3.448 \(\int (a+b \cos (c+d x))^4 \sec ^7(c+d x) \, dx\)

Optimal. Leaf size=222 \[ \frac {7 a^3 b \tan (c+d x) \sec ^4(c+d x)}{15 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {a^2 \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^2}{6 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d} \]

[Out]

1/16*(5*a^4+36*a^2*b^2+8*b^4)*arctanh(sin(d*x+c))/d+4/5*a*b*(4*a^2+5*b^2)*tan(d*x+c)/d+1/16*(5*a^4+36*a^2*b^2+
8*b^4)*sec(d*x+c)*tan(d*x+c)/d+1/24*a^2*(5*a^2+32*b^2)*sec(d*x+c)^3*tan(d*x+c)/d+7/15*a^3*b*sec(d*x+c)^4*tan(d
*x+c)/d+1/6*a^2*(a+b*cos(d*x+c))^2*sec(d*x+c)^5*tan(d*x+c)/d+4/15*a*b*(4*a^2+5*b^2)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.38, antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 8, 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, 3768, 3770} \[ \frac {4 a b \left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {\left (36 a^2 b^2+5 a^4+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {\left (36 a^2 b^2+5 a^4+8 b^4\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {7 a^3 b \tan (c+d x) \sec ^4(c+d x)}{15 d}+\frac {a^2 \tan (c+d x) \sec ^5(c+d x) (a+b \cos (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^7,x]

[Out]

((5*a^4 + 36*a^2*b^2 + 8*b^4)*ArcTanh[Sin[c + d*x]])/(16*d) + (4*a*b*(4*a^2 + 5*b^2)*Tan[c + d*x])/(5*d) + ((5
*a^4 + 36*a^2*b^2 + 8*b^4)*Sec[c + d*x]*Tan[c + d*x])/(16*d) + (a^2*(5*a^2 + 32*b^2)*Sec[c + d*x]^3*Tan[c + d*
x])/(24*d) + (7*a^3*b*Sec[c + d*x]^4*Tan[c + d*x])/(15*d) + (a^2*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^5*Tan[c +
d*x])/(6*d) + (4*a*b*(4*a^2 + 5*b^2)*Tan[c + d*x]^3)/(15*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 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 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))^4 \sec ^7(c+d x) \, dx &=\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{6} \int (a+b \cos (c+d x)) \left (14 a^2 b+a \left (5 a^2+18 b^2\right ) \cos (c+d x)+3 b \left (a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {7 a^3 b \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{30} \int \left (-5 a^2 \left (5 a^2+32 b^2\right )-24 a b \left (4 a^2+5 b^2\right ) \cos (c+d x)-15 b^2 \left (a^2+2 b^2\right ) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a^2 \left (5 a^2+32 b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {7 a^3 b \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{120} \int \left (-96 a b \left (4 a^2+5 b^2\right )-15 \left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a^2 \left (5 a^2+32 b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {7 a^3 b \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {1}{5} \left (4 a b \left (4 a^2+5 b^2\right )\right ) \int \sec ^4(c+d x) \, dx-\frac {1}{8} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {7 a^3 b \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}-\frac {1}{16} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int \sec (c+d x) \, dx-\frac {\left (4 a b \left (4 a^2+5 b^2\right )\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {\left (5 a^4+36 a^2 b^2+8 b^4\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {a^2 \left (5 a^2+32 b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {7 a^3 b \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (a+b \cos (c+d x))^2 \sec ^5(c+d x) \tan (c+d x)}{6 d}+\frac {4 a b \left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 154, normalized size = 0.69 \[ \frac {15 \left (5 a^4+36 a^2 b^2+8 b^4\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (40 a^4 \sec ^5(c+d x)+64 a b \left (5 \left (2 a^2+b^2\right ) \tan ^2(c+d x)+15 \left (a^2+b^2\right )+3 a^2 \tan ^4(c+d x)\right )+10 a^2 \left (5 a^2+36 b^2\right ) \sec ^3(c+d x)+15 \left (5 a^4+36 a^2 b^2+8 b^4\right ) \sec (c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^7,x]

[Out]

(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*Sec[c +
d*x] + 10*a^2*(5*a^2 + 36*b^2)*Sec[c + d*x]^3 + 40*a^4*Sec[c + d*x]^5 + 64*a*b*(15*(a^2 + b^2) + 5*(2*a^2 + b
^2)*Tan[c + d*x]^2 + 3*a^2*Tan[c + d*x]^4)))/(240*d)

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fricas [A]  time = 1.10, size = 217, normalized size = 0.98 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (128 \, {\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{5} + 192 \, a^{3} b \cos \left (d x + c\right ) + 15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 40 \, a^{4} + 64 \, {\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^7,x, algorithm="fricas")

[Out]

1/480*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*
cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(128*(4*a^3*b + 5*a*b^3)*cos(d*x + c)^5 + 192*a^3*b*cos(d*x + c) + 1
5*(5*a^4 + 36*a^2*b^2 + 8*b^4)*cos(d*x + c)^4 + 40*a^4 + 64*(4*a^3*b + 5*a*b^3)*cos(d*x + c)^3 + 10*(5*a^4 + 3
6*a^2*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^6)

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giac [B]  time = 0.78, size = 592, normalized size = 2.67 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, 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 ) - 15 \, {\left (5 \, a^{4} + 36 \, 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 (165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, b^{4} \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 )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^7,x, algorithm="giac")

[Out]

1/240*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*lo
g(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(165*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 90
0*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*b^4*tan(1/2*d*x + 1/2*c)^11 + 25*a
^4*tan(1/2*d*x + 1/2*c)^9 + 2240*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 3520*a*b
^3*tan(1/2*d*x + 1/2*c)^9 - 360*b^4*tan(1/2*d*x + 1/2*c)^9 + 450*a^4*tan(1/2*d*x + 1/2*c)^7 - 4992*a^3*b*tan(1
/2*d*x + 1/2*c)^7 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*b^4*tan(1/2*d
*x + 1/2*c)^7 + 450*a^4*tan(1/2*d*x + 1/2*c)^5 + 4992*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 360*a^2*b^2*tan(1/2*d*x +
1/2*c)^5 + 5760*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 240*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*a^4*tan(1/2*d*x + 1/2*c)^3
- 2240*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3520*a*b^3*tan(1/2*d*x + 1/2*c)^3
- 360*b^4*tan(1/2*d*x + 1/2*c)^3 + 165*a^4*tan(1/2*d*x + 1/2*c) + 960*a^3*b*tan(1/2*d*x + 1/2*c) + 900*a^2*b^
2*tan(1/2*d*x + 1/2*c) + 960*a*b^3*tan(1/2*d*x + 1/2*c) + 120*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^
2 - 1)^6)/d

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maple [A]  time = 0.12, size = 302, normalized size = 1.36 \[ \frac {a^{4} \left (\sec ^{5}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{6 d}+\frac {5 a^{4} \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{24 d}+\frac {5 a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {32 a^{3} b \tan \left (d x +c \right )}{15 d}+\frac {4 a^{3} b \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {16 a^{3} b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {3 a^{2} b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {9 a^{2} b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{4 d}+\frac {9 a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {8 a \,b^{3} \tan \left (d x +c \right )}{3 d}+\frac {4 a \,b^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{4} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*sec(d*x+c)^7,x)

[Out]

1/6*a^4*sec(d*x+c)^5*tan(d*x+c)/d+5/24*a^4*sec(d*x+c)^3*tan(d*x+c)/d+5/16*a^4*sec(d*x+c)*tan(d*x+c)/d+5/16/d*a
^4*ln(sec(d*x+c)+tan(d*x+c))+32/15*a^3*b*tan(d*x+c)/d+4/5*a^3*b*sec(d*x+c)^4*tan(d*x+c)/d+16/15*a^3*b*sec(d*x+
c)^2*tan(d*x+c)/d+3/2/d*a^2*b^2*tan(d*x+c)*sec(d*x+c)^3+9/4/d*a^2*b^2*tan(d*x+c)*sec(d*x+c)+9/4/d*a^2*b^2*ln(s
ec(d*x+c)+tan(d*x+c))+8/3/d*a*b^3*tan(d*x+c)+4/3/d*a*b^3*tan(d*x+c)*sec(d*x+c)^2+1/2/d*b^4*tan(d*x+c)*sec(d*x+
c)+1/2/d*b^4*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.76, size = 275, normalized size = 1.24 \[ \frac {128 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} b + 640 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{3} - 5 \, a^{4} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, a^{2} b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^7,x, algorithm="maxima")

[Out]

1/480*(128*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3*b + 640*(tan(d*x + c)^3 + 3*tan(d*x +
c))*a*b^3 - 5*a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c
)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 180*a^2*b^2*(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)) - 120*b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 4.31, size = 370, normalized size = 1.67 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {5\,a^4}{8}+\frac {9\,a^2\,b^2}{2}+b^4\right )}{d}+\frac {\left (\frac {11\,a^4}{8}-8\,a^3\,b+\frac {15\,a^2\,b^2}{2}-8\,a\,b^3+b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,a^4}{24}+\frac {56\,a^3\,b}{3}-\frac {21\,a^2\,b^2}{2}+\frac {88\,a\,b^3}{3}-3\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {15\,a^4}{4}-\frac {208\,a^3\,b}{5}+3\,a^2\,b^2-48\,a\,b^3+2\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,a^4}{4}+\frac {208\,a^3\,b}{5}+3\,a^2\,b^2+48\,a\,b^3+2\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,a^4}{24}-\frac {56\,a^3\,b}{3}-\frac {21\,a^2\,b^2}{2}-\frac {88\,a\,b^3}{3}-3\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,a^4}{8}+8\,a^3\,b+\frac {15\,a^2\,b^2}{2}+8\,a\,b^3+b^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4/cos(c + d*x)^7,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((5*a^4)/8 + b^4 + (9*a^2*b^2)/2))/d + (tan(c/2 + (d*x)/2)^9*((88*a*b^3)/3 + (56*a^
3*b)/3 + (5*a^4)/24 - 3*b^4 - (21*a^2*b^2)/2) - tan(c/2 + (d*x)/2)^3*((88*a*b^3)/3 + (56*a^3*b)/3 - (5*a^4)/24
+ 3*b^4 + (21*a^2*b^2)/2) + tan(c/2 + (d*x)/2)^5*(48*a*b^3 + (208*a^3*b)/5 + (15*a^4)/4 + 2*b^4 + 3*a^2*b^2)
+ tan(c/2 + (d*x)/2)^7*((15*a^4)/4 - (208*a^3*b)/5 - 48*a*b^3 + 2*b^4 + 3*a^2*b^2) + tan(c/2 + (d*x)/2)*(8*a*b
^3 + 8*a^3*b + (11*a^4)/8 + b^4 + (15*a^2*b^2)/2) + tan(c/2 + (d*x)/2)^11*((11*a^4)/8 - 8*a^3*b - 8*a*b^3 + b^
4 + (15*a^2*b^2)/2))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c
/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

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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))**4*sec(d*x+c)**7,x)

[Out]

Timed out

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