3.236 \(\int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=106 \[ \frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{30 d}+\frac {\sin ^3(c+d x) (a \cos (c+d x)+b)^2}{5 d}+\frac {b \sin ^3(c+d x) (a \cos (c+d x)+b)}{10 d}-\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]

[Out]

1/4*a*b*x-1/4*a*b*cos(d*x+c)*sin(d*x+c)/d+1/30*(4*a^2+b^2)*sin(d*x+c)^3/d+1/10*b*(b+a*cos(d*x+c))*sin(d*x+c)^3
/d+1/5*(b+a*cos(d*x+c))^2*sin(d*x+c)^3/d

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Rubi [A]  time = 0.35, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4397, 2862, 2669, 2635, 8} \[ \frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{30 d}+\frac {\sin ^3(c+d x) (a \cos (c+d x)+b)^2}{5 d}+\frac {b \sin ^3(c+d x) (a \cos (c+d x)+b)}{10 d}-\frac {a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {a b x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

(a*b*x)/4 - (a*b*Cos[c + d*x]*Sin[c + d*x])/(4*d) + ((4*a^2 + b^2)*Sin[c + d*x]^3)/(30*d) + (b*(b + a*Cos[c +
d*x])*Sin[c + d*x]^3)/(10*d) + ((b + a*Cos[c + d*x])^2*Sin[c + d*x]^3)/(5*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2669

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b*(g*Cos[
e + f*x])^(p + 1))/(f*g*(p + 1)), x] + Dist[a, Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])

Rule 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[m, 0] &&  !LtQ[p, -1] && IntegerQ[2*m] &&  !(EqQ[m, 1] && NeQ[c^2 - d^2, 0] && SimplerQ[c + d*x, a + b*x])

Rule 4397

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

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a \sin (c+d x)+b \tan (c+d x))^2 \, dx &=\int \cos (c+d x) (b+a \cos (c+d x))^2 \sin ^2(c+d x) \, dx\\ &=\frac {(b+a \cos (c+d x))^2 \sin ^3(c+d x)}{5 d}+\frac {1}{5} \int (b+a \cos (c+d x)) (2 a+2 b \cos (c+d x)) \sin ^2(c+d x) \, dx\\ &=\frac {b (b+a \cos (c+d x)) \sin ^3(c+d x)}{10 d}+\frac {(b+a \cos (c+d x))^2 \sin ^3(c+d x)}{5 d}+\frac {1}{20} \int \left (10 a b+2 \left (4 a^2+b^2\right ) \cos (c+d x)\right ) \sin ^2(c+d x) \, dx\\ &=\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{30 d}+\frac {b (b+a \cos (c+d x)) \sin ^3(c+d x)}{10 d}+\frac {(b+a \cos (c+d x))^2 \sin ^3(c+d x)}{5 d}+\frac {1}{2} (a b) \int \sin ^2(c+d x) \, dx\\ &=-\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{30 d}+\frac {b (b+a \cos (c+d x)) \sin ^3(c+d x)}{10 d}+\frac {(b+a \cos (c+d x))^2 \sin ^3(c+d x)}{5 d}+\frac {1}{4} (a b) \int 1 \, dx\\ &=\frac {a b x}{4}-\frac {a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {\left (4 a^2+b^2\right ) \sin ^3(c+d x)}{30 d}+\frac {b (b+a \cos (c+d x)) \sin ^3(c+d x)}{10 d}+\frac {(b+a \cos (c+d x))^2 \sin ^3(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 77, normalized size = 0.73 \[ \frac {30 \left (a^2+2 b^2\right ) \sin (c+d x)-5 \left (a^2+4 b^2\right ) \sin (3 (c+d x))-3 a (a \sin (5 (c+d x))-20 b (c+d x)+5 b \sin (4 (c+d x)))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a*Sin[c + d*x] + b*Tan[c + d*x])^2,x]

[Out]

(30*(a^2 + 2*b^2)*Sin[c + d*x] - 5*(a^2 + 4*b^2)*Sin[3*(c + d*x)] - 3*a*(-20*b*(c + d*x) + 5*b*Sin[4*(c + d*x)
] + a*Sin[5*(c + d*x)]))/(240*d)

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fricas [A]  time = 0.59, size = 85, normalized size = 0.80 \[ \frac {15 \, a b d x - {\left (12 \, a^{2} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right ) - 4 \, {\left (a^{2} - 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \, a^{2} - 20 \, b^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/60*(15*a*b*d*x - (12*a^2*cos(d*x + c)^4 + 30*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c) - 4*(a^2 - 5*b^2)*cos(
d*x + c)^2 - 8*a^2 - 20*b^2)*sin(d*x + c))/d

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.09, size = 100, normalized size = 0.94 \[ \frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )+2 a b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{8}+\frac {d x}{8}+\frac {c}{8}\right )+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.37, size = 68, normalized size = 0.64 \[ \frac {80 \, b^{2} \sin \left (d x + c\right )^{3} - 16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} + 15 \, {\left (4 \, d x + 4 \, c - \sin \left (4 \, d x + 4 \, c\right )\right )} a b}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/240*(80*b^2*sin(d*x + c)^3 - 16*(3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^2 + 15*(4*d*x + 4*c - sin(4*d*x + 4*
c))*a*b)/d

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mupad [B]  time = 0.79, size = 101, normalized size = 0.95 \[ \frac {a^2\,\sin \left (c+d\,x\right )}{8\,d}+\frac {b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {a\,b\,x}{4}-\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}-\frac {a^2\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}-\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}-\frac {a\,b\,\sin \left (4\,c+4\,d\,x\right )}{16\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a*sin(c + d*x) + b*tan(c + d*x))^2,x)

[Out]

(a^2*sin(c + d*x))/(8*d) + (b^2*sin(c + d*x))/(4*d) + (a*b*x)/4 - (a^2*sin(3*c + 3*d*x))/(48*d) - (a^2*sin(5*c
+ 5*d*x))/(80*d) - (b^2*sin(3*c + 3*d*x))/(12*d) - (a*b*sin(4*c + 4*d*x))/(16*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \sin {\left (c + d x \right )} + b \tan {\left (c + d x \right )}\right )^{2} \cos ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a*sin(d*x+c)+b*tan(d*x+c))**2,x)

[Out]

Integral((a*sin(c + d*x) + b*tan(c + d*x))**2*cos(c + d*x)**3, x)

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