3.3 \(\int (a+a \sec (c+d x)) \sin ^5(c+d x) \, dx\)

Optimal. Leaf size=87 \[ -\frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^4(c+d x)}{4 d}+\frac{2 a \cos ^3(c+d x)}{3 d}+\frac{a \cos ^2(c+d x)}{d}-\frac{a \cos (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d} \]

[Out]

-((a*Cos[c + d*x])/d) + (a*Cos[c + d*x]^2)/d + (2*a*Cos[c + d*x]^3)/(3*d) - (a*Cos[c + d*x]^4)/(4*d) - (a*Cos[
c + d*x]^5)/(5*d) - (a*Log[Cos[c + d*x]])/d

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Rubi [A]  time = 0.0869581, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3872, 2836, 12, 88} \[ -\frac{a \cos ^5(c+d x)}{5 d}-\frac{a \cos ^4(c+d x)}{4 d}+\frac{2 a \cos ^3(c+d x)}{3 d}+\frac{a \cos ^2(c+d x)}{d}-\frac{a \cos (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[c + d*x])/d) + (a*Cos[c + d*x]^2)/d + (2*a*Cos[c + d*x]^3)/(3*d) - (a*Cos[c + d*x]^4)/(4*d) - (a*Cos[
c + d*x]^5)/(5*d) - (a*Log[Cos[c + d*x]])/d

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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]

Rule 2836

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.0841183, size = 83, normalized size = 0.95 \[ -\frac{5 a \cos (c+d x)}{8 d}+\frac{5 a \cos (3 (c+d x))}{48 d}-\frac{a \cos (5 (c+d x))}{80 d}-\frac{a \left (\frac{1}{4} \cos ^4(c+d x)-\cos ^2(c+d x)+\log (\cos (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*Cos[c + d*x])/(8*d) + (5*a*Cos[3*(c + d*x)])/(48*d) - (a*Cos[5*(c + d*x)])/(80*d) - (a*(-Cos[c + d*x]^2
+ Cos[c + d*x]^4/4 + Log[Cos[c + d*x]]))/d

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Maple [A]  time = 0.086, size = 95, normalized size = 1.1 \begin{align*} -{\frac{8\,a\cos \left ( dx+c \right ) }{15\,d}}-{\frac{a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}-{\frac{4\,a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*a*cos(d*x+c)/d-1/5/d*a*cos(d*x+c)*sin(d*x+c)^4-4/15/d*a*cos(d*x+c)*sin(d*x+c)^2-1/4/d*a*sin(d*x+c)^4-1/2
/d*a*sin(d*x+c)^2-a*ln(cos(d*x+c))/d

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Maxima [A]  time = 1.10881, size = 93, normalized size = 1.07 \begin{align*} -\frac{12 \, a \cos \left (d x + c\right )^{5} + 15 \, a \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, a \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, a \log \left (\cos \left (d x + c\right )\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*a*cos(d*x + c)^5 + 15*a*cos(d*x + c)^4 - 40*a*cos(d*x + c)^3 - 60*a*cos(d*x + c)^2 + 60*a*cos(d*x +
c) + 60*a*log(cos(d*x + c)))/d

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Fricas [A]  time = 1.79931, size = 193, normalized size = 2.22 \begin{align*} -\frac{12 \, a \cos \left (d x + c\right )^{5} + 15 \, a \cos \left (d x + c\right )^{4} - 40 \, a \cos \left (d x + c\right )^{3} - 60 \, a \cos \left (d x + c\right )^{2} + 60 \, a \cos \left (d x + c\right ) + 60 \, a \log \left (-\cos \left (d x + c\right )\right )}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*a*cos(d*x + c)^5 + 15*a*cos(d*x + c)^4 - 40*a*cos(d*x + c)^3 - 60*a*cos(d*x + c)^2 + 60*a*cos(d*x +
c) + 60*a*log(-cos(d*x + c)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*sin(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.42544, size = 271, normalized size = 3.11 \begin{align*} \frac{60 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{201 \, a - \frac{1125 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{2610 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{1970 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{805 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{137 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x +
c) + 1) - 1)) + (201*a - 1125*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2610*a*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2 - 1970*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 805*a*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^
4 - 137*a*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5)/d