3.1.34 \(\int \cos (c+d x) (a+a \sec (c+d x))^4 \, dx\) [34]

Optimal. Leaf size=73 \[ 4 a^4 x+\frac {13 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^4 \sin (c+d x)}{d}+\frac {4 a^4 \tan (c+d x)}{d}+\frac {a^4 \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

4*a^4*x+13/2*a^4*arctanh(sin(d*x+c))/d+a^4*sin(d*x+c)/d+4*a^4*tan(d*x+c)/d+1/2*a^4*sec(d*x+c)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3876, 2717, 3855, 3852, 8, 3853} \begin {gather*} \frac {a^4 \sin (c+d x)}{d}+\frac {4 a^4 \tan (c+d x)}{d}+\frac {13 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^4 \tan (c+d x) \sec (c+d x)}{2 d}+4 a^4 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

4*a^4*x + (13*a^4*ArcTanh[Sin[c + d*x]])/(2*d) + (a^4*Sin[c + d*x])/d + (4*a^4*Tan[c + d*x])/d + (a^4*Sec[c +
d*x]*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3852

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 3853

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 3855

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

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (4 a^4+a^4 \cos (c+d x)+6 a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+a^4 \sec ^3(c+d x)\right ) \, dx\\ &=4 a^4 x+a^4 \int \cos (c+d x) \, dx+a^4 \int \sec ^3(c+d x) \, dx+\left (4 a^4\right ) \int \sec ^2(c+d x) \, dx+\left (6 a^4\right ) \int \sec (c+d x) \, dx\\ &=4 a^4 x+\frac {6 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^4 \sin (c+d x)}{d}+\frac {a^4 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^4 \int \sec (c+d x) \, dx-\frac {\left (4 a^4\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=4 a^4 x+\frac {13 a^4 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^4 \sin (c+d x)}{d}+\frac {4 a^4 \tan (c+d x)}{d}+\frac {a^4 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(73)=146\).
time = 1.39, size = 272, normalized size = 3.73 \begin {gather*} \frac {1}{64} a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (16 x-\frac {26 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {26 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 \cos (d x) \sin (c)}{d}+\frac {4 \cos (c) \sin (d x)}{d}+\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {16 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {1}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {16 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*Sec[(c + d*x)/2]^8*(16*x - (26*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (26*Log
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*Cos[d*x]*Sin[c])/d + (4*Cos[c]*Sin[d*x])/d + 1/(d*(Cos[(c + d*x)
/2] - Sin[(c + d*x)/2])^2) + (16*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))
- 1/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (16*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]))))/64

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 91, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} \tan \left (d x +c \right )+6 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} \left (d x +c \right )+a^{4} \sin \left (d x +c \right )}{d}\)	\(91\)
default	\(\frac {a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} \tan \left (d x +c \right )+6 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} \left (d x +c \right )+a^{4} \sin \left (d x +c \right )}{d}\)	\(91\)
risch	\(4 a^{4} x -\frac {i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left ({\mathrm e}^{3 i \left (d x +c \right )}-8 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-8\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\)	\(141\)
norman	\(\frac {-4 a^{4} x -\frac {11 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {13 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {3 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+8 a^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 a^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 a^{4} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {13 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {13 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 110, normalized size = 1.51 \begin {gather*} \frac {16 \, {\left (d x + c\right )} a^{4} - a^{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 )} + 12 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{4} \sin \left (d x + c\right ) + 16 \, a^{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)*(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/4*(16*(d*x + c)*a^4 - a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) -
1)) + 12*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*a^4*sin(d*x + c) + 16*a^4*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]
time = 3.12, size = 111, normalized size = 1.52 \begin {gather*} \frac {16 \, a^{4} d x \cos \left (d x + c\right )^{2} + 13 \, a^{4} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 13 \, a^{4} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} \cos \left (d x + c\right ) + a^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(16*a^4*d*x*cos(d*x + c)^2 + 13*a^4*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 13*a^4*cos(d*x + c)^2*log(-sin(
d*x + c) + 1) + 2*(2*a^4*cos(d*x + c)^2 + 8*a^4*cos(d*x + c) + a^4)*sin(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{4} \left (\int 4 \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \cos {\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**4*(Integral(4*cos(c + d*x)*sec(c + d*x), x) + Integral(6*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(4*cos(
c + d*x)*sec(c + d*x)**3, x) + Integral(cos(c + d*x)*sec(c + d*x)**4, x) + Integral(cos(c + d*x), x))

________________________________________________________________________________________

Giac [A]
time = 0.49, size = 129, normalized size = 1.77 \begin {gather*} \frac {8 \, {\left (d x + c\right )} a^{4} + 13 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 13 \, a^{4} \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} - \frac {2 \, {\left (7 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, a^{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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*(d*x + c)*a^4 + 13*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 13*a^4*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) - 2*(7*a^4*tan(1/2*d*x + 1/2*c)^3 - 9*a^4*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

________________________________________________________________________________________

Mupad [B]
time = 0.91, size = 115, normalized size = 1.58 \begin {gather*} 4\,a^4\,x+\frac {13\,a^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {5\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-11\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a^4*x + (13*a^4*atanh(tan(c/2 + (d*x)/2)))/d + (2*a^4*tan(c/2 + (d*x)/2)^3 + 5*a^4*tan(c/2 + (d*x)/2)^5 - 11
*a^4*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 + tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^6 - 1))

________________________________________________________________________________________