3.20 \(\int (a+a \sec (c+d x))^2 \sin ^7(c+d x) \, dx\)

Optimal. Leaf size=131 \[ \frac{a^2 \cos ^7(c+d x)}{7 d}+\frac{a^2 \cos ^6(c+d x)}{3 d}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{3 a^2 \cos ^4(c+d x)}{2 d}+\frac{3 a^2 \cos ^2(c+d x)}{d}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

(2*a^2*Cos[c + d*x])/d + (3*a^2*Cos[c + d*x]^2)/d - (3*a^2*Cos[c + d*x]^4)/(2*d) - (2*a^2*Cos[c + d*x]^5)/(5*d
) + (a^2*Cos[c + d*x]^6)/(3*d) + (a^2*Cos[c + d*x]^7)/(7*d) - (2*a^2*Log[Cos[c + d*x]])/d + (a^2*Sec[c + d*x])
/d

________________________________________________________________________________________

Rubi [A]  time = 0.168269, antiderivative size = 131, 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} \[ \frac{a^2 \cos ^7(c+d x)}{7 d}+\frac{a^2 \cos ^6(c+d x)}{3 d}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}-\frac{3 a^2 \cos ^4(c+d x)}{2 d}+\frac{3 a^2 \cos ^2(c+d x)}{d}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{a^2 \sec (c+d x)}{d}-\frac{2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^7,x]

[Out]

(2*a^2*Cos[c + d*x])/d + (3*a^2*Cos[c + d*x]^2)/d - (3*a^2*Cos[c + d*x]^4)/(2*d) - (2*a^2*Cos[c + d*x]^5)/(5*d
) + (a^2*Cos[c + d*x]^6)/(3*d) + (a^2*Cos[c + d*x]^7)/(7*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 ^7(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^5(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{a^2 (-a-x)^3 (-a+x)^5}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^7 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-a-x)^3 (-a+x)^5}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-2 a^6+\frac{a^8}{x^2}-\frac{2 a^7}{x}+6 a^5 x-6 a^3 x^3+2 a^2 x^4+2 a x^5-x^6\right ) \, dx,x,-a \cos (c+d x)\right )}{a^5 d}\\ &=\frac{2 a^2 \cos (c+d x)}{d}+\frac{3 a^2 \cos ^2(c+d x)}{d}-\frac{3 a^2 \cos ^4(c+d x)}{2 d}-\frac{2 a^2 \cos ^5(c+d x)}{5 d}+\frac{a^2 \cos ^6(c+d x)}{3 d}+\frac{a^2 \cos ^7(c+d x)}{7 d}-\frac{2 a^2 \log (\cos (c+d x))}{d}+\frac{a^2 \sec (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.534327, size = 107, normalized size = 0.82 \[ \frac{a^2 \sec (c+d x) (11760 \cos (2 (c+d x))+5250 \cos (3 (c+d x))-588 \cos (4 (c+d x))-770 \cos (5 (c+d x))-48 \cos (6 (c+d x))+70 \cos (7 (c+d x))+15 \cos (8 (c+d x))-70 \cos (c+d x) (384 \log (\cos (c+d x))+5)+25725)}{13440 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^7,x]

[Out]

(a^2*(25725 + 11760*Cos[2*(c + d*x)] + 5250*Cos[3*(c + d*x)] - 588*Cos[4*(c + d*x)] - 770*Cos[5*(c + d*x)] - 4
8*Cos[6*(c + d*x)] + 70*Cos[7*(c + d*x)] + 15*Cos[8*(c + d*x)] - 70*Cos[c + d*x]*(5 + 384*Log[Cos[c + d*x]]))*
Sec[c + d*x])/(13440*d)

________________________________________________________________________________________

Maple [A]  time = 0.044, size = 168, normalized size = 1.3 \begin{align*}{\frac{96\,{a}^{2}\cos \left ( dx+c \right ) }{35\,d}}+{\frac{6\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{7\,d}}+{\frac{36\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{35\,d}}+{\frac{48\,{a}^{2}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{35\,d}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{3\,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 ) ^{8}}{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)^7,x)

[Out]

96/35*a^2*cos(d*x+c)/d+6/7/d*a^2*cos(d*x+c)*sin(d*x+c)^6+36/35/d*a^2*cos(d*x+c)*sin(d*x+c)^4+48/35/d*a^2*cos(d
*x+c)*sin(d*x+c)^2-1/3/d*a^2*sin(d*x+c)^6-1/2/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)^8/cos(d*x+c)

________________________________________________________________________________________

Maxima [A]  time = 1.02748, size = 144, normalized size = 1.1 \begin{align*} \frac{30 \, a^{2} \cos \left (d x + c\right )^{7} + 70 \, a^{2} \cos \left (d x + c\right )^{6} - 84 \, a^{2} \cos \left (d x + c\right )^{5} - 315 \, a^{2} \cos \left (d x + c\right )^{4} + 630 \, a^{2} \cos \left (d x + c\right )^{2} + 420 \, a^{2} \cos \left (d x + c\right ) - 420 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) + \frac{210 \, a^{2}}{\cos \left (d x + c\right )}}{210 \, 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)^7,x, algorithm="maxima")

[Out]

1/210*(30*a^2*cos(d*x + c)^7 + 70*a^2*cos(d*x + c)^6 - 84*a^2*cos(d*x + c)^5 - 315*a^2*cos(d*x + c)^4 + 630*a^
2*cos(d*x + c)^2 + 420*a^2*cos(d*x + c) - 420*a^2*log(cos(d*x + c)) + 210*a^2/cos(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.89551, size = 342, normalized size = 2.61 \begin{align*} \frac{120 \, a^{2} \cos \left (d x + c\right )^{8} + 280 \, a^{2} \cos \left (d x + c\right )^{7} - 336 \, a^{2} \cos \left (d x + c\right )^{6} - 1260 \, a^{2} \cos \left (d x + c\right )^{5} + 2520 \, a^{2} \cos \left (d x + c\right )^{3} + 1680 \, a^{2} \cos \left (d x + c\right )^{2} - 1680 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 875 \, a^{2} \cos \left (d x + c\right ) + 840 \, a^{2}}{840 \, 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)^7,x, algorithm="fricas")

[Out]

1/840*(120*a^2*cos(d*x + c)^8 + 280*a^2*cos(d*x + c)^7 - 336*a^2*cos(d*x + c)^6 - 1260*a^2*cos(d*x + c)^5 + 25
20*a^2*cos(d*x + c)^3 + 1680*a^2*cos(d*x + c)^2 - 1680*a^2*cos(d*x + c)*log(-cos(d*x + c)) - 875*a^2*cos(d*x +
 c) + 840*a^2)/(d*cos(d*x + c))

________________________________________________________________________________________

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)**7,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.42313, size = 432, normalized size = 3.3 \begin{align*} \frac{420 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 420 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{420 \,{\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{357 \, a^{2} - \frac{3759 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{16737 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{42595 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{43855 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{25389 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{8043 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{1089 \, a^{2}{\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{7}}}{210 \, 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)^7,x, algorithm="giac")

[Out]

1/210*(420*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 420*a^2*log(abs(-(cos(d*x + c) - 1)/(cos
(d*x + c) + 1) - 1)) + 420*(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) + (357*a^2 - 3759*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 16737*a^2*(cos(d*x + c) - 1)^2/(cos
(d*x + c) + 1)^2 - 42595*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 43855*a^2*(cos(d*x + c) - 1)^4/(cos(d
*x + c) + 1)^4 - 25389*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 8043*a^2*(cos(d*x + c) - 1)^6/(cos(d*x
+ c) + 1)^6 - 1089*a^2*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^
7)/d