3.1.35 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^4 \, dx\) [35]

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

[Out]

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

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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 = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3876, 2717, 2715, 8, 3855, 3852} \begin {gather*} \frac {4 a^4 \sin (c+d x)}{d}+\frac {a^4 \tan (c+d x)}{d}+\frac {4 a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^4 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {13 a^4 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(241\) vs. \(2(73)=146\).
time = 1.80, size = 241, normalized size = 3.30 \begin {gather*} \frac {1}{64} a^4 (1+\cos (c+d x))^4 \sec ^8\left (\frac {1}{2} (c+d x)\right ) \left (26 x-\frac {16 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {16 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {16 \cos (d x) \sin (c)}{d}+\frac {\cos (2 d x) \sin (2 c)}{d}+\frac {16 \cos (c) \sin (d x)}{d}+\frac {\cos (2 c) \sin (2 d x)}{d}+\frac {4 \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 {4 \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]^2*(a + a*Sec[c + d*x])^4,x]

[Out]

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

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Maple [A]
time = 0.11, size = 82, normalized size = 1.12

method	result	size
derivativedivides	\(\frac {a^{4} \tan \left (d x +c \right )+4 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} \left (d x +c \right )+4 a^{4} \sin \left (d x +c \right )+a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(82\)
default	\(\frac {a^{4} \tan \left (d x +c \right )+4 a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} \left (d x +c \right )+4 a^{4} \sin \left (d x +c \right )+a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(82\)
risch	\(\frac {13 a^{4} x}{2}-\frac {i a^{4} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {2 i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {2 i a^{4} {\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {i a^{4} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i a^{4}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {4 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}\)	\(144\)
norman	\(\frac {-\frac {13 a^{4} x}{2}-\frac {11 a^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {20 a^{4} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{4} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {12 a^{4} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{4} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {13 a^{4} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+13 a^{4} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-13 a^{4} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {13 a^{4} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {13 a^{4} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {4 a^{4} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\)	\(258\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 85, normalized size = 1.16 \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^{4} + 8 \, a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{4} \sin \left (d x + c\right ) + 4 \, 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)^2*(a+a*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^4 + 8*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c)
- 1)) + 16*a^4*sin(d*x + c) + 4*a^4*tan(d*x + c))/d

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Fricas [A]
time = 2.94, size = 105, normalized size = 1.44 \begin {gather*} \frac {13 \, a^{4} d x \cos \left (d x + c\right ) + 4 \, a^{4} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a^{4} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{4} \cos \left (d x + c\right ) + 2 \, a^{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+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/2*(13*a^4*d*x*cos(d*x + c) + 4*a^4*cos(d*x + c)*log(sin(d*x + c) + 1) - 4*a^4*cos(d*x + c)*log(-sin(d*x + c)
+ 1) + (a^4*cos(d*x + c)^2 + 8*a^4*cos(d*x + c) + 2*a^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*} a^{4} \left (\int 4 \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\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)**2*(a+a*sec(d*x+c))**4,x)

[Out]

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

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Giac [A]
time = 0.47, size = 129, normalized size = 1.77 \begin {gather*} \frac {13 \, {\left (d x + c\right )} a^{4} + 8 \, a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, 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)^2*(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.89, size = 117, normalized size = 1.60 \begin {gather*} \frac {13\,a^4\,x}{2}+\frac {8\,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)^2*(a + a/cos(c + d*x))^4,x)

[Out]

(13*a^4*x)/2 + (8*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))

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