3.443 \(\int \cos ^9(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=114 \[ \frac {\left (6 a^2-6 a b+b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin (c+d x)}{d}+\frac {(a-b)^2 \sin ^9(c+d x)}{9 d}-\frac {2 (a-b) (2 a-b) \sin ^7(c+d x)}{7 d}-\frac {2 a (2 a-b) \sin ^3(c+d x)}{3 d} \]

[Out]

a^2*sin(d*x+c)/d-2/3*a*(2*a-b)*sin(d*x+c)^3/d+1/5*(6*a^2-6*a*b+b^2)*sin(d*x+c)^5/d-2/7*(a-b)*(2*a-b)*sin(d*x+c
)^7/d+1/9*(a-b)^2*sin(d*x+c)^9/d

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Rubi [A]  time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3676, 373} \[ \frac {\left (6 a^2-6 a b+b^2\right ) \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin (c+d x)}{d}+\frac {(a-b)^2 \sin ^9(c+d x)}{9 d}-\frac {2 (a-b) (2 a-b) \sin ^7(c+d x)}{7 d}-\frac {2 a (2 a-b) \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^9*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(a^2*Sin[c + d*x])/d - (2*a*(2*a - b)*Sin[c + d*x]^3)/(3*d) + ((6*a^2 - 6*a*b + b^2)*Sin[c + d*x]^5)/(5*d) - (
2*(a - b)*(2*a - b)*Sin[c + d*x]^7)/(7*d) + ((a - b)^2*Sin[c + d*x]^9)/(9*d)

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

\begin {align*} \int \cos ^9(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (1-x^2\right )^2 \left (a-(a-b) x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2-2 a (2 a-b) x^2+\left (6 a^2-6 a b+b^2\right ) x^4-2 \left (2 a^2-3 a b+b^2\right ) x^6+(a-b)^2 x^8\right ) \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {a^2 \sin (c+d x)}{d}-\frac {2 a (2 a-b) \sin ^3(c+d x)}{3 d}+\frac {\left (6 a^2-6 a b+b^2\right ) \sin ^5(c+d x)}{5 d}-\frac {2 (a-b) (2 a-b) \sin ^7(c+d x)}{7 d}+\frac {(a-b)^2 \sin ^9(c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 116, normalized size = 1.02 \[ \frac {630 \left (63 a^2+14 a b+3 b^2\right ) \sin (c+d x)+420 \left (21 a^2-b^2\right ) \sin (3 (c+d x))+252 \left (9 a^2-4 a b-b^2\right ) \sin (5 (c+d x))+35 (a-b)^2 \sin (9 (c+d x))+45 (9 a-b) (a-b) \sin (7 (c+d x))}{80640 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^9*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(630*(63*a^2 + 14*a*b + 3*b^2)*Sin[c + d*x] + 420*(21*a^2 - b^2)*Sin[3*(c + d*x)] + 252*(9*a^2 - 4*a*b - b^2)*
Sin[5*(c + d*x)] + 45*(a - b)*(9*a - b)*Sin[7*(c + d*x)] + 35*(a - b)^2*Sin[9*(c + d*x)])/(80640*d)

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fricas [A]  time = 0.51, size = 117, normalized size = 1.03 \[ \frac {{\left (35 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{8} + 10 \, {\left (4 \, a^{2} + a b - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{6} + 3 \, {\left (16 \, a^{2} + 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (16 \, a^{2} + 4 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 128 \, a^{2} + 32 \, a b + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/315*(35*(a^2 - 2*a*b + b^2)*cos(d*x + c)^8 + 10*(4*a^2 + a*b - 5*b^2)*cos(d*x + c)^6 + 3*(16*a^2 + 4*a*b + b
^2)*cos(d*x + c)^4 + 4*(16*a^2 + 4*a*b + b^2)*cos(d*x + c)^2 + 128*a^2 + 32*a*b + 8*b^2)*sin(d*x + c)/d

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giac [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(cos(d*x+c)^9*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.06, size = 183, normalized size = 1.61 \[ \frac {b^{2} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{6}\left (d x +c \right )\right )}{9}-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{21}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{105}\right )+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{8}\left (d x +c \right )\right )}{9}+\frac {\left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{63}\right )+\frac {a^{2} \left (\frac {128}{35}+\cos ^{8}\left (d x +c \right )+\frac {8 \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (d x +c \right )\right )}{35}\right ) \sin \left (d x +c \right )}{9}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^9*(a+b*tan(d*x+c)^2)^2,x)

[Out]

1/d*(b^2*(-1/9*sin(d*x+c)^3*cos(d*x+c)^6-1/21*sin(d*x+c)*cos(d*x+c)^6+1/105*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2
)*sin(d*x+c))+2*a*b*(-1/9*sin(d*x+c)*cos(d*x+c)^8+1/63*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*s
in(d*x+c))+1/9*a^2*(128/35+cos(d*x+c)^8+8/7*cos(d*x+c)^6+48/35*cos(d*x+c)^4+64/35*cos(d*x+c)^2)*sin(d*x+c))

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maxima [A]  time = 0.49, size = 104, normalized size = 0.91 \[ \frac {35 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{9} - 90 \, {\left (2 \, a^{2} - 3 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{7} + 63 \, {\left (6 \, a^{2} - 6 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} - 210 \, {\left (2 \, a^{2} - a b\right )} \sin \left (d x + c\right )^{3} + 315 \, a^{2} \sin \left (d x + c\right )}{315 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^9*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/315*(35*(a^2 - 2*a*b + b^2)*sin(d*x + c)^9 - 90*(2*a^2 - 3*a*b + b^2)*sin(d*x + c)^7 + 63*(6*a^2 - 6*a*b + b
^2)*sin(d*x + c)^5 - 210*(2*a^2 - a*b)*sin(d*x + c)^3 + 315*a^2*sin(d*x + c))/d

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mupad [B]  time = 12.40, size = 188, normalized size = 1.65 \[ \frac {\frac {63\,a^2\,\sin \left (c+d\,x\right )}{128}+\frac {3\,b^2\,\sin \left (c+d\,x\right )}{128}+\frac {7\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{64}+\frac {9\,a^2\,\sin \left (5\,c+5\,d\,x\right )}{320}+\frac {9\,a^2\,\sin \left (7\,c+7\,d\,x\right )}{1792}+\frac {a^2\,\sin \left (9\,c+9\,d\,x\right )}{2304}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{192}-\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{320}+\frac {b^2\,\sin \left (7\,c+7\,d\,x\right )}{1792}+\frac {b^2\,\sin \left (9\,c+9\,d\,x\right )}{2304}+\frac {7\,a\,b\,\sin \left (c+d\,x\right )}{64}-\frac {a\,b\,\sin \left (5\,c+5\,d\,x\right )}{80}-\frac {5\,a\,b\,\sin \left (7\,c+7\,d\,x\right )}{896}-\frac {a\,b\,\sin \left (9\,c+9\,d\,x\right )}{1152}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^9*(a + b*tan(c + d*x)^2)^2,x)

[Out]

((63*a^2*sin(c + d*x))/128 + (3*b^2*sin(c + d*x))/128 + (7*a^2*sin(3*c + 3*d*x))/64 + (9*a^2*sin(5*c + 5*d*x))
/320 + (9*a^2*sin(7*c + 7*d*x))/1792 + (a^2*sin(9*c + 9*d*x))/2304 - (b^2*sin(3*c + 3*d*x))/192 - (b^2*sin(5*c
+ 5*d*x))/320 + (b^2*sin(7*c + 7*d*x))/1792 + (b^2*sin(9*c + 9*d*x))/2304 + (7*a*b*sin(c + d*x))/64 - (a*b*si
n(5*c + 5*d*x))/80 - (5*a*b*sin(7*c + 7*d*x))/896 - (a*b*sin(9*c + 9*d*x))/1152)/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(cos(d*x+c)**9*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

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