3.57 \(\int \frac{\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=193 \[ \frac{8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}+\frac{4 (9 A-19 B) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(6 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{x (6 A-13 B)}{2 a^3}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(6 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

-((6*A - 13*B)*x)/(2*a^3) + (8*(9*A - 19*B)*Sin[c + d*x])/(15*a^3*d) - ((6*A - 13*B)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) + ((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((6*A - 11*B)*Cos[c + d*x]^
3*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (4*(9*A - 19*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^
3*Cos[c + d*x]))

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Rubi [A]  time = 0.467678, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2977, 2734} \[ \frac{8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}+\frac{4 (9 A-19 B) \sin (c+d x) \cos ^2(c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}-\frac{(6 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{x (6 A-13 B)}{2 a^3}+\frac{(A-B) \sin (c+d x) \cos ^4(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac{(6 A-11 B) \sin (c+d x) \cos ^3(c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

-((6*A - 13*B)*x)/(2*a^3) + (8*(9*A - 19*B)*Sin[c + d*x])/(15*a^3*d) - ((6*A - 13*B)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) + ((A - B)*Cos[c + d*x]^4*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((6*A - 11*B)*Cos[c + d*x]^
3*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) + (4*(9*A - 19*B)*Cos[c + d*x]^2*Sin[c + d*x])/(15*d*(a^3 + a^
3*Cos[c + d*x]))

Rule 2977

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

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\cos ^3(c+d x) (4 a (A-B)-a (2 A-7 B) \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) \left (3 a^2 (6 A-11 B)-a^2 (18 A-43 B) \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \cos (c+d x) \left (8 a^3 (9 A-19 B)-15 a^3 (6 A-13 B) \cos (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(6 A-13 B) x}{2 a^3}+\frac{8 (9 A-19 B) \sin (c+d x)}{15 a^3 d}-\frac{(6 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac{(A-B) \cos ^4(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{(6 A-11 B) \cos ^3(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{4 (9 A-19 B) \cos ^2(c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.791297, size = 435, normalized size = 2.25 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (-600 d x (6 A-13 B) \cos \left (c+\frac{d x}{2}\right )-600 d x (6 A-13 B) \cos \left (\frac{d x}{2}\right )-4500 A \sin \left (c+\frac{d x}{2}\right )+4860 A \sin \left (c+\frac{3 d x}{2}\right )-900 A \sin \left (2 c+\frac{3 d x}{2}\right )+1452 A \sin \left (2 c+\frac{5 d x}{2}\right )+300 A \sin \left (3 c+\frac{5 d x}{2}\right )+60 A \sin \left (3 c+\frac{7 d x}{2}\right )+60 A \sin \left (4 c+\frac{7 d x}{2}\right )-1800 A d x \cos \left (c+\frac{3 d x}{2}\right )-1800 A d x \cos \left (2 c+\frac{3 d x}{2}\right )-360 A d x \cos \left (2 c+\frac{5 d x}{2}\right )-360 A d x \cos \left (3 c+\frac{5 d x}{2}\right )+7020 A \sin \left (\frac{d x}{2}\right )+7560 B \sin \left (c+\frac{d x}{2}\right )-9230 B \sin \left (c+\frac{3 d x}{2}\right )+930 B \sin \left (2 c+\frac{3 d x}{2}\right )-2782 B \sin \left (2 c+\frac{5 d x}{2}\right )-750 B \sin \left (3 c+\frac{5 d x}{2}\right )-105 B \sin \left (3 c+\frac{7 d x}{2}\right )-105 B \sin \left (4 c+\frac{7 d x}{2}\right )+15 B \sin \left (4 c+\frac{9 d x}{2}\right )+15 B \sin \left (5 c+\frac{9 d x}{2}\right )+3900 B d x \cos \left (c+\frac{3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac{5 d x}{2}\right )-12760 B \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^4*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-600*(6*A - 13*B)*d*x*Cos[(d*x)/2] - 600*(6*A - 13*B)*d*x*Cos[c + (d*x)/2] - 1800*
A*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[c + (3*d*x)/2] - 1800*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*B*d*x*Cos[2*
c + (3*d*x)/2] - 360*A*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 360*A*d*x*Cos[3*c + (5*d*x)
/2] + 780*B*d*x*Cos[3*c + (5*d*x)/2] + 7020*A*Sin[(d*x)/2] - 12760*B*Sin[(d*x)/2] - 4500*A*Sin[c + (d*x)/2] +
7560*B*Sin[c + (d*x)/2] + 4860*A*Sin[c + (3*d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] - 900*A*Sin[2*c + (3*d*x)/2] +
930*B*Sin[2*c + (3*d*x)/2] + 1452*A*Sin[2*c + (5*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 300*A*Sin[3*c + (5*d
*x)/2] - 750*B*Sin[3*c + (5*d*x)/2] + 60*A*Sin[3*c + (7*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 60*A*Sin[4*c +
(7*d*x)/2] - 105*B*Sin[4*c + (7*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(
1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.066, size = 292, normalized size = 1.5 \begin{align*}{\frac{A}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}-{\frac{A}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{2\,B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+{\frac{17\,A}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{31\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}A}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-7\,{\frac{B \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-5\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) A}{d{a}^{3}}}+13\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) B}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/d/a^3*B*tan(1/2*d*x+1/2*c)^5-1/2/d/a^3*tan(1/2*d*x+1/2*c)^3*A+2/3/d/a^3
*B*tan(1/2*d*x+1/2*c)^3+17/4/d/a^3*A*tan(1/2*d*x+1/2*c)-31/4/d/a^3*B*tan(1/2*d*x+1/2*c)+2/d/a^3/(1+tan(1/2*d*x
+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A-7/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)^3+2/d/a^3/(1+tan(1
/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)-5/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)-6/d/a^3*arctan
(tan(1/2*d*x+1/2*c))*A+13/d/a^3*arctan(tan(1/2*d*x+1/2*c))*B

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Maxima [A]  time = 1.5081, size = 435, normalized size = 2.25 \begin{align*} -\frac{B{\left (\frac{60 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{780 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, A{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \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{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
- 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3) - 3*A*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x +
c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(
d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

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Fricas [A]  time = 1.47336, size = 497, normalized size = 2.58 \begin{align*} -\frac{15 \,{\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (6 \, A - 13 \, B\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (6 \, A - 13 \, B\right )} d x -{\left (15 \, B \cos \left (d x + c\right )^{4} + 15 \,{\left (2 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (234 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (114 \, A - 239 \, B\right )} \cos \left (d x + c\right ) + 144 \, A - 304 \, B\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(15*(6*A - 13*B)*d*x*cos(d*x + c)^3 + 45*(6*A - 13*B)*d*x*cos(d*x + c)^2 + 45*(6*A - 13*B)*d*x*cos(d*x +
c) + 15*(6*A - 13*B)*d*x - (15*B*cos(d*x + c)^4 + 15*(2*A - 3*B)*cos(d*x + c)^3 + (234*A - 479*B)*cos(d*x + c
)^2 + 3*(114*A - 239*B)*cos(d*x + c) + 144*A - 304*B)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x +
c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [A]  time = 27.8623, size = 966, normalized size = 5.01 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((-180*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60
*a**3*d) - 360*A*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*
a**3*d) - 180*A*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 3*A*tan(c/2
+ d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 24*A*tan(c/2 + d*x
/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 198*A*tan(c/2 + d*x/2)**
5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 600*A*tan(c/2 + d*x/2)**3/(60
*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 375*A*tan(c/2 + d*x/2)/(60*a**3*d*
tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*B*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*t
an(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 780*B*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*ta
n(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*B*d*x/(60*a**3*d*tan(c/2 + d*x/2)**4 + 1
20*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*B*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3
*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 34*B*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan
(c/2 + d*x/2)**2 + 60*a**3*d) - 388*B*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2
+ d*x/2)**2 + 60*a**3*d) - 1310*B*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*
x/2)**2 + 60*a**3*d) - 765*B*tan(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**4 + 120*a**3*d*tan(c/2 + d*x/2)**2
+ 60*a**3*d), Ne(d, 0)), (x*(A + B*cos(c))*cos(c)**4/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.17815, size = 270, normalized size = 1.4 \begin{align*} -\frac{\frac{30 \,{\left (d x + c\right )}{\left (6 \, A - 13 \, B\right )}}{a^{3}} - \frac{60 \,{\left (2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 7 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 5 \, 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} a^{3}} - \frac{3 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 30 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 40 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 255 \, A a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(d*x + c)*(6*A - 13*B)/a^3 - 60*(2*A*tan(1/2*d*x + 1/2*c)^3 - 7*B*tan(1/2*d*x + 1/2*c)^3 + 2*A*tan(1
/2*d*x + 1/2*c) - 5*B*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2
*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 30*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 40*B*a^12*tan(1/2*d*x + 1/2*c)^3
+ 255*A*a^12*tan(1/2*d*x + 1/2*c) - 465*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d