3.5.81 \(\int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \, dx\) [481]

Optimal. Leaf size=108 \[ \frac {1}{2} a^2 \left (a^2+12 b^2\right ) x+\frac {4 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {3 a^3 b \sin (c+d x)}{d}+\frac {a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 108, 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 = {3926, 4161, 4132, 8, 4130, 3855} \begin {gather*} \frac {3 a^3 b \sin (c+d x)}{d}-\frac {b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\frac {1}{2} a^2 x \left (a^2+12 b^2\right )+\frac {a^2 \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac {4 a b^3 \tanh ^{-1}(\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3926

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

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4161

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f
*x])^n/(f*(n + 2))), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1
) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C
, n}, x] &&  !LtQ[n, -1]

Rubi steps

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

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Mathematica [A]
time = 0.71, size = 119, normalized size = 1.10 \begin {gather*} \frac {2 a \left (a \left (a^2+12 b^2\right ) (c+d x)-8 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+16 a^3 b \sin (c+d x)+a^4 \sin (2 (c+d x))+4 b^4 \tan (c+d x)}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 87, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 b \,a^{3} \sin \left (d x +c \right )+6 b^{2} a^{2} \left (d x +c \right )+4 b^{3} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \tan \left (d x +c \right )}{d}\)	\(87\)
default	\(\frac {a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 b \,a^{3} \sin \left (d x +c \right )+6 b^{2} a^{2} \left (d x +c \right )+4 b^{3} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+b^{4} \tan \left (d x +c \right )}{d}\)	\(87\)
risch	\(\frac {a^{4} x}{2}+6 a^{2} b^{2} x -\frac {i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i b \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i b \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i b^{4}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {4 b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {4 b^{3} a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(157\)
norman	\(\frac {\left (-\frac {1}{2} a^{4}-6 b^{2} a^{2}\right ) x +\left (-\frac {1}{2} a^{4}-6 b^{2} a^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{4}+6 b^{2} a^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {1}{2} a^{4}+6 b^{2} a^{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-a^{4}-12 b^{2} a^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{4}+12 b^{2} a^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (3 a^{4}-2 b^{4}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (a^{4}-8 b \,a^{3}+2 b^{4}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (a^{4}+8 b \,a^{3}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{3} \left (a -4 b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (4 b +a \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4 b^{3} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {4 b^{3} a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 90, normalized size = 0.83 \begin {gather*} \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 24 \, {\left (d x + c\right )} a^{2} b^{2} + 8 \, a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{3} b \sin \left (d x + c\right ) + 4 \, b^{4} \tan \left (d x + c\right )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.49, size = 116, normalized size = 1.07 \begin {gather*} \frac {4 \, a b^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a b^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} + 12 \, a^{2} b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{3} b \cos \left (d x + c\right ) + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 170, normalized size = 1.57 \begin {gather*} \frac {8 \, a b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, a b^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, b^{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 (a^{4} + 12 \, a^{2} b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{3} b \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.03, size = 150, normalized size = 1.39 \begin {gather*} \frac {a^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 {b^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 {a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {4\,a^3\,b\,\sin \left (c+d\,x\right )}{d}+\frac {8\,a\,b^3\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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