3.40 \(\int (a+a \sec (c+d x))^3 \sin ^5(c+d x) \, dx\)

Optimal. Leaf size=134 \[ -\frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{3 a^3 \cos ^4(c+d x)}{4 d}-\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{5 a^3 \cos ^2(c+d x)}{2 d}+\frac{5 a^3 \cos (c+d x)}{d}+\frac{a^3 \sec ^2(c+d x)}{2 d}+\frac{3 a^3 \sec (c+d x)}{d}-\frac{a^3 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.166756, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3872, 2836, 12, 88} \[ -\frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{3 a^3 \cos ^4(c+d x)}{4 d}-\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{5 a^3 \cos ^2(c+d x)}{2 d}+\frac{5 a^3 \cos (c+d x)}{d}+\frac{a^3 \sec ^2(c+d x)}{2 d}+\frac{3 a^3 \sec (c+d x)}{d}-\frac{a^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

(5*a^3*Cos[c + d*x])/d + (5*a^3*Cos[c + d*x]^2)/(2*d) - (a^3*Cos[c + d*x]^3)/(3*d) - (3*a^3*Cos[c + d*x]^4)/(4
*d) - (a^3*Cos[c + d*x]^5)/(5*d) - (a^3*Log[Cos[c + d*x]])/d + (3*a^3*Sec[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(
2*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))^3 \sin ^5(c+d x) \, dx &=-\int (-a-a \cos (c+d x))^3 \sin ^2(c+d x) \tan ^3(c+d x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{a^3 (-a-x)^2 (-a+x)^5}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-a-x)^2 (-a+x)^5}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-5 a^4-\frac{a^7}{x^3}+\frac{3 a^6}{x^2}-\frac{a^5}{x}+5 a^3 x+a^2 x^2-3 a x^3+x^4\right ) \, dx,x,-a \cos (c+d x)\right )}{a^2 d}\\ &=\frac{5 a^3 \cos (c+d x)}{d}+\frac{5 a^3 \cos ^2(c+d x)}{2 d}-\frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{3 a^3 \cos ^4(c+d x)}{4 d}-\frac{a^3 \cos ^5(c+d x)}{5 d}-\frac{a^3 \log (\cos (c+d x))}{d}+\frac{3 a^3 \sec (c+d x)}{d}+\frac{a^3 \sec ^2(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.628204, size = 108, normalized size = 0.81 \[ -\frac{a^3 \sec ^2(c+d x) (-12350 \cos (c+d x)-2074 \cos (3 (c+d x))-330 \cos (4 (c+d x))+82 \cos (5 (c+d x))+45 \cos (6 (c+d x))+6 \cos (7 (c+d x))+960 \log (\cos (c+d x))+15 \cos (2 (c+d x)) (64 \log (\cos (c+d x))+31)-120)}{1920 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^5,x]

[Out]

-(a^3*(-120 - 12350*Cos[c + d*x] - 2074*Cos[3*(c + d*x)] - 330*Cos[4*(c + d*x)] + 82*Cos[5*(c + d*x)] + 45*Cos
[6*(c + d*x)] + 6*Cos[7*(c + d*x)] + 960*Log[Cos[c + d*x]] + 15*Cos[2*(c + d*x)]*(31 + 64*Log[Cos[c + d*x]]))*
Sec[c + d*x]^2)/(1920*d)

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Maple [A]  time = 0.047, size = 155, normalized size = 1.2 \begin{align*}{\frac{112\,{a}^{3}\cos \left ( dx+c \right ) }{15\,d}}+{\frac{14\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{56\,{a}^{3}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{{a}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x)

[Out]

112/15*a^3*cos(d*x+c)/d+14/5/d*a^3*cos(d*x+c)*sin(d*x+c)^4+56/15/d*a^3*cos(d*x+c)*sin(d*x+c)^2-1/4/d*a^3*sin(d
*x+c)^4-1/2/d*a^3*sin(d*x+c)^2-a^3*ln(cos(d*x+c))/d+3/d*a^3*sin(d*x+c)^6/cos(d*x+c)+1/2/d*a^3*sin(d*x+c)^6/cos
(d*x+c)^2

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Maxima [A]  time = 1.01187, size = 143, normalized size = 1.07 \begin{align*} -\frac{12 \, a^{3} \cos \left (d x + c\right )^{5} + 45 \, a^{3} \cos \left (d x + c\right )^{4} + 20 \, a^{3} \cos \left (d x + c\right )^{3} - 150 \, a^{3} \cos \left (d x + c\right )^{2} - 300 \, a^{3} \cos \left (d x + c\right ) + 60 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) - \frac{30 \,{\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/60*(12*a^3*cos(d*x + c)^5 + 45*a^3*cos(d*x + c)^4 + 20*a^3*cos(d*x + c)^3 - 150*a^3*cos(d*x + c)^2 - 300*a^
3*cos(d*x + c) + 60*a^3*log(cos(d*x + c)) - 30*(6*a^3*cos(d*x + c) + a^3)/cos(d*x + c)^2)/d

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Fricas [A]  time = 1.8304, size = 346, normalized size = 2.58 \begin{align*} -\frac{96 \, a^{3} \cos \left (d x + c\right )^{7} + 360 \, a^{3} \cos \left (d x + c\right )^{6} + 160 \, a^{3} \cos \left (d x + c\right )^{5} - 1200 \, a^{3} \cos \left (d x + c\right )^{4} - 2400 \, a^{3} \cos \left (d x + c\right )^{3} + 480 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) + 465 \, a^{3} \cos \left (d x + c\right )^{2} - 1440 \, a^{3} \cos \left (d x + c\right ) - 240 \, a^{3}}{480 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/480*(96*a^3*cos(d*x + c)^7 + 360*a^3*cos(d*x + c)^6 + 160*a^3*cos(d*x + c)^5 - 1200*a^3*cos(d*x + c)^4 - 24
00*a^3*cos(d*x + c)^3 + 480*a^3*cos(d*x + c)^2*log(-cos(d*x + c)) + 465*a^3*cos(d*x + c)^2 - 1440*a^3*cos(d*x
+ c) - 240*a^3)/(d*cos(d*x + c)^2)

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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))**3*sin(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.29833, size = 401, normalized size = 2.99 \begin{align*} \frac{60 \, a^{3} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{3} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{30 \,{\left (15 \, a^{3} + \frac{14 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{3 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}} - \frac{399 \, a^{3} - \frac{1395 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{390 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{650 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{565 \, a^{3}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{137 \, a^{3}{\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))^3*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/60*(60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^3*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + 30*(15*a^3 + 14*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^3*(cos(d*x + c) - 1)^2/(co
s(d*x + c) + 1)^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2 - (399*a^3 - 1395*a^3*(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) + 390*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 650*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c
) + 1)^3 - 565*a^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 137*a^3*(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