\(\int \cos ^7(c+d x) (a+b \tan ^2(c+d x))^2 \, dx\) [442]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \sin (c+d x)}{d}-\frac {a (3 a-2 b) \sin ^3(c+d x)}{3 d}+\frac {(a-b) (3 a-b) \sin ^5(c+d x)}{5 d}-\frac {(a-b)^2 \sin ^7(c+d x)}{7 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3757, 380} \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {a^2 \sin (c+d x)}{d}-\frac {(a-b)^2 \sin ^7(c+d x)}{7 d}+\frac {(a-b) (3 a-b) \sin ^5(c+d x)}{5 d}-\frac {a (3 a-2 b) \sin ^3(c+d x)}{3 d} \]

[In]

Int[Cos[c + d*x]^7*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

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

Rule 380

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 3757

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.90 \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {105 a^2 \sin (c+d x)-35 a (3 a-2 b) \sin ^3(c+d x)+21 \left (3 a^2-4 a b+b^2\right ) \sin ^5(c+d x)-15 (a-b)^2 \sin ^7(c+d x)}{105 d} \]

[In]

Integrate[Cos[c + d*x]^7*(a + b*Tan[c + d*x]^2)^2,x]

[Out]

(105*a^2*Sin[c + d*x] - 35*a*(3*a - 2*b)*Sin[c + d*x]^3 + 21*(3*a^2 - 4*a*b + b^2)*Sin[c + d*x]^5 - 15*(a - b)
^2*Sin[c + d*x]^7)/(105*d)

Maple [A] (verified)

Time = 47.25 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.78

method	result	size
derivativedivides	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}{7}-\frac {3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{35}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{35}\right )+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )+\frac {a^{2} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(153\)
default	\(\frac {b^{2} \left (-\frac {\sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{4}}{7}-\frac {3 \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4}}{35}+\frac {\left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{35}\right )+2 a b \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )^{6}}{7}+\frac {\left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{35}\right )+\frac {a^{2} \left (\frac {16}{5}+\cos \left (d x +c \right )^{6}+\frac {6 \cos \left (d x +c \right )^{4}}{5}+\frac {8 \cos \left (d x +c \right )^{2}}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(153\)
risch	\(\frac {35 a^{2} \sin \left (d x +c \right )}{64 d}+\frac {5 \sin \left (d x +c \right ) a b}{32 d}+\frac {3 \sin \left (d x +c \right ) b^{2}}{64 d}+\frac {\sin \left (7 d x +7 c \right ) a^{2}}{448 d}-\frac {\sin \left (7 d x +7 c \right ) a b}{224 d}+\frac {\sin \left (7 d x +7 c \right ) b^{2}}{448 d}+\frac {7 \sin \left (5 d x +5 c \right ) a^{2}}{320 d}-\frac {3 \sin \left (5 d x +5 c \right ) a b}{160 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2}}{320 d}+\frac {7 \sin \left (3 d x +3 c \right ) a^{2}}{64 d}-\frac {\sin \left (3 d x +3 c \right ) a b}{96 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2}}{64 d}\)	\(193\)

[In]

int(cos(d*x+c)^7*(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(b^2*(-1/7*sin(d*x+c)^3*cos(d*x+c)^4-3/35*sin(d*x+c)*cos(d*x+c)^4+1/35*(2+cos(d*x+c)^2)*sin(d*x+c))+2*a*b*
(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))+1/7*a^2*(16/5+cos(d*x+c)^6+
6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {{\left (15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{6} + 6 \, {\left (3 \, a^{2} + a b - 4 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (24 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 48 \, a^{2} + 16 \, a b + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{105 \, d} \]

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/105*(15*(a^2 - 2*a*b + b^2)*cos(d*x + c)^6 + 6*(3*a^2 + a*b - 4*b^2)*cos(d*x + c)^4 + (24*a^2 + 8*a*b + 3*b^
2)*cos(d*x + c)^2 + 48*a^2 + 16*a*b + 6*b^2)*sin(d*x + c)/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**7*(a+b*tan(d*x+c)**2)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=-\frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{7} - 21 \, {\left (3 \, a^{2} - 4 \, a b + b^{2}\right )} \sin \left (d x + c\right )^{5} + 35 \, {\left (3 \, a^{2} - 2 \, a b\right )} \sin \left (d x + c\right )^{3} - 105 \, a^{2} \sin \left (d x + c\right )}{105 \, d} \]

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

-1/105*(15*(a^2 - 2*a*b + b^2)*sin(d*x + c)^7 - 21*(3*a^2 - 4*a*b + b^2)*sin(d*x + c)^5 + 35*(3*a^2 - 2*a*b)*s
in(d*x + c)^3 - 105*a^2*sin(d*x + c))/d

Giac [F(-1)]

Timed out. \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)^7*(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 12.05 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.86 \[ \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx=\frac {\frac {35\,a^2\,\sin \left (c+d\,x\right )}{64}+\frac {3\,b^2\,\sin \left (c+d\,x\right )}{64}+\frac {7\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{64}+\frac {7\,a^2\,\sin \left (5\,c+5\,d\,x\right )}{320}+\frac {a^2\,\sin \left (7\,c+7\,d\,x\right )}{448}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{64}-\frac {b^2\,\sin \left (5\,c+5\,d\,x\right )}{320}+\frac {b^2\,\sin \left (7\,c+7\,d\,x\right )}{448}+\frac {5\,a\,b\,\sin \left (c+d\,x\right )}{32}-\frac {a\,b\,\sin \left (3\,c+3\,d\,x\right )}{96}-\frac {3\,a\,b\,\sin \left (5\,c+5\,d\,x\right )}{160}-\frac {a\,b\,\sin \left (7\,c+7\,d\,x\right )}{224}}{d} \]

[In]

int(cos(c + d*x)^7*(a + b*tan(c + d*x)^2)^2,x)

[Out]

((35*a^2*sin(c + d*x))/64 + (3*b^2*sin(c + d*x))/64 + (7*a^2*sin(3*c + 3*d*x))/64 + (7*a^2*sin(5*c + 5*d*x))/3
20 + (a^2*sin(7*c + 7*d*x))/448 - (b^2*sin(3*c + 3*d*x))/64 - (b^2*sin(5*c + 5*d*x))/320 + (b^2*sin(7*c + 7*d*
x))/448 + (5*a*b*sin(c + d*x))/32 - (a*b*sin(3*c + 3*d*x))/96 - (3*a*b*sin(5*c + 5*d*x))/160 - (a*b*sin(7*c +
7*d*x))/224)/d