3.271 \(\int \cos ^m(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx\)

Optimal. Leaf size=155 \[ -\frac{b \left (3 a^2-b^2\right ) \cos ^m(c+d x)}{d m}-\frac{a \left (a^2-3 b^2\right ) \cos ^{m+1}(c+d x)}{d (m+1)}+\frac{3 a^2 b \cos ^{m+2}(c+d x)}{d (m+2)}+\frac{a^3 \cos ^{m+3}(c+d x)}{d (m+3)}+\frac{3 a b^2 \cos ^{m-1}(c+d x)}{d (1-m)}+\frac{b^3 \cos ^{m-2}(c+d x)}{d (2-m)} \]

[Out]

(b^3*Cos[c + d*x]^(-2 + m))/(d*(2 - m)) + (3*a*b^2*Cos[c + d*x]^(-1 + m))/(d*(1 - m)) - (b*(3*a^2 - b^2)*Cos[c
+ d*x]^m)/(d*m) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^(1 + m))/(d*(1 + m)) + (3*a^2*b*Cos[c + d*x]^(2 + m))/(d*(2 +
m)) + (a^3*Cos[c + d*x]^(3 + m))/(d*(3 + m))

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Rubi [A]  time = 0.384805, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {4397, 2837, 948} \[ -\frac{b \left (3 a^2-b^2\right ) \cos ^m(c+d x)}{d m}-\frac{a \left (a^2-3 b^2\right ) \cos ^{m+1}(c+d x)}{d (m+1)}+\frac{3 a^2 b \cos ^{m+2}(c+d x)}{d (m+2)}+\frac{a^3 \cos ^{m+3}(c+d x)}{d (m+3)}+\frac{3 a b^2 \cos ^{m-1}(c+d x)}{d (1-m)}+\frac{b^3 \cos ^{m-2}(c+d x)}{d (2-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^3*Cos[c + d*x]^(-2 + m))/(d*(2 - m)) + (3*a*b^2*Cos[c + d*x]^(-1 + m))/(d*(1 - m)) - (b*(3*a^2 - b^2)*Cos[c
+ d*x]^m)/(d*m) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^(1 + m))/(d*(1 + m)) + (3*a^2*b*Cos[c + d*x]^(2 + m))/(d*(2 +
m)) + (a^3*Cos[c + d*x]^(3 + m))/(d*(3 + m))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rubi steps

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

Mathematica [A]  time = 1.51235, size = 246, normalized size = 1.59 \[ \frac{\cos ^{m+1}(c+d x) (a+b \sec (c+d x))^3 \left (-a m \left (m^3-m^2-4 m+4\right ) \left (a^2 (m+9)-12 b^2 (m+3)\right ) \cos ^3(c+d x)+\left (m^3-2 m^2-m+2\right ) \cos ^2(c+d x) \left (2 b (m+3) \left (2 b^2 (m+2)-3 a^2 (m+4)\right )+6 a^2 b m (m+3) \cos (2 (c+d x))+a^3 m (m+2) \cos (3 (c+d x))\right )-12 a b^2 m \left (m^4+4 m^3-m^2-16 m-12\right ) \cos (c+d x)-4 b^3 m \left (m^4+5 m^3+5 m^2-5 m-6\right )\right )}{4 d (m-2) (m-1) m (m+1) (m+2) (m+3) (a \cos (c+d x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^(1 + m)*(-4*b^3*m*(-6 - 5*m + 5*m^2 + 5*m^3 + m^4) - 12*a*b^2*m*(-12 - 16*m - m^2 + 4*m^3 + m^4)
*Cos[c + d*x] - a*m*(4 - 4*m - m^2 + m^3)*(-12*b^2*(3 + m) + a^2*(9 + m))*Cos[c + d*x]^3 + (2 - m - 2*m^2 + m^
3)*Cos[c + d*x]^2*(2*b*(3 + m)*(2*b^2*(2 + m) - 3*a^2*(4 + m)) + 6*a^2*b*m*(3 + m)*Cos[2*(c + d*x)] + a^3*m*(2
+ m)*Cos[3*(c + d*x)]))*(a + b*Sec[c + d*x])^3)/(4*d*(-2 + m)*(-1 + m)*m*(1 + m)*(2 + m)*(3 + m)*(b + a*Cos[c
+ d*x])^3)

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Maple [F]  time = 3.616, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{m} \left ( a\sin \left ( dx+c \right ) +b\tan \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

[Out]

int(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^3,x)

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Maxima [A]  time = 1.12904, size = 243, normalized size = 1.57 \begin{align*} \frac{\frac{{\left ({\left (m + 1\right )} \cos \left (d x + c\right )^{3} -{\left (m + 3\right )} \cos \left (d x + c\right )\right )} a^{3} \cos \left (d x + c\right )^{m}}{m^{2} + 4 \, m + 3} + \frac{3 \,{\left (m \cos \left (d x + c\right )^{2} - m - 2\right )} a^{2} b \cos \left (d x + c\right )^{m}}{m^{2} + 2 \, m} + \frac{3 \,{\left ({\left (m - 1\right )} \cos \left (d x + c\right )^{2} - m - 1\right )} a b^{2} \cos \left (d x + c\right )^{m}}{{\left (m^{2} - 1\right )} \cos \left (d x + c\right )} + \frac{{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} - m\right )} b^{3} \cos \left (d x + c\right )^{m}}{{\left (m^{2} - 2 \, m\right )} \cos \left (d x + c\right )^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

(((m + 1)*cos(d*x + c)^3 - (m + 3)*cos(d*x + c))*a^3*cos(d*x + c)^m/(m^2 + 4*m + 3) + 3*(m*cos(d*x + c)^2 - m
- 2)*a^2*b*cos(d*x + c)^m/(m^2 + 2*m) + 3*((m - 1)*cos(d*x + c)^2 - m - 1)*a*b^2*cos(d*x + c)^m/((m^2 - 1)*cos
(d*x + c)) + ((m - 2)*cos(d*x + c)^2 - m)*b^3*cos(d*x + c)^m/((m^2 - 2*m)*cos(d*x + c)^2))/d

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Fricas [B]  time = 0.612053, size = 876, normalized size = 5.65 \begin{align*} -\frac{{\left (b^{3} m^{5} + 5 \, b^{3} m^{4} + 5 \, b^{3} m^{3} -{\left (a^{3} m^{5} - 5 \, a^{3} m^{3} + 4 \, a^{3} m\right )} \cos \left (d x + c\right )^{5} - 5 \, b^{3} m^{2} - 3 \,{\left (a^{2} b m^{5} + a^{2} b m^{4} - 7 \, a^{2} b m^{3} - a^{2} b m^{2} + 6 \, a^{2} b m\right )} \cos \left (d x + c\right )^{4} - 6 \, b^{3} m +{\left ({\left (a^{3} - 3 \, a b^{2}\right )} m^{5} + 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} m^{4} - 7 \,{\left (a^{3} - 3 \, a b^{2}\right )} m^{3} - 8 \,{\left (a^{3} - 3 \, a b^{2}\right )} m^{2} + 12 \,{\left (a^{3} - 3 \, a b^{2}\right )} m\right )} \cos \left (d x + c\right )^{3} +{\left ({\left (3 \, a^{2} b - b^{3}\right )} m^{5} + 3 \,{\left (3 \, a^{2} b - b^{3}\right )} m^{4} - 5 \,{\left (3 \, a^{2} b - b^{3}\right )} m^{3} + 36 \, a^{2} b - 12 \, b^{3} - 15 \,{\left (3 \, a^{2} b - b^{3}\right )} m^{2} + 4 \,{\left (3 \, a^{2} b - b^{3}\right )} m\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (a b^{2} m^{5} + 4 \, a b^{2} m^{4} - a b^{2} m^{3} - 16 \, a b^{2} m^{2} - 12 \, a b^{2} m\right )} \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{{\left (d m^{6} + 3 \, d m^{5} - 5 \, d m^{4} - 15 \, d m^{3} + 4 \, d m^{2} + 12 \, d m\right )} \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^m*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-(b^3*m^5 + 5*b^3*m^4 + 5*b^3*m^3 - (a^3*m^5 - 5*a^3*m^3 + 4*a^3*m)*cos(d*x + c)^5 - 5*b^3*m^2 - 3*(a^2*b*m^5
+ a^2*b*m^4 - 7*a^2*b*m^3 - a^2*b*m^2 + 6*a^2*b*m)*cos(d*x + c)^4 - 6*b^3*m + ((a^3 - 3*a*b^2)*m^5 + 2*(a^3 -
3*a*b^2)*m^4 - 7*(a^3 - 3*a*b^2)*m^3 - 8*(a^3 - 3*a*b^2)*m^2 + 12*(a^3 - 3*a*b^2)*m)*cos(d*x + c)^3 + ((3*a^2*
b - b^3)*m^5 + 3*(3*a^2*b - b^3)*m^4 - 5*(3*a^2*b - b^3)*m^3 + 36*a^2*b - 12*b^3 - 15*(3*a^2*b - b^3)*m^2 + 4*
(3*a^2*b - b^3)*m)*cos(d*x + c)^2 + 3*(a*b^2*m^5 + 4*a*b^2*m^4 - a*b^2*m^3 - 16*a*b^2*m^2 - 12*a*b^2*m)*cos(d*
x + c))*cos(d*x + c)^m/((d*m^6 + 3*d*m^5 - 5*d*m^4 - 15*d*m^3 + 4*d*m^2 + 12*d*m)*cos(d*x + c)^2)

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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)**m*(a*sin(d*x+c)+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)^m*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out