3.435 \(\int (a+b \cos (c+d x))^3 \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=109 \[ \frac {a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{3 d}+\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{6 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d} \]

[Out]

1/2*b*(3*a^2+2*b^2)*arctanh(sin(d*x+c))/d+1/3*a*(2*a^2+9*b^2)*tan(d*x+c)/d+7/6*a^2*b*sec(d*x+c)*tan(d*x+c)/d+1
/3*a^2*(a+b*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2792, 3021, 2748, 3767, 8, 3770} \[ \frac {a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{3 d}+\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {7 a^2 b \tan (c+d x) \sec (c+d x)}{6 d}+\frac {a^2 \tan (c+d x) \sec ^2(c+d x) (a+b \cos (c+d x))}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4,x]

[Out]

(b*(3*a^2 + 2*b^2)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*a^2 + 9*b^2)*Tan[c + d*x])/(3*d) + (7*a^2*b*Sec[c + d*
x]*Tan[c + d*x])/(6*d) + (a^2*(a + b*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 ^4(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int \left (7 a^2 b+a \left (2 a^2+9 b^2\right ) \cos (c+d x)+b \left (a^2+3 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {7 a^2 b \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (2 a \left (2 a^2+9 b^2\right )+3 b \left (3 a^2+2 b^2\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {7 a^2 b \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (b \left (3 a^2+2 b^2\right )\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (a \left (2 a^2+9 b^2\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {7 a^2 b \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac {\left (a \left (2 a^2+9 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a \left (2 a^2+9 b^2\right ) \tan (c+d x)}{3 d}+\frac {7 a^2 b \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x)) \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.26, size = 70, normalized size = 0.64 \[ \frac {\left (9 a^2 b+6 b^3\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) \left (2 a^2 \tan ^2(c+d x)+6 a^2+9 a b \sec (c+d x)+18 b^2\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4,x]

[Out]

((9*a^2*b + 6*b^3)*ArcTanh[Sin[c + d*x]] + a*Tan[c + d*x]*(6*a^2 + 18*b^2 + 9*a*b*Sec[c + d*x] + 2*a^2*Tan[c +
d*x]^2))/(6*d)

________________________________________________________________________________________

fricas [A]  time = 1.08, size = 126, normalized size = 1.16 \[ \frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (9 \, a^{2} b \cos \left (d x + c\right ) + 2 \, a^{3} + 2 \, {\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(3*(3*a^2*b + 2*b^3)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(3*a^2*b + 2*b^3)*cos(d*x + c)^3*log(-sin(d
*x + c) + 1) + 2*(9*a^2*b*cos(d*x + c) + 2*a^3 + 2*(2*a^3 + 9*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x
+ c)^3)

________________________________________________________________________________________

giac [B]  time = 0.55, size = 205, normalized size = 1.88 \[ \frac {3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(3*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) - 2*(6*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*a*b^2*tan(1/2*d*x + 1/2*c)^5 -
4*a^3*tan(1/2*d*x + 1/2*c)^3 - 36*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*a^3*tan(1/2*d*x + 1/2*c) + 9*a^2*b*tan(1/2*
d*x + 1/2*c) + 18*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

________________________________________________________________________________________

maple [A]  time = 0.10, size = 118, normalized size = 1.08 \[ \frac {2 a^{3} \tan \left (d x +c \right )}{3 d}+\frac {a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 b^{2} a \tan \left (d x +c \right )}{d}+\frac {b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.83, size = 113, normalized size = 1.04 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 9 \, a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, a b^{2} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3 - 9*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c
) + 1) + log(sin(d*x + c) - 1)) + 6*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*a*b^2*tan(d*x + c
))/d

________________________________________________________________________________________

mupad [B]  time = 2.63, size = 157, normalized size = 1.44 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,a^2\,b+2\,b^3\right )}{d}-\frac {\left (2\,a^3-3\,a^2\,b+6\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,a^3}{3}-12\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^3+3\,a^2\,b+6\,a\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^3/cos(c + d*x)^4,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*(3*a^2*b + 2*b^3))/d - (tan(c/2 + (d*x)/2)^5*(6*a*b^2 - 3*a^2*b + 2*a^3) - tan(c/2
+ (d*x)/2)^3*(12*a*b^2 + (4*a^3)/3) + tan(c/2 + (d*x)/2)*(6*a*b^2 + 3*a^2*b + 2*a^3))/(d*(3*tan(c/2 + (d*x)/2)
^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________