3.443 \(\int (a+b \cos (c+d x))^4 \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=114 \[ \frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}-\frac {2 a b \left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {b^2 \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} b^2 x \left (12 a^2+b^2\right )+\frac {a^2 \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

[Out]

1/2*b^2*(12*a^2+b^2)*x+4*a^3*b*arctanh(sin(d*x+c))/d-2*a*b*(a^2-2*b^2)*sin(d*x+c)/d-1/2*b^2*(2*a^2-b^2)*cos(d*
x+c)*sin(d*x+c)/d+a^2*(a+b*cos(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.23, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2792, 3033, 3023, 2735, 3770} \[ -\frac {2 a b \left (a^2-2 b^2\right ) \sin (c+d x)}{d}-\frac {b^2 \left (2 a^2-b^2\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} b^2 x \left (12 a^2+b^2\right )+\frac {4 a^3 b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x) (a+b \cos (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2,x]

[Out]

(b^2*(12*a^2 + b^2)*x)/2 + (4*a^3*b*ArcTanh[Sin[c + d*x]])/d - (2*a*b*(a^2 - 2*b^2)*Sin[c + d*x])/d - (b^2*(2*
a^2 - b^2)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a^2*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/d

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 3023

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

Rule 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

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

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Mathematica [A]  time = 0.66, size = 119, normalized size = 1.04 \[ \frac {4 a^4 \tan (c+d x)+2 b \left (-8 a^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+b \left (12 a^2+b^2\right ) (c+d x)\right )+16 a b^3 \sin (c+d x)+b^4 \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^4*Sec[c + d*x]^2,x]

[Out]

(2*b*(b*(12*a^2 + b^2)*(c + d*x) - 8*a^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*a^3*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + 16*a*b^3*Sin[c + d*x] + b^4*Sin[2*(c + d*x)] + 4*a^4*Tan[c + d*x])/(4*d)

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fricas [A]  time = 1.22, size = 116, normalized size = 1.02 \[ \frac {4 \, a^{3} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a^{3} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (12 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (b^{4} \cos \left (d x + c\right )^{2} + 8 \, a b^{3} \cos \left (d x + c\right ) + 2 \, a^{4}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/2*(4*a^3*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 4*a^3*b*cos(d*x + c)*log(-sin(d*x + c) + 1) + (12*a^2*b^2 +
b^4)*d*x*cos(d*x + c) + (b^4*cos(d*x + c)^2 + 8*a*b^3*cos(d*x + c) + 2*a^4)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 0.55, size = 170, normalized size = 1.49 \[ \frac {8 \, a^{3} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, a^{3} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + {\left (12 \, a^{2} b^{2} + b^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^2,x, algorithm="giac")

[Out]

1/2*(8*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*a^3*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 4*a^4*tan(1/2*d
*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (12*a^2*b^2 + b^4)*(d*x + c) + 2*(8*a*b^3*tan(1/2*d*x + 1/2*c)^3 -
b^4*tan(1/2*d*x + 1/2*c)^3 + 8*a*b^3*tan(1/2*d*x + 1/2*c) + b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2
+ 1)^2)/d

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maple [A]  time = 0.09, size = 109, normalized size = 0.96 \[ \frac {a^{4} \tan \left (d x +c \right )}{d}+\frac {4 a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+6 a^{2} b^{2} x +\frac {6 a^{2} b^{2} c}{d}+\frac {4 a \,b^{3} \sin \left (d x +c \right )}{d}+\frac {b^{4} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{4} x}{2}+\frac {b^{4} c}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^4*sec(d*x+c)^2,x)

[Out]

a^4*tan(d*x+c)/d+4/d*a^3*b*ln(sec(d*x+c)+tan(d*x+c))+6*a^2*b^2*x+6/d*a^2*b^2*c+4/d*a*b^3*sin(d*x+c)+1/2/d*b^4*
cos(d*x+c)*sin(d*x+c)+1/2*b^4*x+1/2/d*b^4*c

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maxima [A]  time = 0.63, size = 90, normalized size = 0.79 \[ \frac {24 \, {\left (d x + c\right )} a^{2} b^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4} + 8 \, a^{3} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a b^{3} \sin \left (d x + c\right ) + 4 \, a^{4} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^4*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

1/4*(24*(d*x + c)*a^2*b^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*b^4 + 8*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 16*a*b^3*sin(d*x + c) + 4*a^4*tan(d*x + c))/d

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mupad [B]  time = 0.71, size = 150, normalized size = 1.32 \[ \frac {b^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {12\,a^2\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {b^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {8\,a^3\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^4/cos(c + d*x)^2,x)

[Out]

(b^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a^4*sin(c + d*x))/(d*cos(c + d*x)) + (12*a^2*b^2*atan(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*a*b^3*sin(c + d*x))/d + (b^4*cos(c + d*x)*sin(c + d*x))/(2*d) +
(8*a^3*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \cos {\left (c + d x \right )}\right )^{4} \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**4*sec(d*x+c)**2,x)

[Out]

Integral((a + b*cos(c + d*x))**4*sec(c + d*x)**2, x)

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