3.446 \(\int (a+b \cos (c+d x))^4 \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=154 \[ \frac {5 a^3 b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}+\frac {\left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

1/8*(3*a^4+24*a^2*b^2+8*b^4)*arctanh(sin(d*x+c))/d+4/3*a*b*(2*a^2+3*b^2)*tan(d*x+c)/d+1/8*a^2*(3*a^2+22*b^2)*s
ec(d*x+c)*tan(d*x+c)/d+5/6*a^3*b*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/
d

________________________________________________________________________________________

Rubi [A]  time = 0.34, antiderivative size = 154, 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, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {\left (24 a^2 b^2+3 a^4+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {5 a^3 b \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {a^2 \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5,x]

[Out]

((3*a^4 + 24*a^2*b^2 + 8*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a*b*(2*a^2 + 3*b^2)*Tan[c + d*x])/(3*d) + (a^2
*(3*a^2 + 22*b^2)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (5*a^3*b*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + (a^2*(a + b
*Cos[c + d*x])^2*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 3031

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

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))^4 \sec ^5(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x)) \left (10 a^2 b+3 a \left (a^2+4 b^2\right ) \cos (c+d x)+b \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{12} \int \left (-3 a^2 \left (3 a^2+22 b^2\right )-16 a b \left (2 a^2+3 b^2\right ) \cos (c+d x)-3 b^2 \left (a^2+4 b^2\right ) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-32 a b \left (2 a^2+3 b^2\right )-3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} \left (4 a b \left (2 a^2+3 b^2\right )\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (-3 a^4-24 a^2 b^2-8 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (4 a b \left (2 a^2+3 b^2\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {\left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {4 a b \left (2 a^2+3 b^2\right ) \tan (c+d x)}{3 d}+\frac {a^2 \left (3 a^2+22 b^2\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {5 a^3 b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.49, size = 101, normalized size = 0.66 \[ \frac {3 \left (3 a^4+24 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) \left (6 a^3 \sec ^3(c+d x)+32 b \left (3 \left (a^2+b^2\right )+a^2 \tan ^2(c+d x)\right )+9 a \left (a^2+8 b^2\right ) \sec (c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^5,x]

[Out]

(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*ArcTanh[Sin[c + d*x]] + a*Tan[c + d*x]*(9*a*(a^2 + 8*b^2)*Sec[c + d*x] + 6*a^3
*Sec[c + d*x]^3 + 32*b*(3*(a^2 + b^2) + a^2*Tan[c + d*x]^2)))/(24*d)

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 163, normalized size = 1.06 \[ \frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, a^{3} b \cos \left (d x + c\right ) + 6 \, a^{4} + 32 \, {\left (2 \, a^{3} b + 3 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (a^{4} + 8 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*cos
(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a^3*b*cos(d*x + c) + 6*a^4 + 32*(2*a^3*b + 3*a*b^3)*cos(d*x + c)^3
+ 9*(a^4 + 8*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

giac [B]  time = 0.70, size = 360, normalized size = 2.34 \[ \frac {3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, a^{4} + 24 \, a^{2} b^{2} + 8 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 288 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 96 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*a^4 + 24*a^2*b^2 + 8*b^4)*log(a
bs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*a^4*tan(1/2*d*x + 1/2*c)^7 - 96*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 72*a^2*b^
2*tan(1/2*d*x + 1/2*c)^7 - 96*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*a^4*tan(1/2*d*x + 1/2*c)^5 + 160*a^3*b*tan(1/2*
d*x + 1/2*c)^5 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 288*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 9*a^4*tan(1/2*d*x + 1/
2*c)^3 - 160*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 288*a*b^3*tan(1/2*d*x + 1/2*c)
^3 + 15*a^4*tan(1/2*d*x + 1/2*c) + 96*a^3*b*tan(1/2*d*x + 1/2*c) + 72*a^2*b^2*tan(1/2*d*x + 1/2*c) + 96*a*b^3*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

________________________________________________________________________________________

maple [A]  time = 0.11, size = 188, normalized size = 1.22 \[ \frac {a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 a^{3} b \tan \left (d x +c \right )}{3 d}+\frac {4 a^{3} b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {3 a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {3 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a \,b^{3} \tan \left (d x +c \right )}{d}+\frac {b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.98, size = 187, normalized size = 1.21 \[ \frac {64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} b - 3 \, a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 192 \, a b^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^3*b - 3*a^4*(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)) - 72*a^2*b^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*b^4*(log(sin(d*x + c) + 1) - log(sin(
d*x + c) - 1)) + 192*a*b^3*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.29, size = 245, normalized size = 1.59 \[ \frac {\left (\frac {5\,a^4}{4}-8\,a^3\,b+6\,a^2\,b^2-8\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^4}{4}+\frac {40\,a^3\,b}{3}-6\,a^2\,b^2+24\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,a^4}{4}-\frac {40\,a^3\,b}{3}-6\,a^2\,b^2-24\,a\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^4}{4}+8\,a^3\,b+6\,a^2\,b^2+8\,a\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,a^4}{4}+6\,a^2\,b^2+2\,b^4\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4/cos(c + d*x)^5,x)

[Out]

(tan(c/2 + (d*x)/2)*(8*a*b^3 + 8*a^3*b + (5*a^4)/4 + 6*a^2*b^2) - tan(c/2 + (d*x)/2)^7*(8*a*b^3 + 8*a^3*b - (5
*a^4)/4 - 6*a^2*b^2) - tan(c/2 + (d*x)/2)^3*(24*a*b^3 + (40*a^3*b)/3 - (3*a^4)/4 + 6*a^2*b^2) + tan(c/2 + (d*x
)/2)^5*(24*a*b^3 + (40*a^3*b)/3 + (3*a^4)/4 - 6*a^2*b^2))/(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)) + (atanh(tan(c/2 + (d*x)/2))*((3*a^4)/4 + 2*b^4 + 6*a^2*
b^2))/d

________________________________________________________________________________________

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))**4*sec(d*x+c)**5,x)

[Out]

Timed out

________________________________________________________________________________________