3.436 \(\int (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=133 \[ \frac {b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a \left (a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{4 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d} \]

[Out]

3/8*a*(a^2+4*b^2)*arctanh(sin(d*x+c))/d+b*(2*a^2+b^2)*tan(d*x+c)/d+3/8*a*(a^2+4*b^2)*sec(d*x+c)*tan(d*x+c)/d+3
/4*a^2*b*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a^2*(a+b*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.20, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2792, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 a \left (a^2+4 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {3 a^2 b \tan (c+d x) \sec ^2(c+d x)}{4 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^5,x]

[Out]

(3*a*(a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) + (b*(2*a^2 + b^2)*Tan[c + d*x])/d + (3*a*(a^2 + 4*b^2)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (3*a^2*b*Sec[c + d*x]^2*Tan[c + d*x])/(4*d) + (a^2*(a + b*Cos[c + d*x])*Sec[c + d
*x]^3*Tan[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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 (a+b \cos (c+d x))^3 \sec ^5(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \left (9 a^2 b+3 a \left (a^2+4 b^2\right ) \cos (c+d x)+2 b \left (a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {3 a^2 b \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int \left (9 a \left (a^2+4 b^2\right )+12 b \left (2 a^2+b^2\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {3 a^2 b \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\left (b \left (2 a^2+b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{4} \left (3 a \left (a^2+4 b^2\right )\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {3 a \left (a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^2 b \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (3 a \left (a^2+4 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac {\left (b \left (2 a^2+b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac {3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {b \left (2 a^2+b^2\right ) \tan (c+d x)}{d}+\frac {3 a \left (a^2+4 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^2 b \sec ^2(c+d x) \tan (c+d x)}{4 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 90, normalized size = 0.68 \[ \frac {3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (2 a^3 \sec ^3(c+d x)+8 b \left (a^2 \tan ^2(c+d x)+3 a^2+b^2\right )+3 a \left (a^2+4 b^2\right ) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^5,x]

[Out]

(3*a*(a^2 + 4*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*a*(a^2 + 4*b^2)*Sec[c + d*x] + 2*a^3*Sec[c + d*x]^3
+ 8*b*(3*a^2 + b^2 + a^2*Tan[c + d*x]^2)))/(8*d)

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fricas [A]  time = 1.04, size = 140, normalized size = 1.05 \[ \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a^{2} b \cos \left (d x + c\right ) + 8 \, {\left (2 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, a^{3} + 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.69, size = 330, normalized size = 2.48 \[ \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/8*(3*(a^3 + 4*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(a^3 + 4*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) + 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 8*b
^3*tan(1/2*d*x + 1/2*c)^7 + 3*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 12*a*b^2*tan(1/2*
d*x + 1/2*c)^5 + 24*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*a^2*b*tan(1/2*d*x + 1/2*c)^
3 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*b^3*tan(1/2*d*x + 1/2*c)^3 + 5*a^3*tan(1/2*d*x + 1/2*c) + 24*a^2*b*ta
n(1/2*d*x + 1/2*c) + 12*a*b^2*tan(1/2*d*x + 1/2*c) + 8*b^3*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 = 0.11, size = 160, normalized size = 1.20 \[ \frac {a^{3} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a^{2} b \tan \left (d x +c \right )}{d}+\frac {a^{2} b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{d}+\frac {3 b^{2} a \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x)

[Out]

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

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maxima [A]  time = 0.80, size = 158, normalized size = 1.19 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b - a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, b^{3} \tan \left (d x + c\right )}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/16*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b - a^3*(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)) - 12*a*b^2*(2*sin(d*x + c)/(sin(d
*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 16*b^3*tan(d*x + c))/d

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mupad [B]  time = 4.22, size = 224, normalized size = 1.68 \[ \frac {\left (\frac {5\,a^3}{4}-6\,a^2\,b+3\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^3}{4}+10\,a^2\,b-3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^3}{4}-10\,a^2\,b-3\,a\,b^2-6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^3}{4}+6\,a^2\,b+3\,a\,b^2+2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2+4\,b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^3/cos(c + d*x)^5,x)

[Out]

(tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 6*a^2*b + (5*a^3)/4 - 2*b^3) - tan(c/2 + (d*x)/2)^3*(3*a*b^2 + 10*a^2*b - (3*
a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)^5*(10*a^2*b - 3*a*b^2 + (3*a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)*(3*a*b^2
+ 6*a^2*b + (5*a^3)/4 + 2*b^3))/(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*a*atanh(tan(c/2 + (d*x)/2))*(a^2 + 4*b^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((a+b*cos(d*x+c))**3*sec(d*x+c)**5,x)

[Out]

Timed out

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