3.225 \(\int (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=85 \[ \frac {3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 B \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan ^3(c+d x)}{3 d}+\frac {C \tan (c+d x)}{d} \]

[Out]

3/8*B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d+3/8*B*sec(d*x+c)*tan(d*x+c)/d+1/4*B*sec(d*x+c)^3*tan(d*x+c)/d+1/3*C
*tan(d*x+c)^3/d

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Rubi [A]  time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3010, 2748, 3768, 3770, 3767} \[ \frac {3 B \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 B \tan (c+d x) \sec (c+d x)}{8 d}+\frac {C \tan ^3(c+d x)}{3 d}+\frac {C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(3*B*ArcTanh[Sin[c + d*x]])/(8*d) + (C*Tan[c + d*x])/d + (3*B*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (B*Sec[c + d*
x]^3*Tan[c + d*x])/(4*d) + (C*Tan[c + d*x]^3)/(3*d)

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 76, normalized size = 0.89 \[ \frac {B \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 B \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d}+\frac {C \left (\frac {1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*B*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x]))/(8*d) + (C*(
Tan[c + d*x] + Tan[c + d*x]^3/3))/d

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fricas [A]  time = 0.46, size = 99, normalized size = 1.16 \[ \frac {9 \, B \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, B \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, C \cos \left (d x + c\right )^{3} + 9 \, B \cos \left (d x + c\right )^{2} + 8 \, C \cos \left (d x + c\right ) + 6 \, B\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/48*(9*B*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*B*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*C*cos(d*x +
c)^3 + 9*B*cos(d*x + c)^2 + 8*C*cos(d*x + c) + 6*B)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.60, size = 164, normalized size = 1.93 \[ \frac {9 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/24*(9*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*B*tan(1/2*d*x +
1/2*c)^7 - 24*C*tan(1/2*d*x + 1/2*c)^7 + 9*B*tan(1/2*d*x + 1/2*c)^5 + 40*C*tan(1/2*d*x + 1/2*c)^5 + 9*B*tan(1/
2*d*x + 1/2*c)^3 - 40*C*tan(1/2*d*x + 1/2*c)^3 + 15*B*tan(1/2*d*x + 1/2*c) + 24*C*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.28, size = 92, normalized size = 1.08 \[ \frac {2 C \tan \left (d x +c \right )}{3 d}+\frac {C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

2/3*C*tan(d*x+c)/d+1/3/d*C*tan(d*x+c)*sec(d*x+c)^2+1/4*B*sec(d*x+c)^3*tan(d*x+c)/d+3/8*B*sec(d*x+c)*tan(d*x+c)
/d+3/8/d*B*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.48, size = 95, normalized size = 1.12 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C - 3 \, B {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C - 3*B*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*
sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 3.53, size = 150, normalized size = 1.76 \[ \frac {\left (\frac {5\,B}{4}-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,B}{4}+\frac {10\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,B}{4}-\frac {10\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,B}{4}+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^6,x)

[Out]

(tan(c/2 + (d*x)/2)^7*((5*B)/4 - 2*C) + tan(c/2 + (d*x)/2)^3*((3*B)/4 - (10*C)/3) + tan(c/2 + (d*x)/2)^5*((3*B
)/4 + (10*C)/3) + tan(c/2 + (d*x)/2)*((5*B)/4 + 2*C))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*
tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*B*atanh(tan(c/2 + (d*x)/2)))/(4*d)

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sympy [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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

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