3.21 \(\int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx\)

Optimal. Leaf size=112 \[ -\frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cos ^4(c+d x)}{2 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos ^2(c+d x)}{d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.157634, antiderivative size = 112, 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} \[ -\frac{a^2 \cos ^5(c+d x)}{5 d}-\frac{a^2 \cos ^4(c+d x)}{2 d}+\frac{a^2 \cos ^3(c+d x)}{3 d}+\frac{2 a^2 \cos ^2(c+d x)}{d}+\frac{a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^5,x]

[Out]

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

Mathematica [A]  time = 0.288114, size = 87, normalized size = 0.78 \[ -\frac{a^2 \sec (c+d x) (-275 \cos (2 (c+d x))-165 \cos (3 (c+d x))-2 \cos (4 (c+d x))+15 \cos (5 (c+d x))+3 \cos (6 (c+d x))+30 \cos (c+d x) (32 \log (\cos (c+d x))-3)-750)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^5,x]

[Out]

-(a^2*(-750 - 275*Cos[2*(c + d*x)] - 165*Cos[3*(c + d*x)] - 2*Cos[4*(c + d*x)] + 15*Cos[5*(c + d*x)] + 3*Cos[6
*(c + d*x)] + 30*Cos[c + d*x]*(-3 + 32*Log[Cos[c + d*x]]))*Sec[c + d*x])/(480*d)

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Maple [A]  time = 0.043, size = 130, normalized size = 1.2 \begin{align*}{\frac{32\,{a}^{2}\cos \left ( dx+c \right ) }{15\,d}}+{\frac{4\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{16\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{15\,d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{2\,d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}-2\,{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{d\cos \left ( dx+c \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^5,x)

[Out]

32/15*a^2*cos(d*x+c)/d+4/5/d*a^2*cos(d*x+c)*sin(d*x+c)^4+16/15/d*a^2*cos(d*x+c)*sin(d*x+c)^2-1/2/d*a^2*sin(d*x
+c)^4-1/d*a^2*sin(d*x+c)^2-2*a^2*ln(cos(d*x+c))/d+1/d*a^2*sin(d*x+c)^6/cos(d*x+c)

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Maxima [A]  time = 1.00812, size = 127, normalized size = 1.13 \begin{align*} -\frac{6 \, a^{2} \cos \left (d x + c\right )^{5} + 15 \, a^{2} \cos \left (d x + c\right )^{4} - 10 \, a^{2} \cos \left (d x + c\right )^{3} - 60 \, a^{2} \cos \left (d x + c\right )^{2} - 30 \, a^{2} \cos \left (d x + c\right ) + 60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac{30 \, a^{2}}{\cos \left (d x + c\right )}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

-1/30*(6*a^2*cos(d*x + c)^5 + 15*a^2*cos(d*x + c)^4 - 10*a^2*cos(d*x + c)^3 - 60*a^2*cos(d*x + c)^2 - 30*a^2*c
os(d*x + c) + 60*a^2*log(cos(d*x + c)) - 30*a^2/cos(d*x + c))/d

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Fricas [A]  time = 1.7875, size = 301, normalized size = 2.69 \begin{align*} -\frac{48 \, a^{2} \cos \left (d x + c\right )^{6} + 120 \, a^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{4} - 480 \, a^{2} \cos \left (d x + c\right )^{3} - 240 \, a^{2} \cos \left (d x + c\right )^{2} + 480 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + 195 \, a^{2} \cos \left (d x + c\right ) - 240 \, a^{2}}{240 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

-1/240*(48*a^2*cos(d*x + c)^6 + 120*a^2*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^4 - 480*a^2*cos(d*x + c)^3 - 240*
a^2*cos(d*x + c)^2 + 480*a^2*cos(d*x + c)*log(-cos(d*x + c)) + 195*a^2*cos(d*x + c) - 240*a^2)/(d*cos(d*x + c)
)

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

[Out]

Timed out

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Giac [B]  time = 1.48864, size = 365, normalized size = 3.26 \begin{align*} \frac{60 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{60 \,{\left (2 \, a^{2} + \frac{a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1} + \frac{69 \, a^{2} - \frac{525 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{1650 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{1610 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{745 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{137 \, a^{2}{\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}}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^5,x, algorithm="giac")

[Out]

1/30*(60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*
x + c) + 1) - 1)) + 60*(2*a^2 + a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x + c) +
1) + 1) + (69*a^2 - 525*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1650*a^2*(cos(d*x + c) - 1)^2/(cos(d*x +
c) + 1)^2 - 1610*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 745*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) +
1)^4 - 137*a^2*(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