sec \left (d x +c \right )^{4}}+\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{2}}}{d a}\)	\(49\)
norman	\(\frac {\frac {4}{15 a d}+\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\)	\(83\)
risch	\(\frac {\cos \left (d x +c \right )}{8 a d}-\frac {\cos \left (5 d x +5 c \right )}{80 a d}+\frac {\cos \left (4 d x +4 c \right )}{32 a d}+\frac {\cos \left (3 d x +3 c \right )}{48 a d}-\frac {\cos \left (2 d x +2 c \right )}{8 a d}\)	\(84\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^5/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(-1/5/sec(d*x+c)^5+1/4/sec(d*x+c)^4+1/3/sec(d*x+c)^3-1/2/sec(d*x+c)^2)

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Maxima [A]
time = 0.28, size = 49, normalized size = 0.89 \begin {gather*} -\frac {12 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right )^{2}}{60 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(12*cos(d*x + c)^5 - 15*cos(d*x + c)^4 - 20*cos(d*x + c)^3 + 30*cos(d*x + c)^2)/(a*d)

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Fricas [A]
time = 2.44, size = 49, normalized size = 0.89 \begin {gather*} -\frac {12 \, \cos \left (d x + c\right )^{5} - 15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right )^{2}}{60 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(12*cos(d*x + c)^5 - 15*cos(d*x + c)^4 - 20*cos(d*x + c)^3 + 30*cos(d*x + c)^2)/(a*d)

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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(d*x+c)**5/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]
time = 0.43, size = 97, normalized size = 1.76 \begin {gather*} \frac {4 \, {\left (\frac {5 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {10 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - 1\right )}}{15 \, a d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^5/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

4/15*(5*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 10*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 30*(cos(d*x + c
) - 1)^3/(cos(d*x + c) + 1)^3 - 1)/(a*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)

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Mupad [B]
time = 0.07, size = 58, normalized size = 1.05 \begin {gather*} -\frac {\frac {{\cos \left (c+d\,x\right )}^2}{2\,a}-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}-\frac {{\cos \left (c+d\,x\right )}^4}{4\,a}+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^5/(a + a/cos(c + d*x)),x)

[Out]

-(cos(c + d*x)^2/(2*a) - cos(c + d*x)^3/(3*a) - cos(c + d*x)^4/(4*a) + cos(c + d*x)^5/(5*a))/d

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