3.1 \(\int (a+a \sec (c+d x)) \sin ^9(c+d x) \, dx\)

Optimal. Leaf size=152 \[ -\frac{a \cos ^9(c+d x)}{9 d}-\frac{a \cos ^8(c+d x)}{8 d}+\frac{4 a \cos ^7(c+d x)}{7 d}+\frac{2 a \cos ^6(c+d x)}{3 d}-\frac{6 a \cos ^5(c+d x)}{5 d}-\frac{3 a \cos ^4(c+d x)}{2 d}+\frac{4 a \cos ^3(c+d x)}{3 d}+\frac{2 a \cos ^2(c+d x)}{d}-\frac{a \cos (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d} \]

[Out]

-((a*Cos[c + d*x])/d) + (2*a*Cos[c + d*x]^2)/d + (4*a*Cos[c + d*x]^3)/(3*d) - (3*a*Cos[c + d*x]^4)/(2*d) - (6*
a*Cos[c + d*x]^5)/(5*d) + (2*a*Cos[c + d*x]^6)/(3*d) + (4*a*Cos[c + d*x]^7)/(7*d) - (a*Cos[c + d*x]^8)/(8*d) -
(a*Cos[c + d*x]^9)/(9*d) - (a*Log[Cos[c + d*x]])/d

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Rubi [A]  time = 0.107119, antiderivative size = 152, 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} \[ -\frac{a \cos ^9(c+d x)}{9 d}-\frac{a \cos ^8(c+d x)}{8 d}+\frac{4 a \cos ^7(c+d x)}{7 d}+\frac{2 a \cos ^6(c+d x)}{3 d}-\frac{6 a \cos ^5(c+d x)}{5 d}-\frac{3 a \cos ^4(c+d x)}{2 d}+\frac{4 a \cos ^3(c+d x)}{3 d}+\frac{2 a \cos ^2(c+d x)}{d}-\frac{a \cos (c+d x)}{d}-\frac{a \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])*Sin[c + d*x]^9,x]

[Out]

-((a*Cos[c + d*x])/d) + (2*a*Cos[c + d*x]^2)/d + (4*a*Cos[c + d*x]^3)/(3*d) - (3*a*Cos[c + d*x]^4)/(2*d) - (6*
a*Cos[c + d*x]^5)/(5*d) + (2*a*Cos[c + d*x]^6)/(3*d) + (4*a*Cos[c + d*x]^7)/(7*d) - (a*Cos[c + d*x]^8)/(8*d) -
(a*Cos[c + d*x]^9)/(9*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 ^9(c+d x) \, dx &=-\int (-a-a \cos (c+d x)) \sin ^8(c+d x) \tan (c+d x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{a (-a-x)^4 (-a+x)^5}{x} \, dx,x,-a \cos (c+d x)\right )}{a^9 d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(-a-x)^4 (-a+x)^5}{x} \, dx,x,-a \cos (c+d x)\right )}{a^8 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^8-\frac{a^9}{x}+4 a^7 x-4 a^6 x^2-6 a^5 x^3+6 a^4 x^4+4 a^3 x^5-4 a^2 x^6-a x^7+x^8\right ) \, dx,x,-a \cos (c+d x)\right )}{a^8 d}\\ &=-\frac{a \cos (c+d x)}{d}+\frac{2 a \cos ^2(c+d x)}{d}+\frac{4 a \cos ^3(c+d x)}{3 d}-\frac{3 a \cos ^4(c+d x)}{2 d}-\frac{6 a \cos ^5(c+d x)}{5 d}+\frac{2 a \cos ^6(c+d x)}{3 d}+\frac{4 a \cos ^7(c+d x)}{7 d}-\frac{a \cos ^8(c+d x)}{8 d}-\frac{a \cos ^9(c+d x)}{9 d}-\frac{a \log (\cos (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.210393, size = 106, normalized size = 0.7 \[ -\frac{a \left (10080 \cos ^8(c+d x)-53760 \cos ^6(c+d x)+120960 \cos ^4(c+d x)-161280 \cos ^2(c+d x)+39690 \cos (c+d x)-8820 \cos (3 (c+d x))+2268 \cos (5 (c+d x))-405 \cos (7 (c+d x))+35 \cos (9 (c+d x))+80640 \log (\cos (c+d x))\right )}{80640 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])*Sin[c + d*x]^9,x]

[Out]

-(a*(39690*Cos[c + d*x] - 161280*Cos[c + d*x]^2 + 120960*Cos[c + d*x]^4 - 53760*Cos[c + d*x]^6 + 10080*Cos[c +
d*x]^8 - 8820*Cos[3*(c + d*x)] + 2268*Cos[5*(c + d*x)] - 405*Cos[7*(c + d*x)] + 35*Cos[9*(c + d*x)] + 80640*L
og[Cos[c + d*x]]))/(80640*d)

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Maple [A]  time = 0.092, size = 163, normalized size = 1.1 \begin{align*} -{\frac{128\,a\cos \left ( dx+c \right ) }{315\,d}}-{\frac{\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{8}a}{9\,d}}-{\frac{8\,a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{63\,d}}-{\frac{16\,a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{105\,d}}-{\frac{64\,a\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{315\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{8\,d}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{6\,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)^9,x)

[Out]

-128/315*a*cos(d*x+c)/d-1/9/d*cos(d*x+c)*sin(d*x+c)^8*a-8/63/d*a*cos(d*x+c)*sin(d*x+c)^6-16/105/d*a*cos(d*x+c)
*sin(d*x+c)^4-64/315/d*a*cos(d*x+c)*sin(d*x+c)^2-1/8/d*a*sin(d*x+c)^8-1/6/d*a*sin(d*x+c)^6-1/4/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 = 1.10328, size = 153, normalized size = 1.01 \begin{align*} -\frac{280 \, a \cos \left (d x + c\right )^{9} + 315 \, a \cos \left (d x + c\right )^{8} - 1440 \, a \cos \left (d x + c\right )^{7} - 1680 \, a \cos \left (d x + c\right )^{6} + 3024 \, a \cos \left (d x + c\right )^{5} + 3780 \, a \cos \left (d x + c\right )^{4} - 3360 \, a \cos \left (d x + c\right )^{3} - 5040 \, a \cos \left (d x + c\right )^{2} + 2520 \, a \cos \left (d x + c\right ) + 2520 \, a \log \left (\cos \left (d x + c\right )\right )}{2520 \, 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)^9,x, algorithm="maxima")

[Out]

-1/2520*(280*a*cos(d*x + c)^9 + 315*a*cos(d*x + c)^8 - 1440*a*cos(d*x + c)^7 - 1680*a*cos(d*x + c)^6 + 3024*a*
cos(d*x + c)^5 + 3780*a*cos(d*x + c)^4 - 3360*a*cos(d*x + c)^3 - 5040*a*cos(d*x + c)^2 + 2520*a*cos(d*x + c) +
2520*a*log(cos(d*x + c)))/d

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Fricas [A]  time = 1.89077, size = 339, normalized size = 2.23 \begin{align*} -\frac{280 \, a \cos \left (d x + c\right )^{9} + 315 \, a \cos \left (d x + c\right )^{8} - 1440 \, a \cos \left (d x + c\right )^{7} - 1680 \, a \cos \left (d x + c\right )^{6} + 3024 \, a \cos \left (d x + c\right )^{5} + 3780 \, a \cos \left (d x + c\right )^{4} - 3360 \, a \cos \left (d x + c\right )^{3} - 5040 \, a \cos \left (d x + c\right )^{2} + 2520 \, a \cos \left (d x + c\right ) + 2520 \, a \log \left (-\cos \left (d x + c\right )\right )}{2520 \, 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)^9,x, algorithm="fricas")

[Out]

-1/2520*(280*a*cos(d*x + c)^9 + 315*a*cos(d*x + c)^8 - 1440*a*cos(d*x + c)^7 - 1680*a*cos(d*x + c)^6 + 3024*a*
cos(d*x + c)^5 + 3780*a*cos(d*x + c)^4 - 3360*a*cos(d*x + c)^3 - 5040*a*cos(d*x + c)^2 + 2520*a*cos(d*x + c) +
2520*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)**9,x)

[Out]

Timed out

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Giac [B]  time = 1.49467, size = 396, normalized size = 2.61 \begin{align*} \frac{2520 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2520 \, a \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac{9177 \, a - \frac{87633 \, a{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac{375732 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{953988 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{1594782 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{1336734 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{781956 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{302004 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{69201 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{7129 \, a{\left (\cos \left (d x + c\right ) - 1\right )}^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{9}}}{2520 \, 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)^9,x, algorithm="giac")

[Out]

1/2520*(2520*a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 2520*a*log(abs(-(cos(d*x + c) - 1)/(cos(
d*x + c) + 1) - 1)) + (9177*a - 87633*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 375732*a*(cos(d*x + c) - 1)^2/
(cos(d*x + c) + 1)^2 - 953988*a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 1594782*a*(cos(d*x + c) - 1)^4/(co
s(d*x + c) + 1)^4 - 1336734*a*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 781956*a*(cos(d*x + c) - 1)^6/(cos(d
*x + c) + 1)^6 - 302004*a*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 69201*a*(cos(d*x + c) - 1)^8/(cos(d*x +
c) + 1)^8 - 7129*a*(cos(d*x + c) - 1)^9/(cos(d*x + c) + 1)^9)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^9)/d