3.19 \(\int (a+a \sec (c+d x))^2 \sin ^9(c+d x) \, dx\)

Optimal. Leaf size=183 \[ -\frac {a^2 \cos ^9(c+d x)}{9 d}-\frac {a^2 \cos ^8(c+d x)}{4 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}+\frac {4 a^2 \cos ^6(c+d x)}{3 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {3 a^2 \cos ^4(c+d x)}{d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {4 a^2 \cos ^2(c+d x)}{d}+\frac {3 a^2 \cos (c+d x)}{d}+\frac {a^2 \sec (c+d x)}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d} \]

[Out]

3*a^2*cos(d*x+c)/d+4*a^2*cos(d*x+c)^2/d-2/3*a^2*cos(d*x+c)^3/d-3*a^2*cos(d*x+c)^4/d-2/5*a^2*cos(d*x+c)^5/d+4/3
*a^2*cos(d*x+c)^6/d+3/7*a^2*cos(d*x+c)^7/d-1/4*a^2*cos(d*x+c)^8/d-1/9*a^2*cos(d*x+c)^9/d-2*a^2*ln(cos(d*x+c))/
d+a^2*sec(d*x+c)/d

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Rubi [A]  time = 0.19, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2836, 12, 88} \[ -\frac {a^2 \cos ^9(c+d x)}{9 d}-\frac {a^2 \cos ^8(c+d x)}{4 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}+\frac {4 a^2 \cos ^6(c+d x)}{3 d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}-\frac {3 a^2 \cos ^4(c+d x)}{d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}+\frac {4 a^2 \cos ^2(c+d x)}{d}+\frac {3 a^2 \cos (c+d x)}{d}+\frac {a^2 \sec (c+d x)}{d}-\frac {2 a^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^9,x]

[Out]

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

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]))

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 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]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^2 \sin ^9(c+d x) \, dx &=\int (-a-a \cos (c+d x))^2 \sin ^7(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {a^2 (-a-x)^4 (-a+x)^6}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^9 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(-a-x)^4 (-a+x)^6}{x^2} \, dx,x,-a \cos (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a^8+\frac {a^{10}}{x^2}-\frac {2 a^9}{x}+8 a^7 x+2 a^6 x^2-12 a^5 x^3+2 a^4 x^4+8 a^3 x^5-3 a^2 x^6-2 a x^7+x^8\right ) \, dx,x,-a \cos (c+d x)\right )}{a^7 d}\\ &=\frac {3 a^2 \cos (c+d x)}{d}+\frac {4 a^2 \cos ^2(c+d x)}{d}-\frac {2 a^2 \cos ^3(c+d x)}{3 d}-\frac {3 a^2 \cos ^4(c+d x)}{d}-\frac {2 a^2 \cos ^5(c+d x)}{5 d}+\frac {4 a^2 \cos ^6(c+d x)}{3 d}+\frac {3 a^2 \cos ^7(c+d x)}{7 d}-\frac {a^2 \cos ^8(c+d x)}{4 d}-\frac {a^2 \cos ^9(c+d x)}{9 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {a^2 \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.94, size = 127, normalized size = 0.69 \[ -\frac {a^2 \sec (c+d x) (-361620 \cos (2 (c+d x))-134820 \cos (3 (c+d x))+29232 \cos (4 (c+d x))+24780 \cos (5 (c+d x))-1458 \cos (6 (c+d x))-3885 \cos (7 (c+d x))-380 \cos (8 (c+d x))+315 \cos (9 (c+d x))+70 \cos (10 (c+d x))+210 \cos (c+d x) (3072 \log (\cos (c+d x))[phone])}{322560 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sin[c + d*x]^9,x]

[Out]

-1/322560*(a^2*(-714420 - 361620*Cos[2*(c + d*x)] - 134820*Cos[3*(c + d*x)] + 29232*Cos[4*(c + d*x)] + 24780*C
os[5*(c + d*x)] - 1458*Cos[6*(c + d*x)] - 3885*Cos[7*(c + d*x)] - 380*Cos[8*(c + d*x)] + 315*Cos[9*(c + d*x)]
+ 70*Cos[10*(c + d*x)] + 210*Cos[c + d*x]*(205 + 3072*Log[Cos[c + d*x]]))*Sec[c + d*x])/d

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fricas [A]  time = 0.57, size = 167, normalized size = 0.91 \[ -\frac {17920 \, a^{2} \cos \left (d x + c\right )^{10} + 40320 \, a^{2} \cos \left (d x + c\right )^{9} - 69120 \, a^{2} \cos \left (d x + c\right )^{8} - 215040 \, a^{2} \cos \left (d x + c\right )^{7} + 64512 \, a^{2} \cos \left (d x + c\right )^{6} + 483840 \, a^{2} \cos \left (d x + c\right )^{5} + 107520 \, a^{2} \cos \left (d x + c\right )^{4} - 645120 \, a^{2} \cos \left (d x + c\right )^{3} - 483840 \, a^{2} \cos \left (d x + c\right )^{2} + 322560 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + 197295 \, a^{2} \cos \left (d x + c\right ) - 161280 \, a^{2}}{161280 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^9,x, algorithm="fricas")

[Out]

-1/161280*(17920*a^2*cos(d*x + c)^10 + 40320*a^2*cos(d*x + c)^9 - 69120*a^2*cos(d*x + c)^8 - 215040*a^2*cos(d*
x + c)^7 + 64512*a^2*cos(d*x + c)^6 + 483840*a^2*cos(d*x + c)^5 + 107520*a^2*cos(d*x + c)^4 - 645120*a^2*cos(d
*x + c)^3 - 483840*a^2*cos(d*x + c)^2 + 322560*a^2*cos(d*x + c)*log(-cos(d*x + c)) + 197295*a^2*cos(d*x + c) -
161280*a^2)/(d*cos(d*x + c))

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giac [B]  time = 1.93, size = 370, normalized size = 2.02 \[ \frac {2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 2520 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {2520 \, {\left (2 \, a^{2} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1} + \frac {1457 \, a^{2} - \frac {20673 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {123012 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {421428 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {949662 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {1009134 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {666036 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {276804 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {66681 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {7129 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{9}}}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^9,x, algorithm="giac")

[Out]

1/1260*(2520*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 2520*a^2*log(abs(-(cos(d*x + c) - 1)/(
cos(d*x + c) + 1) - 1)) + 2520*(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) + (1457*a^2 - 20673*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 123012*a^2*(cos(d*x + c) - 1)
^2/(cos(d*x + c) + 1)^2 - 421428*a^2*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 949662*a^2*(cos(d*x + c) - 1)
^4/(cos(d*x + c) + 1)^4 - 1009134*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 666036*a^2*(cos(d*x + c) - 1
)^6/(cos(d*x + c) + 1)^6 - 276804*a^2*(cos(d*x + c) - 1)^7/(cos(d*x + c) + 1)^7 + 66681*a^2*(cos(d*x + c) - 1)
^8/(cos(d*x + c) + 1)^8 - 7129*a^2*(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

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maple [A]  time = 0.74, size = 206, normalized size = 1.13 \[ \frac {1024 a^{2} \cos \left (d x +c \right )}{315 d}+\frac {8 a^{2} \left (\sin ^{8}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{9 d}+\frac {64 a^{2} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{63 d}+\frac {128 a^{2} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{105 d}+\frac {512 a^{2} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{315 d}-\frac {a^{2} \left (\sin ^{8}\left (d x +c \right )\right )}{4 d}-\frac {a^{2} \left (\sin ^{6}\left (d x +c \right )\right )}{3 d}-\frac {a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d}-\frac {a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{d}-\frac {2 a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2*sin(d*x+c)^9,x)

[Out]

1024/315*a^2*cos(d*x+c)/d+8/9/d*a^2*sin(d*x+c)^8*cos(d*x+c)+64/63/d*a^2*cos(d*x+c)*sin(d*x+c)^6+128/105/d*a^2*
cos(d*x+c)*sin(d*x+c)^4+512/315/d*a^2*cos(d*x+c)*sin(d*x+c)^2-1/4/d*a^2*sin(d*x+c)^8-1/3/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)^10/cos(d*x+c)

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maxima [A]  time = 0.33, size = 146, normalized size = 0.80 \[ -\frac {140 \, a^{2} \cos \left (d x + c\right )^{9} + 315 \, a^{2} \cos \left (d x + c\right )^{8} - 540 \, a^{2} \cos \left (d x + c\right )^{7} - 1680 \, a^{2} \cos \left (d x + c\right )^{6} + 504 \, a^{2} \cos \left (d x + c\right )^{5} + 3780 \, a^{2} \cos \left (d x + c\right )^{4} + 840 \, a^{2} \cos \left (d x + c\right )^{3} - 5040 \, a^{2} \cos \left (d x + c\right )^{2} - 3780 \, a^{2} \cos \left (d x + c\right ) + 2520 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {1260 \, a^{2}}{\cos \left (d x + c\right )}}{1260 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2*sin(d*x+c)^9,x, algorithm="maxima")

[Out]

-1/1260*(140*a^2*cos(d*x + c)^9 + 315*a^2*cos(d*x + c)^8 - 540*a^2*cos(d*x + c)^7 - 1680*a^2*cos(d*x + c)^6 +
504*a^2*cos(d*x + c)^5 + 3780*a^2*cos(d*x + c)^4 + 840*a^2*cos(d*x + c)^3 - 5040*a^2*cos(d*x + c)^2 - 3780*a^2
*cos(d*x + c) + 2520*a^2*log(cos(d*x + c)) - 1260*a^2/cos(d*x + c))/d

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mupad [B]  time = 0.99, size = 146, normalized size = 0.80 \[ -\frac {\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^3}{3}-\frac {a^2}{\cos \left (c+d\,x\right )}-4\,a^2\,{\cos \left (c+d\,x\right )}^2-3\,a^2\,\cos \left (c+d\,x\right )+3\,a^2\,{\cos \left (c+d\,x\right )}^4+\frac {2\,a^2\,{\cos \left (c+d\,x\right )}^5}{5}-\frac {4\,a^2\,{\cos \left (c+d\,x\right )}^6}{3}-\frac {3\,a^2\,{\cos \left (c+d\,x\right )}^7}{7}+\frac {a^2\,{\cos \left (c+d\,x\right )}^8}{4}+\frac {a^2\,{\cos \left (c+d\,x\right )}^9}{9}+2\,a^2\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^9*(a + a/cos(c + d*x))^2,x)

[Out]

-((2*a^2*cos(c + d*x)^3)/3 - a^2/cos(c + d*x) - 4*a^2*cos(c + d*x)^2 - 3*a^2*cos(c + d*x) + 3*a^2*cos(c + d*x)
^4 + (2*a^2*cos(c + d*x)^5)/5 - (4*a^2*cos(c + d*x)^6)/3 - (3*a^2*cos(c + d*x)^7)/7 + (a^2*cos(c + d*x)^8)/4 +
(a^2*cos(c + d*x)^9)/9 + 2*a^2*log(cos(c + d*x)))/d

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**2*sin(d*x+c)**9,x)

[Out]

Timed out

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