∫ cos6(c + dx)(a + b tan2(c + dx)) dx

3.436 \(\int \cos ^6(c+d x) (a+b \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=87 \[ \frac {(a-b) \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {(5 a+b) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {(5 a+b) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x (5 a+b) \]

[Out]

1/16*(5*a+b)*x+1/16*(5*a+b)*cos(d*x+c)*sin(d*x+c)/d+1/24*(5*a+b)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*(a-b)*cos(d*x+c

)^5*sin(d*x+c)/d

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Rubi [A] time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3675, 385, 199, 203} \[ \frac {(a-b) \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {(5 a+b) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {(5 a+b) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x (5 a+b) \]

Antiderivative was successfully veriﬁed.

[In]

Int[Cos[c + d*x]^6*(a + b*Tan[c + d*x]^2),x]

[Out]

((5*a + b)*x)/16 + ((5*a + b)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + ((5*a + b)*Cos[c + d*x]^3*Sin[c + d*x])/(24*

d) + ((a - b)*Cos[c + d*x]^5*Sin[c + d*x])/(6*d)

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +

1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In

tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]

)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;

FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +

1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /

; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[

{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)

^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p

] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \cos ^6(c+d x) \left (a+b \tan ^2(c+d x)\right ) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {a+b x^2}{\left (1+x^2\right )^4} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac {(a-b) \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {(5 a+b) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3} \, dx,x,\tan (c+d x)\right )}{6 d}\\ &=\frac {(5 a+b) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {(a-b) \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {(5 a+b) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=\frac {(5 a+b) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {(5 a+b) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {(a-b) \cos ^5(c+d x) \sin (c+d x)}{6 d}+\frac {(5 a+b) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=\frac {1}{16} (5 a+b) x+\frac {(5 a+b) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {(5 a+b) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac {(a-b) \cos ^5(c+d x) \sin (c+d x)}{6 d}\\ \end {align*}

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Mathematica [A] time = 0.21, size = 74, normalized size = 0.85 \[ \frac {3 (15 a+b) \sin (2 (c+d x))+(9 a-3 b) \sin (4 (c+d x))+a \sin (6 (c+d x))+60 a c+60 a d x-b \sin (6 (c+d x))+12 b d x}{192 d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Tan[c + d*x]^2),x]

[Out]

(60*a*c + 60*a*d*x + 12*b*d*x + 3*(15*a + b)*Sin[2*(c + d*x)] + (9*a - 3*b)*Sin[4*(c + d*x)] + a*Sin[6*(c + d*

x)] - b*Sin[6*(c + d*x)])/(192*d)

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fricas [A] time = 0.53, size = 66, normalized size = 0.76 \[ \frac {3 \, {\left (5 \, a + b\right )} d x + {\left (8 \, {\left (a - b\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (5 \, a + b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, a + b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(3*(5*a + b)*d*x + (8*(a - b)*cos(d*x + c)^5 + 2*(5*a + b)*cos(d*x + c)^3 + 3*(5*a + b)*cos(d*x + c))*sin

(d*x + c))/d

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*tan(d*x+c)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/