3.78 \(\int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx\)

Optimal. Leaf size=126 \[ \frac {664 \sin (c+d x)}{105 a^4 d}-\frac {4 \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {88 \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {4 x}{a^4}-\frac {12 \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

-4*x/a^4+664/105*sin(d*x+c)/a^4/d-88/105*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2-4*sin(d*x+c)/a^4/d/(1+sec(d*x+c))-1
/7*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-12/35*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.30, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3817, 4020, 3787, 2637, 8} \[ \frac {664 \sin (c+d x)}{105 a^4 d}-\frac {4 \sin (c+d x)}{a^4 d (\sec (c+d x)+1)}-\frac {88 \sin (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}-\frac {4 x}{a^4}-\frac {12 \sin (c+d x)}{35 a d (a \sec (c+d x)+a)^3}-\frac {\sin (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*x)/a^4 + (664*Sin[c + d*x])/(105*a^4*d) - (88*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*Sin[c +
d*x])/(a^4*d*(1 + Sec[c + d*x])) - Sin[c + d*x]/(7*d*(a + a*Sec[c + d*x])^4) - (12*Sin[c + d*x])/(35*a*d*(a +
a*Sec[c + d*x])^3)

Rule 8

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

Rule 2637

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

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3817

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

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^4} \, dx &=-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {\int \frac {\cos (c+d x) (-8 a+4 a \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) \left (-52 a^2+36 a^2 \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {88 \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) \left (-244 a^3+176 a^3 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {88 \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {\int \cos (c+d x) \left (-664 a^4+420 a^4 \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {88 \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {4 \int 1 \, dx}{a^4}+\frac {664 \int \cos (c+d x) \, dx}{105 a^4}\\ &=-\frac {4 x}{a^4}+\frac {664 \sin (c+d x)}{105 a^4 d}-\frac {88 \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {\sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {12 \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3}-\frac {4 \sin (c+d x)}{d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 0.46, size = 263, normalized size = 2.09 \[ -\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (46130 \sin \left (c+\frac {d x}{2}\right )-46116 \sin \left (c+\frac {3 d x}{2}\right )+18060 \sin \left (2 c+\frac {3 d x}{2}\right )-19292 \sin \left (2 c+\frac {5 d x}{2}\right )+2100 \sin \left (3 c+\frac {5 d x}{2}\right )-3791 \sin \left (3 c+\frac {7 d x}{2}\right )-735 \sin \left (4 c+\frac {7 d x}{2}\right )-105 \sin \left (4 c+\frac {9 d x}{2}\right )-105 \sin \left (5 c+\frac {9 d x}{2}\right )+29400 d x \cos \left (c+\frac {d x}{2}\right )+17640 d x \cos \left (c+\frac {3 d x}{2}\right )+17640 d x \cos \left (2 c+\frac {3 d x}{2}\right )+5880 d x \cos \left (2 c+\frac {5 d x}{2}\right )+5880 d x \cos \left (3 c+\frac {5 d x}{2}\right )+840 d x \cos \left (3 c+\frac {7 d x}{2}\right )+840 d x \cos \left (4 c+\frac {7 d x}{2}\right )-60830 \sin \left (\frac {d x}{2}\right )+29400 d x \cos \left (\frac {d x}{2}\right )\right )}{26880 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/26880*(Sec[c/2]*Sec[(c + d*x)/2]^7*(29400*d*x*Cos[(d*x)/2] + 29400*d*x*Cos[c + (d*x)/2] + 17640*d*x*Cos[c +
(3*d*x)/2] + 17640*d*x*Cos[2*c + (3*d*x)/2] + 5880*d*x*Cos[2*c + (5*d*x)/2] + 5880*d*x*Cos[3*c + (5*d*x)/2] +
840*d*x*Cos[3*c + (7*d*x)/2] + 840*d*x*Cos[4*c + (7*d*x)/2] - 60830*Sin[(d*x)/2] + 46130*Sin[c + (d*x)/2] - 4
6116*Sin[c + (3*d*x)/2] + 18060*Sin[2*c + (3*d*x)/2] - 19292*Sin[2*c + (5*d*x)/2] + 2100*Sin[3*c + (5*d*x)/2]
- 3791*Sin[3*c + (7*d*x)/2] - 735*Sin[4*c + (7*d*x)/2] - 105*Sin[4*c + (9*d*x)/2] - 105*Sin[5*c + (9*d*x)/2]))
/(a^4*d)

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fricas [A]  time = 0.89, size = 162, normalized size = 1.29 \[ -\frac {420 \, d x \cos \left (d x + c\right )^{4} + 1680 \, d x \cos \left (d x + c\right )^{3} + 2520 \, d x \cos \left (d x + c\right )^{2} + 1680 \, d x \cos \left (d x + c\right ) + 420 \, d x - {\left (105 \, \cos \left (d x + c\right )^{4} + 1184 \, \cos \left (d x + c\right )^{3} + 2636 \, \cos \left (d x + c\right )^{2} + 2236 \, \cos \left (d x + c\right ) + 664\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

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]

-1/105*(420*d*x*cos(d*x + c)^4 + 1680*d*x*cos(d*x + c)^3 + 2520*d*x*cos(d*x + c)^2 + 1680*d*x*cos(d*x + c) + 4
20*d*x - (105*cos(d*x + c)^4 + 1184*cos(d*x + c)^3 + 2636*cos(d*x + c)^2 + 2236*cos(d*x + c) + 664)*sin(d*x +
c))/(a^4*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [A]  time = 0.50, size = 112, normalized size = 0.89 \[ -\frac {\frac {3360 \, {\left (d x + c\right )}}{a^{4}} - \frac {1680 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} + \frac {15 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5145 \, a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

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]

-1/840*(3360*(d*x + c)/a^4 - 1680*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^4) + (15*a^24*tan(1/2*d
*x + 1/2*c)^7 - 147*a^24*tan(1/2*d*x + 1/2*c)^5 + 805*a^24*tan(1/2*d*x + 1/2*c)^3 - 5145*a^24*tan(1/2*d*x + 1/
2*c))/a^28)/d

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maple [A]  time = 0.60, size = 126, normalized size = 1.00 \[ -\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{56 d \,a^{4}}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {23 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/56/d/a^4*tan(1/2*d*x+1/2*c)^7+7/40/d/a^4*tan(1/2*d*x+1/2*c)^5-23/24/d/a^4*tan(1/2*d*x+1/2*c)^3+49/8/d/a^4*t
an(1/2*d*x+1/2*c)+2/d/a^4*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-8/d/a^4*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.80, size = 158, normalized size = 1.25 \[ \frac {\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}}{840 \, d} \]

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]

1/840*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x
+ c)/(cos(d*x + c) + 1) - 805*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 -
15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4)/d

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mupad [B]  time = 0.75, size = 137, normalized size = 1.09 \[ -\frac {15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-192\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+1144\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-6112\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-1680\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+3360\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (c+d\,x\right )}{840\,a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \]

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]

-(15*sin(c/2 + (d*x)/2) - 192*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 1144*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d
*x)/2) - 6112*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 1680*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) + 3360*co
s(c/2 + (d*x)/2)^7*(c + d*x))/(840*a^4*d*cos(c/2 + (d*x)/2)^7)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\cos {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

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]

Integral(cos(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)/a**4

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