3.529 \(\int \cos ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=114 \[ \frac{\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac{\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac{\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac{3 a b \cos ^5(c+d x)}{20 d}-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d} \]

[Out]

(-3*a*b*Cos[c + d*x]^5)/(20*d) + ((4*a^2 + b^2)*Sin[c + d*x])/(4*d) - ((4*a^2 + b^2)*Sin[c + d*x]^3)/(6*d) + (
(4*a^2 + b^2)*Sin[c + d*x]^5)/(20*d) - (b*Cos[c + d*x]^5*(a + b*Tan[c + d*x]))/(4*d)

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Rubi [A]  time = 0.102317, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3508, 3486, 2633} \[ \frac{\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac{\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac{\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac{3 a b \cos ^5(c+d x)}{20 d}-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(-3*a*b*Cos[c + d*x]^5)/(20*d) + ((4*a^2 + b^2)*Sin[c + d*x])/(4*d) - ((4*a^2 + b^2)*Sin[c + d*x]^3)/(6*d) + (
(4*a^2 + b^2)*Sin[c + d*x]^5)/(20*d) - (b*Cos[c + d*x]^5*(a + b*Tan[c + d*x]))/(4*d)

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \tan (c+d x))^2 \, dx &=-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{1}{4} \int \cos ^5(c+d x) \left (-4 a^2-b^2-3 a b \tan (c+d x)\right ) \, dx\\ &=-\frac{3 a b \cos ^5(c+d x)}{20 d}-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{1}{4} \left (-4 a^2-b^2\right ) \int \cos ^5(c+d x) \, dx\\ &=-\frac{3 a b \cos ^5(c+d x)}{20 d}-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}-\frac{\left (4 a^2+b^2\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=-\frac{3 a b \cos ^5(c+d x)}{20 d}+\frac{\left (4 a^2+b^2\right ) \sin (c+d x)}{4 d}-\frac{\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{6 d}+\frac{\left (4 a^2+b^2\right ) \sin ^5(c+d x)}{20 d}-\frac{b \cos ^5(c+d x) (a+b \tan (c+d x))}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.210422, size = 116, normalized size = 1.02 \[ \frac{150 a^2 \sin (c+d x)+25 a^2 \sin (3 (c+d x))+3 a^2 \sin (5 (c+d x))-60 a b \cos (c+d x)-30 a b \cos (3 (c+d x))-6 a b \cos (5 (c+d x))+30 b^2 \sin (c+d x)-5 b^2 \sin (3 (c+d x))-3 b^2 \sin (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(-60*a*b*Cos[c + d*x] - 30*a*b*Cos[3*(c + d*x)] - 6*a*b*Cos[5*(c + d*x)] + 150*a^2*Sin[c + d*x] + 30*b^2*Sin[c
+ d*x] + 25*a^2*Sin[3*(c + d*x)] - 5*b^2*Sin[3*(c + d*x)] + 3*a^2*Sin[5*(c + d*x)] - 3*b^2*Sin[5*(c + d*x)])/
(240*d)

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Maple [A]  time = 0.055, size = 88, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({b}^{2} \left ( -{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{ \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{15}} \right ) -{\frac{2\,ab \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x)

[Out]

1/d*(b^2*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(2+cos(d*x+c)^2)*sin(d*x+c))-2/5*a*b*cos(d*x+c)^5+1/5*a^2*(8/3+cos
(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]  time = 1.36548, size = 104, normalized size = 0.91 \begin{align*} -\frac{6 \, a b \cos \left (d x + c\right )^{5} -{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2} +{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} b^{2}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/15*(6*a*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^2 + (3*sin(d*x + c)^5
- 5*sin(d*x + c)^3)*b^2)/d

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Fricas [A]  time = 1.90536, size = 169, normalized size = 1.48 \begin{align*} -\frac{6 \, a b \cos \left (d x + c\right )^{5} -{\left (3 \,{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} +{\left (4 \, a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/15*(6*a*b*cos(d*x + c)^5 - (3*(a^2 - b^2)*cos(d*x + c)^4 + (4*a^2 + b^2)*cos(d*x + c)^2 + 8*a^2 + 2*b^2)*si
n(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)**5*(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError