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

Optimal. Leaf size=86 \[ \frac{a^2 \sin (c+d x)}{d}-\frac{(a-b)^2 \sin ^7(c+d x)}{7 d}+\frac{(a-b) (3 a-b) \sin ^5(c+d x)}{5 d}-\frac{a (3 a-2 b) \sin ^3(c+d x)}{3 d} \]

[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)

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Rubi [A]  time = 0.0804554, antiderivative size = 86, normalized size of antiderivative = 1., 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} \[ \frac{a^2 \sin (c+d x)}{d}-\frac{(a-b)^2 \sin ^7(c+d x)}{7 d}+\frac{(a-b) (3 a-b) \sin ^5(c+d x)}{5 d}-\frac{a (3 a-2 b) \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[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 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]

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]

Rubi steps

\begin{align*} \int \cos ^7(c+d x) \left (a+b \tan ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{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{\operatorname{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]  time = 0.366442, size = 77, normalized size = 0.9 \[ \frac{21 \left (3 a^2-4 a b+b^2\right ) \sin ^5(c+d x)+105 a^2 \sin (c+d x)-15 (a-b)^2 \sin ^7(c+d x)-35 a (3 a-2 b) \sin ^3(c+d x)}{105 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.056, size = 153, normalized size = 1.8 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{7}}-{\frac{3\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{35}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{35}} \right ) +2\,ab \left ( -1/7\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{6}+1/35\, \left ( 8/3+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) +{\frac{{a}^{2}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

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Maxima [A]  time = 1.09244, size = 109, normalized size = 1.27 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.52151, size = 231, normalized size = 2.69 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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

[Out]

Timed out

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

[Out]

Timed out