3.5.82 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx\) [482]

Optimal. Leaf size=115 \[ 2 a b \left (a^2+2 b^2\right ) x+\frac {b^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d} \]

[Out]

2*a*b*(a^2+2*b^2)*x+b^4*arctanh(sin(d*x+c))/d+1/3*a^2*(2*a^2+17*b^2)*sin(d*x+c)/d+4/3*a^3*b*cos(d*x+c)*sin(d*x
+c)/d+1/3*a^2*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]
time = 0.17, antiderivative size = 115, 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, 4159, 4132, 8, 4130, 3855} \begin {gather*} \frac {4 a^3 b \sin (c+d x) \cos (c+d x)}{3 d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+2 a b x \left (a^2+2 b^2\right )+\frac {a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {b^4 \tanh ^{-1}(\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*(a^2 + 2*b^2)*x + (b^4*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Sin[c + d*x])/(3*d) + (4*a^3*b*C
os[c + d*x]*Sin[c + d*x])/(3*d) + (a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*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 4159

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[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (8 a^2 b+a \left (2 a^2+9 b^2\right ) \sec (c+d x)+3 b^3 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac {1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2+17 b^2\right )-12 a b \left (a^2+2 b^2\right ) \sec (c+d x)-6 b^4 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac {1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2+17 b^2\right )-6 b^4 \sec ^2(c+d x)\right ) \, dx+\left (2 a b \left (a^2+2 b^2\right )\right ) \int 1 \, dx\\ &=2 a b \left (a^2+2 b^2\right ) x+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+b^4 \int \sec (c+d x) \, dx\\ &=2 a b \left (a^2+2 b^2\right ) x+\frac {b^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac {4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac {a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 128, normalized size = 1.11 \begin {gather*} \frac {24 a b \left (a^2+2 b^2\right ) (c+d x)-12 b^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 a^2 \left (a^2+8 b^2\right ) \sin (c+d x)+12 a^3 b \sin (2 (c+d x))+a^4 \sin (3 (c+d x))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a*b*(a^2 + 2*b^2)*(c + d*x) - 12*b^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^4*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + 9*a^2*(a^2 + 8*b^2)*Sin[c + d*x] + 12*a^3*b*Sin[2*(c + d*x)] + a^4*Sin[3*(c + d*x)])/(
12*d)

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Maple [A]
time = 0.11, size = 98, normalized size = 0.85

method	result	size
derivativedivides	\(\frac {\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 b \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 b^{2} a^{2} \sin \left (d x +c \right )+4 b^{3} a \left (d x +c \right )+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(98\)
default	\(\frac {\frac {a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 b \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+6 b^{2} a^{2} \sin \left (d x +c \right )+4 b^{3} a \left (d x +c \right )+b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(98\)
risch	\(2 a^{3} b x +4 a \,b^{3} x -\frac {3 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} b^{2} a^{2}}{d}+\frac {3 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} b^{2} a^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{4}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{4}}{d}+\frac {a^{4} \sin \left (3 d x +3 c \right )}{12 d}+\frac {b \,a^{3} \sin \left (2 d x +2 c \right )}{d}\)	\(169\)
norman	\(\frac {\left (-2 b \,a^{3}-4 b^{3} a \right ) x +\left (-6 b \,a^{3}-12 b^{3} a \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 b \,a^{3}+4 b^{3} a \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 b \,a^{3}+12 b^{3} a \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 a^{2} \left (a^{2}-2 b a -6 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (a^{2}-2 b a +6 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (a^{2}+2 b a -6 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (a^{2}+2 b a +6 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (7 a^{2}-18 b a +18 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{2} \left (7 a^{2}+18 b a +18 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(360\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 102, normalized size = 0.89 \begin {gather*} -\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b - 24 \, {\left (d x + c\right )} a b^{3} - 3 \, b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, a^{2} b^{2} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3*b - 24*(d*x + c)*a*b^3
- 3*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*a^2*b^2*sin(d*x + c))/d

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Fricas [A]
time = 3.43, size = 98, normalized size = 0.85 \begin {gather*} \frac {3 \, b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \, {\left (a^{3} b + 2 \, a b^{3}\right )} d x + 2 \, {\left (a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{3} b \cos \left (d x + c\right ) + 2 \, a^{4} + 18 \, a^{2} b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*b^4*log(sin(d*x + c) + 1) - 3*b^4*log(-sin(d*x + c) + 1) + 12*(a^3*b + 2*a*b^3)*d*x + 2*(a^4*cos(d*x +
c)^2 + 6*a^3*b*cos(d*x + c) + 2*a^4 + 18*a^2*b^2)*sin(d*x + c))/d

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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 ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 212, normalized size = 1.84 \begin {gather*} \frac {3 \, b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, b^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 6 \, {\left (a^{3} b + 2 \, a b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, a^{2} 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}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.11, size = 158, normalized size = 1.37 \begin {gather*} \frac {3\,a^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,b^4\,\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}+\frac {a^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {8\,a\,b^3\,\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^3\,b\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a^4*sin(c + d*x))/(4*d) + (2*b^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (a^4*sin(3*c + 3*d*x))/(
12*d) + (a^3*b*sin(2*c + 2*d*x))/d + (6*a^2*b^2*sin(c + d*x))/d + (8*a*b^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (
d*x)/2)))/d + (4*a^3*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d

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