3.429 \(\int \cos ^5(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=54 \[ \frac {(a-b) \sin ^5(c+d x)}{5 d}-\frac {(2 a-b) \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3676, 373} \[ \frac {(a-b) \sin ^5(c+d x)}{5 d}-\frac {(2 a-b) \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2),x]

[Out]

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

Rule 373

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n
)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]

Rule 3676

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 -
ff^2*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 52, normalized size = 0.96 \[ \frac {\sin (c+d x) (4 (7 a-2 b) \cos (2 (c+d x))+3 (a-b) \cos (4 (c+d x))+89 a+11 b)}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Tan[c + d*x]^2),x]

[Out]

((89*a + 11*b + 4*(7*a - 2*b)*Cos[2*(c + d*x)] + 3*(a - b)*Cos[4*(c + d*x)])*Sin[c + d*x])/(120*d)

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fricas [A]  time = 0.55, size = 47, normalized size = 0.87 \[ \frac {{\left (3 \, {\left (a - b\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} + 8 \, a + 2 \, b\right )} \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/15*(3*(a - b)*cos(d*x + c)^4 + (4*a + b)*cos(d*x + c)^2 + 8*a + 2*b)*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)^5*(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.80, size = 72, normalized size = 1.33 \[ \frac {\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+b \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*tan(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.71, size = 47, normalized size = 0.87 \[ \frac {3 \, {\left (a - b\right )} \sin \left (d x + c\right )^{5} - 5 \, {\left (2 \, a - b\right )} \sin \left (d x + c\right )^{3} + 15 \, a \sin \left (d x + c\right )}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.07, size = 71, normalized size = 1.31 \[ \frac {\frac {5\,a\,\sin \left (c+d\,x\right )}{8}+\frac {b\,\sin \left (c+d\,x\right )}{8}+\frac {5\,a\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {a\,\sin \left (5\,c+5\,d\,x\right )}{80}-\frac {b\,\sin \left (3\,c+3\,d\,x\right )}{48}-\frac {b\,\sin \left (5\,c+5\,d\,x\right )}{80}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + b*tan(c + d*x)^2),x)

[Out]

((5*a*sin(c + d*x))/8 + (b*sin(c + d*x))/8 + (5*a*sin(3*c + 3*d*x))/48 + (a*sin(5*c + 5*d*x))/80 - (b*sin(3*c
+ 3*d*x))/48 - (b*sin(5*c + 5*d*x))/80)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (c + d x \right )}\right ) \cos ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+b*tan(d*x+c)**2),x)

[Out]

Integral((a + b*tan(c + d*x)**2)*cos(c + d*x)**5, x)

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