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

Optimal. Leaf size=120 \[ \frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {b \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {3 a^2 b \cos ^4(c+d x)}{4 d}-\frac {3 a b^2 \cos (c+d x)}{d}-\frac {b^3 \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4397, 2837, 12, 894} \[ -\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {b \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}+\frac {3 a^2 b \cos ^4(c+d x)}{4 d}+\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {3 a b^2 \cos (c+d x)}{d}-\frac {b^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b^2*Cos[c + d*x])/d - (b*(3*a^2 - b^2)*Cos[c + d*x]^2)/(2*d) - (a*(a^2 - 3*b^2)*Cos[c + d*x]^3)/(3*d) +
(3*a^2*b*Cos[c + d*x]^4)/(4*d) + (a^3*Cos[c + d*x]^5)/(5*d) - (b^3*Log[Cos[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 ^2(c+d x) (a \sin (c+d x)+b \tan (c+d x))^3 \, dx &=\int (b+a \cos (c+d x))^3 \sin ^2(c+d x) \tan (c+d x) \, dx\\ &=-\frac {\operatorname {Subst}\left (\int \frac {a (b+x)^3 \left (a^2-x^2\right )}{x} \, 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} \, dx,x,a \cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (3 a^2 b^2+\frac {a^2 b^3}{x}+b \left (3 a^2-b^2\right ) x+\left (a^2-3 b^2\right ) x^2-3 b x^3-x^4\right ) \, dx,x,a \cos (c+d x)\right )}{a^2 d}\\ &=-\frac {3 a b^2 \cos (c+d x)}{d}-\frac {b \left (3 a^2-b^2\right ) \cos ^2(c+d x)}{2 d}-\frac {a \left (a^2-3 b^2\right ) \cos ^3(c+d x)}{3 d}+\frac {3 a^2 b \cos ^4(c+d x)}{4 d}+\frac {a^3 \cos ^5(c+d x)}{5 d}-\frac {b^3 \log (\cos (c+d x))}{d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 106, normalized size = 0.88 \[ -\frac {-\frac {1}{5} a^3 \cos ^5(c+d x)+\frac {1}{3} a \left (a^2-3 b^2\right ) \cos ^3(c+d x)+\frac {1}{2} b \left (3 a^2-b^2\right ) \cos ^2(c+d x)-\frac {3}{4} a^2 b \cos ^4(c+d x)+3 a b^2 \cos (c+d x)+b^3 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a*b^2*Cos[c + d*x] + (b*(3*a^2 - b^2)*Cos[c + d*x]^2)/2 + (a*(a^2 - 3*b^2)*Cos[c + d*x]^3)/3 - (3*a^2*b*C
os[c + d*x]^4)/4 - (a^3*Cos[c + d*x]^5)/5 + b^3*Log[Cos[c + d*x]])/d)

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fricas [A]  time = 0.69, size = 101, normalized size = 0.84 \[ \frac {12 \, a^{3} \cos \left (d x + c\right )^{5} + 45 \, a^{2} b \cos \left (d x + c\right )^{4} - 180 \, a b^{2} \cos \left (d x + c\right ) - 20 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} - 60 \, b^{3} \log \left (-\cos \left (d x + c\right )\right ) - 30 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^3*cos(d*x + c)^5 + 45*a^2*b*cos(d*x + c)^4 - 180*a*b^2*cos(d*x + c) - 20*(a^3 - 3*a*b^2)*cos(d*x +
c)^3 - 60*b^3*log(-cos(d*x + c)) - 30*(3*a^2*b - b^3)*cos(d*x + c)^2)/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)^2*(a*sin(d*x+c)+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.11, size = 128, normalized size = 1.07 \[ -\frac {a^{3} \left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{3}\left (d x +c \right )\right )}{5 d}-\frac {2 a^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{15 d}+\frac {3 a^{2} b \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {\cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right ) a \,b^{2}}{d}-\frac {2 a \,b^{2} \cos \left (d x +c \right )}{d}-\frac {b^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {b^{3} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/d*a^3*sin(d*x+c)^2*cos(d*x+c)^3-2/15*a^3*cos(d*x+c)^3/d+3/4/d*a^2*b*sin(d*x+c)^4-1/d*cos(d*x+c)*sin(d*x+c
)^2*a*b^2-2*a*b^2*cos(d*x+c)/d-1/2/d*b^3*sin(d*x+c)^2-b^3*ln(cos(d*x+c))/d

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maxima [A]  time = 0.46, size = 94, normalized size = 0.78 \[ \frac {45 \, a^{2} b \sin \left (d x + c\right )^{4} + 4 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} + 60 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a b^{2} - 30 \, {\left (\sin \left (d x + c\right )^{2} + \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} b^{3}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(45*a^2*b*sin(d*x + c)^4 + 4*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 + 60*(cos(d*x + c)^3 - 3*cos(d*x +
c))*a*b^2 - 30*(sin(d*x + c)^2 + log(sin(d*x + c)^2 - 1))*b^3)/d

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mupad [B]  time = 0.79, size = 237, normalized size = 1.98 \[ \frac {40\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3\,d}-\frac {4\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}-\frac {16\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d}+\frac {32\,a^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{5\,d}-\frac {2\,b^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{d}+\frac {2\,b^3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {2\,b^3\,\mathrm {atanh}\left (\frac {1}{{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}-1\right )}{d}-\frac {12\,a\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {12\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{d}+\frac {8\,a\,b^2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}-\frac {24\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{d}+\frac {12\,a^2\,b\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(40*a^3*cos(c/2 + (d*x)/2)^6)/(3*d) - (4*a^3*cos(c/2 + (d*x)/2)^4)/d - (16*a^3*cos(c/2 + (d*x)/2)^8)/d + (32*a
^3*cos(c/2 + (d*x)/2)^10)/(5*d) - (2*b^3*cos(c/2 + (d*x)/2)^2)/d + (2*b^3*cos(c/2 + (d*x)/2)^4)/d + (2*b^3*ata
nh(1/cos(c/2 + (d*x)/2)^2 - 1))/d - (12*a*b^2*cos(c/2 + (d*x)/2)^4)/d + (12*a^2*b*cos(c/2 + (d*x)/2)^4)/d + (8
*a*b^2*cos(c/2 + (d*x)/2)^6)/d - (24*a^2*b*cos(c/2 + (d*x)/2)^6)/d + (12*a^2*b*cos(c/2 + (d*x)/2)^8)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(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)**2, x)

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