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

Optimal. Leaf size=112 \[ \frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a^2 b \cos ^3(c+d x)}{d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4397, 2837, 12, 894} \[ -\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}+\frac {a^2 b \cos ^3(c+d x)}{d}+\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(3*a^2 - b^2)*Cos[c + d*x])/d) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^2)/(2*d) + (a^2*b*Cos[c + d*x]^3)/d + (a^3
*Cos[c + d*x]^4)/(4*d) - (3*a*b^2*Log[Cos[c + d*x]])/d + (b^3*Sec[c + d*x])/d

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

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] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 (c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx &=\int (b+a \cos (c+d x))^3 \sin (c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a^2 (b+x)^3 \left (a^2-x^2\right )}{x^2} \, dx,x,a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(b+x)^3 \left (a^2-x^2\right )}{x^2} \, dx,x,a \cos (c+d x)\right )}{a d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (3 a^2 b \left (1-\frac {b^2}{3 a^2}\right )+\frac {a^2 b^3}{x^2}+\frac {3 a^2 b^2}{x}+\left (a^2-3 b^2\right ) x-3 b x^2-x^3\right ) \, dx,x,a \cos (c+d x)\right )}{a d}\\ &=-\frac {b \left (3 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {a^2 b \cos ^3(c+d x)}{d}+\frac {a^3 \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \log (\cos (c+d x))}{d}+\frac {b^3 \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 98, normalized size = 0.88 \[ \frac {-4 \left (a^3-6 a b^2\right ) \cos (2 (c+d x))+a^3 \cos (4 (c+d x))+8 b \left (4 b^2-9 a^2\right ) \cos (c+d x)+8 a^2 b \cos (3 (c+d x))-96 a b^2 \log (\cos (c+d x))+32 b^3 \sec (c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*(-9*a^2 + 4*b^2)*Cos[c + d*x] - 4*(a^3 - 6*a*b^2)*Cos[2*(c + d*x)] + 8*a^2*b*Cos[3*(c + d*x)] + a^3*Cos[4
*(c + d*x)] - 96*a*b^2*Log[Cos[c + d*x]] + 32*b^3*Sec[c + d*x])/(32*d)

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fricas [A]  time = 0.61, size = 128, normalized size = 1.14 \[ \frac {8 \, a^{3} \cos \left (d x + c\right )^{5} + 32 \, a^{2} b \cos \left (d x + c\right )^{4} - 96 \, a b^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) - 16 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 32 \, b^{3} - 32 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (5 \, a^{3} - 24 \, a b^{2}\right )} \cos \left (d x + c\right )}{32 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(8*a^3*cos(d*x + c)^5 + 32*a^2*b*cos(d*x + c)^4 - 96*a*b^2*cos(d*x + c)*log(-cos(d*x + c)) - 16*(a^3 - 3*
a*b^2)*cos(d*x + c)^3 + 32*b^3 - 32*(3*a^2*b - b^3)*cos(d*x + c)^2 + (5*a^3 - 24*a*b^2)*cos(d*x + c))/(d*cos(d
*x + c))

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

[Out]

Timed out

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maple [A]  time = 0.08, size = 147, normalized size = 1.31 \[ \frac {a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {\cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{d}-\frac {2 a^{2} b \cos \left (d x +c \right )}{d}-\frac {3 a \,b^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 a \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {b^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{d}+\frac {2 b^{3} \cos \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 87, normalized size = 0.78 \[ \frac {a^{3} \sin \left (d x + c\right )^{4} + 4 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} b - 6 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} a b^{2} + 4 \, b^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^3*sin(d*x + c)^4 + 4*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^2*b - 6*(sin(d*x + c)^2 + log(sin(d*x + c)^2 -
1))*a*b^2 + 4*b^3*(1/cos(d*x + c) + cos(d*x + c)))/d

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mupad [B]  time = 4.18, size = 225, normalized size = 2.01 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (4\,a^3+4\,a^2\,b-6\,a\,b^2+12\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (12\,a^2\,b+6\,a\,b^2-12\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (-4\,a^3+12\,a^2\,b+6\,a\,b^2+4\,b^3\right )-4\,a^2\,b+4\,b^3+6\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {6\,a\,b^2\,\mathrm {atanh}\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^4*(4*a^2*b - 6*a*b^2 + 4*a^3 + 12*b^3) - tan(c/2 + (d*x)/2)^2*(6*a*b^2 + 12*a^2*b - 12*b^3
) + tan(c/2 + (d*x)/2)^6*(6*a*b^2 + 12*a^2*b - 4*a^3 + 4*b^3) - 4*a^2*b + 4*b^3 + 6*a*b^2*tan(c/2 + (d*x)/2)^8
)/(d*(3*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 - tan(
c/2 + (d*x)/2)^10 + 1)) + (6*a*b^2*atanh(tan(c/2 + (d*x)/2)^2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

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