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

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

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 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]))

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 4397

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

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*}

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Mathematica [A]  time = 1.45, size = 246, normalized size = 1.59 \[ \frac {\cos ^{m+1}(c+d x) (a+b \sec (c+d x))^3 \left (-a m \left (m^3-m^2-4 m+4\right ) \left (a^2 (m+9)-12 b^2 (m+3)\right ) \cos ^3(c+d x)+\left (m^3-2 m^2-m+2\right ) \cos ^2(c+d x) \left (a^3 m (m+2) \cos (3 (c+d x))+2 b (m+3) \left (2 b^2 (m+2)-3 a^2 (m+4)\right )+6 a^2 b m (m+3) \cos (2 (c+d x))\right )-12 a b^2 m \left (m^4+4 m^3-m^2-16 m-12\right ) \cos (c+d x)-4 b^3 m \left (m^4+5 m^3+5 m^2-5 m-6\right )\right )}{4 d (m-2) (m-1) m (m+1) (m+2) (m+3) (a \cos (c+d x)+b)^3} \]

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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fricas [B]  time = 0.53, size = 412, normalized size = 2.66 \[ -\frac {{\left (b^{3} m^{5} + 5 \, b^{3} m^{4} + 5 \, b^{3} m^{3} - {\left (a^{3} m^{5} - 5 \, a^{3} m^{3} + 4 \, a^{3} m\right )} \cos \left (d x + c\right )^{5} - 5 \, b^{3} m^{2} - 3 \, {\left (a^{2} b m^{5} + a^{2} b m^{4} - 7 \, a^{2} b m^{3} - a^{2} b m^{2} + 6 \, a^{2} b m\right )} \cos \left (d x + c\right )^{4} - 6 \, b^{3} m + {\left ({\left (a^{3} - 3 \, a b^{2}\right )} m^{5} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} m^{4} - 7 \, {\left (a^{3} - 3 \, a b^{2}\right )} m^{3} - 8 \, {\left (a^{3} - 3 \, a b^{2}\right )} m^{2} + 12 \, {\left (a^{3} - 3 \, a b^{2}\right )} m\right )} \cos \left (d x + c\right )^{3} + {\left ({\left (3 \, a^{2} b - b^{3}\right )} m^{5} + 3 \, {\left (3 \, a^{2} b - b^{3}\right )} m^{4} - 5 \, {\left (3 \, a^{2} b - b^{3}\right )} m^{3} + 36 \, a^{2} b - 12 \, b^{3} - 15 \, {\left (3 \, a^{2} b - b^{3}\right )} m^{2} + 4 \, {\left (3 \, a^{2} b - b^{3}\right )} m\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a b^{2} m^{5} + 4 \, a b^{2} m^{4} - a b^{2} m^{3} - 16 \, a b^{2} m^{2} - 12 \, a b^{2} m\right )} \cos \left (d x + c\right )\right )} \cos \left (d x + c\right )^{m}}{{\left (d m^{6} + 3 \, d m^{5} - 5 \, d m^{4} - 15 \, d m^{3} + 4 \, d m^{2} + 12 \, d m\right )} \cos \left (d x + c\right )^{2}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Modg
cd: no suitable evaluation pointindex.cc index_m operator + Error: Bad Argument ValueUnable to check sign: (2*
pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sig
n: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to che
ck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable
to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)U
nable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nost
ep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/
t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(
-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_noste
p/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t
_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (
2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check s
ign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to c
heck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Evaluation time: 61.5Done

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maple [F]  time = 4.64, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{m}\left (d x +c \right )\right ) \left (a \sin \left (d x +c \right )+b \tan \left (d x +c \right )\right )^{3}\, dx \]

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 = 0.36, size = 180, normalized size = 1.16 \[ \frac {\frac {{\left ({\left (m + 1\right )} \cos \left (d x + c\right )^{3} - {\left (m + 3\right )} \cos \left (d x + c\right )\right )} a^{3} \cos \left (d x + c\right )^{m}}{m^{2} + 4 \, m + 3} + \frac {3 \, {\left (m \cos \left (d x + c\right )^{2} - m - 2\right )} a^{2} b \cos \left (d x + c\right )^{m}}{m^{2} + 2 \, m} + \frac {3 \, {\left ({\left (m - 1\right )} \cos \left (d x + c\right )^{2} - m - 1\right )} a b^{2} \cos \left (d x + c\right )^{m}}{{\left (m^{2} - 1\right )} \cos \left (d x + c\right )} + \frac {{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} - m\right )} b^{3} \cos \left (d x + c\right )^{m}}{{\left (m^{2} - 2 \, m\right )} \cos \left (d x + c\right )^{2}}}{d} \]

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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mupad [B]  time = 7.51, size = 861, normalized size = 5.55 \[ \frac {{\left (\frac {1}{2}\right )}^m\,{\left ({\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\right )}^m\,\left (\frac {a^3\,\left (\frac {m^4}{8}-\frac {5\,m^2}{8}+\frac {1}{2}\right )}{d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {a^3\,{\mathrm {e}}^{c\,10{}\mathrm {i}+d\,x\,10{}\mathrm {i}}\,\left (m^4-5\,m^2+4\right )}{8\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}-\frac {a\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\left (-m^3+m^2+4\,m-4\right )\,\left (a^2\,m+12\,b^2\,m-7\,a^2+36\,b^2\right )}{8\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}-\frac {a\,{\mathrm {e}}^{c\,8{}\mathrm {i}+d\,x\,8{}\mathrm {i}}\,\left (-m^3+m^2+4\,m-4\right )\,\left (a^2\,m+12\,b^2\,m-7\,a^2+36\,b^2\right )}{8\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {3\,a^2\,b\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (m^4+m^3-7\,m^2-m+6\right )}{4\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {3\,a^2\,b\,{\mathrm {e}}^{c\,9{}\mathrm {i}+d\,x\,9{}\mathrm {i}}\,\left (m^4+m^3-7\,m^2-m+6\right )}{4\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}-\frac {a\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\left (m^2-4\right )\,\left (a^2\,m^2+12\,a^2\,m-13\,a^2+6\,b^2\,m^2+60\,b^2\,m+126\,b^2\right )}{4\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}-\frac {a\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,\left (m^2-4\right )\,\left (a^2\,m^2+12\,a^2\,m-13\,a^2+6\,b^2\,m^2+60\,b^2\,m+126\,b^2\right )}{4\,d\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {b\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\left (b^2\,m-6\,a^2+2\,b^2\right )\,\left (m^4+m^3-7\,m^2-m+6\right )}{d\,m\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {b\,{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\left (b^2\,m-6\,a^2+2\,b^2\right )\,\left (m^4+m^3-7\,m^2-m+6\right )}{d\,m\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}+\frac {b\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\left (-m^3-3\,m^2+m+3\right )\,\left (3\,a^2\,m^2+18\,a^2\,m-48\,a^2+4\,b^2\,m^2+16\,b^2\,m+16\,b^2\right )}{2\,d\,m\,\left (m^5+3\,m^4-5\,m^3-15\,m^2+4\,m+12\right )}\right )}{{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}+2\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}+{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1/2)^m*(exp(- c*1i - d*x*1i) + exp(c*1i + d*x*1i))^m*((a^3*(m^4/8 - (5*m^2)/8 + 1/2))/(d*(4*m - 15*m^2 - 5*m
^3 + 3*m^4 + m^5 + 12)) + (a^3*exp(c*10i + d*x*10i)*(m^4 - 5*m^2 + 4))/(8*d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^
5 + 12)) - (a*exp(c*2i + d*x*2i)*(4*m + m^2 - m^3 - 4)*(a^2*m + 12*b^2*m - 7*a^2 + 36*b^2))/(8*d*(4*m - 15*m^2
- 5*m^3 + 3*m^4 + m^5 + 12)) - (a*exp(c*8i + d*x*8i)*(4*m + m^2 - m^3 - 4)*(a^2*m + 12*b^2*m - 7*a^2 + 36*b^2
))/(8*d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12)) + (3*a^2*b*exp(c*1i + d*x*1i)*(m^3 - 7*m^2 - m + m^4 + 6))/
(4*d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12)) + (3*a^2*b*exp(c*9i + d*x*9i)*(m^3 - 7*m^2 - m + m^4 + 6))/(4*
d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12)) - (a*exp(c*4i + d*x*4i)*(m^2 - 4)*(12*a^2*m + 60*b^2*m - 13*a^2 +
126*b^2 + a^2*m^2 + 6*b^2*m^2))/(4*d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12)) - (a*exp(c*6i + d*x*6i)*(m^2
- 4)*(12*a^2*m + 60*b^2*m - 13*a^2 + 126*b^2 + a^2*m^2 + 6*b^2*m^2))/(4*d*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5
+ 12)) + (b*exp(c*3i + d*x*3i)*(b^2*m - 6*a^2 + 2*b^2)*(m^3 - 7*m^2 - m + m^4 + 6))/(d*m*(4*m - 15*m^2 - 5*m^3
+ 3*m^4 + m^5 + 12)) + (b*exp(c*7i + d*x*7i)*(b^2*m - 6*a^2 + 2*b^2)*(m^3 - 7*m^2 - m + m^4 + 6))/(d*m*(4*m -
15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12)) + (b*exp(c*5i + d*x*5i)*(m - 3*m^2 - m^3 + 3)*(18*a^2*m + 16*b^2*m - 48*a
^2 + 16*b^2 + 3*a^2*m^2 + 4*b^2*m^2))/(2*d*m*(4*m - 15*m^2 - 5*m^3 + 3*m^4 + m^5 + 12))))/(exp(c*3i + d*x*3i)
+ 2*exp(c*5i + d*x*5i) + exp(c*7i + d*x*7i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{3} \cos ^{m}{\left (c + d x \right )}\, dx \]

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]

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

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