3.542 \(\int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac{\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac{\sin (c+d x) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{5 d} \]

[Out]

(-2*(4*a^2 + b^2)*Cos[c + d*x]*(b - a*Tan[c + d*x]))/(15*d) - (Cos[c + d*x]^3*(b - 4*a*Tan[c + d*x])*(a + b*Ta
n[c + d*x])^2)/(15*d) + (Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(5*d)

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Rubi [A]  time = 0.0946877, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3512, 737, 805, 637} \[ -\frac{2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac{\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac{\sin (c+d x) \cos ^4(c+d x) (a+b \tan (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(-2*(4*a^2 + b^2)*Cos[c + d*x]*(b - a*Tan[c + d*x]))/(15*d) - (Cos[c + d*x]^3*(b - 4*a*Tan[c + d*x])*(a + b*Ta
n[c + d*x])^2)/(15*d) + (Cos[c + d*x]^4*Sin[c + d*x]*(a + b*Tan[c + d*x])^3)/(5*d)

Rule 3512

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

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*x)*(a + c*x^2
)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 805

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[(m*(c*d*f + a*e*g))/(2*a*c*(p + 1)), Int[(d + e*
x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[Simplif
y[m + 2*p + 3], 0] && LtQ[p, -1]

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rubi steps

\begin{align*} \int \cos ^5(c+d x) (a+b \tan (c+d x))^3 \, dx &=\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(a+x)^3}{\left (1+\frac{x^2}{b^2}\right )^{7/2}} \, dx,x,b \tan (c+d x)\right )}{b d}\\ &=\frac{\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}-\frac{\left (\cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{(-4 a-x) (a+x)^2}{\left (1+\frac{x^2}{b^2}\right )^{5/2}} \, dx,x,b \tan (c+d x)\right )}{5 b d}\\ &=-\frac{\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac{\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}+\frac{\left (2 \left (4 a^2+b^2\right ) \cos (c+d x) \sqrt{\sec ^2(c+d x)}\right ) \operatorname{Subst}\left (\int \frac{a+x}{\left (1+\frac{x^2}{b^2}\right )^{3/2}} \, dx,x,b \tan (c+d x)\right )}{15 b d}\\ &=-\frac{2 \left (4 a^2+b^2\right ) \cos (c+d x) (b-a \tan (c+d x))}{15 d}-\frac{\cos ^3(c+d x) (b-4 a \tan (c+d x)) (a+b \tan (c+d x))^2}{15 d}+\frac{\cos ^4(c+d x) \sin (c+d x) (a+b \tan (c+d x))^3}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.691361, size = 150, normalized size = 1.43 \[ \frac{-30 b \left (3 a^2+b^2\right ) \cos (c+d x)-5 \left (9 a^2 b+b^3\right ) \cos (3 (c+d x))-9 a^2 b \cos (5 (c+d x))+150 a^3 \sin (c+d x)+25 a^3 \sin (3 (c+d x))+3 a^3 \sin (5 (c+d x))+90 a b^2 \sin (c+d x)-15 a b^2 \sin (3 (c+d x))-9 a b^2 \sin (5 (c+d x))+3 b^3 \cos (5 (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x])^3,x]

[Out]

(-30*b*(3*a^2 + b^2)*Cos[c + d*x] - 5*(9*a^2*b + b^3)*Cos[3*(c + d*x)] - 9*a^2*b*Cos[5*(c + d*x)] + 3*b^3*Cos[
5*(c + d*x)] + 150*a^3*Sin[c + d*x] + 90*a*b^2*Sin[c + d*x] + 25*a^3*Sin[3*(c + d*x)] - 15*a*b^2*Sin[3*(c + d*
x)] + 3*a^3*Sin[5*(c + d*x)] - 9*a*b^2*Sin[5*(c + d*x)])/(240*d)

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Maple [A]  time = 0.065, size = 125, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ( -{\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}}-{\frac{2\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{15}} \right ) +3\,a{b}^{2} \left ( -1/5\,\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) \right ) -{\frac{3\,b{a}^{2} \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}+{\frac{{a}^{3}\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))^3,x)

[Out]

1/d*(b^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)+3*a*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))-3/5*b*a^2*cos(d*x+c)^5+1/5*a^3*(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 = 1.13995, size = 144, normalized size = 1.37 \begin{align*} -\frac{9 \, a^{2} 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^{3} + 3 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a b^{2} -{\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} b^{3}}{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))^3,x, algorithm="maxima")

[Out]

-1/15*(9*a^2*b*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3 + 3*(3*sin(d*x +
c)^5 - 5*sin(d*x + c)^3)*a*b^2 - (3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*b^3)/d

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Fricas [A]  time = 1.77587, size = 230, normalized size = 2.19 \begin{align*} -\frac{5 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{5} -{\left (3 \,{\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} + 8 \, a^{3} + 6 \, a b^{2} +{\left (4 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{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))^3,x, algorithm="fricas")

[Out]

-1/15*(5*b^3*cos(d*x + c)^3 + 3*(3*a^2*b - b^3)*cos(d*x + c)^5 - (3*(a^3 - 3*a*b^2)*cos(d*x + c)^4 + 8*a^3 + 6
*a*b^2 + (4*a^3 + 3*a*b^2)*cos(d*x + c)^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)**5*(a+b*tan(d*x+c))**3,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)^5*(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out