3.432 \(\int \frac {\cos ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=127 \[ \frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d}-\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}+\frac {\cos ^3(c+d x)}{3 b d} \]

[Out]

-1/2*a*(2*a^2-3*b^2)*x/b^4+2*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/b^4/d+1/3*cos(d*
x+c)^3/b/d-1/2*cos(d*x+c)*(2*a^2-2*b^2-a*b*sin(d*x+c))/b^3/d

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Rubi [A]  time = 0.25, antiderivative size = 127, 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 = {2695, 2865, 2735, 2660, 618, 204} \[ \frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}-\frac {a x \left (2 a^2-3 b^2\right )}{2 b^4}+\frac {\cos ^3(c+d x)}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(2*a^2 - 3*b^2)*x)/(2*b^4) + (2*(a^2 - b^2)^(3/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*d
) + Cos[c + d*x]^3/(3*b*d) - (Cos[c + d*x]*(2*(a^2 - b^2) - a*b*Sin[c + d*x]))/(2*b^3*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\cos ^3(c+d x)}{3 b d}+\frac {\int \frac {\cos ^2(c+d x) (b+a \sin (c+d x))}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}+\frac {\int \frac {-b \left (a^2-2 b^2\right )-a \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}+\frac {\left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^4}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}+\frac {\left (2 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}-\frac {\left (4 \left (a^2-b^2\right )^2\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^4 d}\\ &=-\frac {a \left (2 a^2-3 b^2\right ) x}{2 b^4}+\frac {2 \left (a^2-b^2\right )^{3/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cos (c+d x) \left (2 \left (a^2-b^2\right )-a b \sin (c+d x)\right )}{2 b^3 d}\\ \end {align*}

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Mathematica [B]  time = 4.51, size = 428, normalized size = 3.37 \[ \frac {\cos ^3(c+d x) \left (\sqrt {a+b} \left (\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (\sqrt {a-b} \sqrt {1-\sin (c+d x)} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}} \left (6 a^2-3 a b \sin (c+d x)+2 b^2 \sin ^2(c+d x)-8 b^2\right )-6 \sqrt {b} \left (-2 a^2+a b+2 b^2\right ) \sinh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {2} \sqrt {b}}\right )\right )-12 \sqrt {a-b} \left (a^2-b^2\right ) \sqrt {1-\sin (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {\frac {b (\sin (c+d x)+1)}{b-a}}}{\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )+12 (a+b) (a-b)^2 \sqrt {1-\sin (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )}{6 b^2 d (a-b)^{3/2} \sqrt {a+b} (1-\sin (c+d x))^{3/2} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (-\frac {b (\sin (c+d x)+1)}{a-b}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*(12*(a - b)^2*(a + b)*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[a +
b]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))])]*Sqrt[1 - Sin[c + d*x]] + Sqrt[a + b]*(-12*Sqrt[a - b]*(a^2 - b^2
)*ArcTanh[Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]/Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]]*Sqrt[1 - Sin[c + d*x
]] + Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*(-6*Sqrt[b]*(-2*a^2 + a*b + 2*b^2)*ArcSinh[(Sqrt[a - b]*Sqrt[-((
b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[2]*Sqrt[b])] + Sqrt[a - b]*Sqrt[1 - Sin[c + d*x]]*Sqrt[(b*(1 + Sin[c +
d*x]))/(-a + b)]*(6*a^2 - 8*b^2 - 3*a*b*Sin[c + d*x] + 2*b^2*Sin[c + d*x]^2)))))/(6*(a - b)^(3/2)*b^2*Sqrt[a +
b]*d*(1 - Sin[c + d*x])^(3/2)*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*(-((b*(1 + Sin[c + d*x]))/(a - b)))^(3
/2))

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fricas [A]  time = 0.50, size = 332, normalized size = 2.61 \[ \left [\frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 3 \, {\left (a^{2} - b^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) - 6 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{6 \, b^{4} d}, \frac {2 \, b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 6 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) - 6 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{6 \, b^{4} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)*sin(d*x + c) - 3*(2*a^3 - 3*a*b^2)*d*x - 3*(a^2 - b^2)*sqrt(
-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*sin(d*x + c
) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) - 6*(a^2*b - b^3)
*cos(d*x + c))/(b^4*d), 1/6*(2*b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)*sin(d*x + c) - 3*(2*a^3 - 3*a*b^2)*d*
x - 6*(a^2 - b^2)^(3/2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) - 6*(a^2*b - b^3)*cos(d*x
+ c))/(b^4*d)]

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giac [A]  time = 2.81, size = 226, normalized size = 1.78 \[ -\frac {\frac {3 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} - 8 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(2*a^3 - 3*a*b^2)*(d*x + c)/b^4 - 12*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a)
+ arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*b^4) + 2*(3*a*b*tan(1/2*d*x + 1/2*c)^
5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 - 12*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(
1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 - 8*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d

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maple [B]  time = 0.15, size = 450, normalized size = 3.54 \[ -\frac {a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {8}{3 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4}}+\frac {3 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{4}}{d \,b^{4} \sqrt {a^{2}-b^{2}}}-\frac {4 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{d \,b^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*
c)^4*a^2+4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4-4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+
1/2*c)^2*a^2+4/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/
2*d*x+1/2*c)-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2+8/3/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3-2/d/b^4*arctan(tan(1/2*
d*x+1/2*c))*a^3+3/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))+2/d/b^4/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c
)+2*b)/(a^2-b^2)^(1/2))*a^4-4/d/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*a
^2+2/d/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 6.12, size = 364, normalized size = 2.87 \[ \frac {\frac {5\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{12}}{b\,d}+\frac {3\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {a\,\sin \left (2\,c+2\,d\,x\right )}{4}}{b^2\,d}-\frac {a^2\,\cos \left (c+d\,x\right )}{b^3\,d}-\frac {2\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b^4\,d}+\frac {2\,\mathrm {atanh}\left (\frac {2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}+a\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^2-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^4+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^5}\right )\,\sqrt {-{\left (a+b\right )}^3\,{\left (a-b\right )}^3}}{b^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((5*cos(c + d*x))/4 + cos(3*c + 3*d*x)/12)/(b*d) + (3*a*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (a*sin(2
*c + 2*d*x))/4)/(b^2*d) - (a^2*cos(c + d*x))/(b^3*d) - (2*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b^
4*d) + (2*atanh((2*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - a^2*sin(c/2 + (d*x)/2)*(
b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2))/(
a^5*cos(c/2 + (d*x)/2) + 2*b^5*sin(c/2 + (d*x)/2) + a*b^4*cos(c/2 + (d*x)/2) + 2*a^4*b*sin(c/2 + (d*x)/2) - 2*
a^3*b^2*cos(c/2 + (d*x)/2) - 4*a^2*b^3*sin(c/2 + (d*x)/2)))*(-(a + b)^3*(a - b)^3)^(1/2))/(b^4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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