3.445 \(\int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=128 \[ -\frac {6 a \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {3 x \left (2 a^2-b^2\right )}{2 b^4}+\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))} \]

[Out]

3/2*(2*a^2-b^2)*x/b^4+3/2*cos(d*x+c)*(2*a-b*sin(d*x+c))/b^3/d-cos(d*x+c)^3/b/d/(a+b*sin(d*x+c))-6*a*arctan((b+
a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))*(a^2-b^2)^(1/2)/b^4/d

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Rubi [A]  time = 0.21, antiderivative size = 128, 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 = {2693, 2865, 2735, 2660, 618, 204} \[ -\frac {6 a \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d}+\frac {3 x \left (2 a^2-b^2\right )}{2 b^4}+\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(2*a^2 - b^2)*x)/(2*b^4) - (6*a*Sqrt[a^2 - b^2]*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*d) +
(3*Cos[c + d*x]*(2*a - b*Sin[c + d*x]))/(2*b^3*d) - Cos[c + d*x]^3/(b*d*(a + b*Sin[c + d*x]))

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 2693

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 + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && 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))^2} \, dx &=-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {3 \int \frac {\cos ^2(c+d x) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {3 \int \frac {-a b-\left (2 a^2-b^2\right ) \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3}\\ &=\frac {3 \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\left (3 a \left (a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^4}\\ &=\frac {3 \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}-\frac {\left (6 a \left (a^2-b^2\right )\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 {3 \left (2 a^2-b^2\right ) x}{2 b^4}+\frac {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}+\frac {\left (12 a \left (a^2-b^2\right )\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 {3 \left (2 a^2-b^2\right ) x}{2 b^4}-\frac {6 a \sqrt {a^2-b^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 {3 \cos (c+d x) (2 a-b \sin (c+d x))}{2 b^3 d}-\frac {\cos ^3(c+d x)}{b d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 5.11, size = 448, normalized size = 3.50 \[ \frac {\cos ^3(c+d x) \left (\sqrt {a+b} \left (\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (\sqrt {a-b} \sqrt {1-\sin (c+d x)} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}} \left (-6 a^2-3 a b \sin (c+d x)+b^2 \sin ^2(c+d x)+2 b^2\right )+6 \sqrt {b} (b-2 a) (a+b \sin (c+d x)) \sinh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {2} \sqrt {b}}\right )\right )+12 a \sqrt {a-b} \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x)) \tanh ^{-1}\left (\frac {\sqrt {\frac {b (\sin (c+d x)+1)}{b-a}}}{\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )-12 a (a-b) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x)) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )}{2 b^2 d (a-b)^{3/2} \sqrt {a+b} (1-\sin (c+d x))^{3/2} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (-\frac {b (\sin (c+d x)+1)}{a-b}\right )^{3/2} (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4/(a + b*Sin[c + d*x])^2,x]

[Out]

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

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fricas [A]  time = 0.48, size = 411, normalized size = 3.21 \[ \left [\frac {b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x + 3 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 3 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) + 3 \, {\left (a b^{2} \cos \left (d x + c\right ) + {\left (2 \, a^{2} b - b^{3}\right )} d x\right )} \sin \left (d x + c\right )}{2 \, {\left (b^{5} d \sin \left (d x + c\right ) + a b^{4} d\right )}}, \frac {b^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, a^{3} - a b^{2}\right )} d x + 6 \, {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + 3 \, {\left (2 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right ) + 3 \, {\left (a b^{2} \cos \left (d x + c\right ) + {\left (2 \, a^{2} b - b^{3}\right )} d x\right )} \sin \left (d x + c\right )}{2 \, {\left (b^{5} d \sin \left (d x + c\right ) + a b^{4} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

[1/2*(b^3*cos(d*x + c)^3 + 3*(2*a^3 - a*b^2)*d*x + 3*(a*b*sin(d*x + c) + a^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)) + 3*(2*a^2*b - b^3)*cos(d*x + c) + 3*(a*b^2*
cos(d*x + c) + (2*a^2*b - b^3)*d*x)*sin(d*x + c))/(b^5*d*sin(d*x + c) + a*b^4*d), 1/2*(b^3*cos(d*x + c)^3 + 3*
(2*a^3 - a*b^2)*d*x + 6*(a*b*sin(d*x + c) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)
*cos(d*x + c))) + 3*(2*a^2*b - b^3)*cos(d*x + c) + 3*(a*b^2*cos(d*x + c) + (2*a^2*b - b^3)*d*x)*sin(d*x + c))/
(b^5*d*sin(d*x + c) + a*b^4*d)]

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giac [A]  time = 1.40, size = 235, normalized size = 1.84 \[ \frac {\frac {3 \, {\left (2 \, a^{2} - b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {12 \, {\left (a^{3} - a b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{3}} + \frac {4 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3} - a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a b^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/2*(3*(2*a^2 - b^2)*(d*x + c)/b^4 - 12*(a^3 - a*b^2)*(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*(b*tan(1/2*d*x + 1/2*c)^3 + 4*a*tan(1/2*d*x
+ 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 4*a)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^3) + 4*(a^2*b*tan(1/2*d*x + 1/2*
c) - b^3*tan(1/2*d*x + 1/2*c) + a^3 - a*b^2)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)*a*b^3)
)/d

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maple [B]  time = 0.24, size = 385, normalized size = 3.01 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {4 a}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{4}}-\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}+\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a}+\frac {2 a^{2}}{d \,b^{3} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {2}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}-\frac {6 a \sqrt {a^{2}-b^{2}}\, \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a+b*sin(d*x+c))^2,x)

[Out]

1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^
2*a-1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)+4/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^2*a+6/d/b^4*arctan(
tan(1/2*d*x+1/2*c))*a^2-3/d/b^2*arctan(tan(1/2*d*x+1/2*c))+2/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c
)*b+a)*a*tan(1/2*d*x+1/2*c)-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)/a*tan(1/2*d*x+1/2*c)+2/d/b^3
/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)*a^2-2/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)
-6/d/b^4*a*(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))^2,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 = 6.40, size = 601, normalized size = 4.70 \[ \frac {\frac {2\,\left (3\,a^2-b^2\right )}{b^3}+\frac {6\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{b^3}+\frac {6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-b^2\right )}{b^3}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,a^2-2\,b^2\right )}{a\,b^2}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2-b^2\right )}{a\,b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (3\,a^2-2\,b^2\right )}{a\,b^2}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}+\frac {6\,a\,\mathrm {atanh}\left (\frac {432\,a^3\,\sqrt {b^2-a^2}}{432\,a^3\,b-\frac {432\,a^5}{b}-864\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+864\,a^2\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {864\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{432\,a^3-\frac {432\,a^5}{b^2}+864\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\frac {864\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}}+\frac {432\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{432\,a^5+864\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b-432\,a^3\,b^2-864\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^3}\right )\,\sqrt {b^2-a^2}}{b^4\,d}-\frac {\mathrm {atan}\left (\frac {432\,a^5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{432\,a^5-648\,a^3\,b^2+216\,a\,b^4}-\frac {648\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{216\,a\,b^2-648\,a^3+\frac {432\,a^5}{b^2}}+\frac {216\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{216\,a-\frac {648\,a^3}{b^2}+\frac {432\,a^5}{b^4}}\right )\,\left (a^2\,2{}\mathrm {i}-b^2\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{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))^2,x)

[Out]

((2*(3*a^2 - b^2))/b^3 + (6*a^2*tan(c/2 + (d*x)/2)^4)/b^3 + (6*tan(c/2 + (d*x)/2)^2*(2*a^2 - b^2))/b^3 + (tan(
c/2 + (d*x)/2)*(9*a^2 - 2*b^2))/(a*b^2) + (4*tan(c/2 + (d*x)/2)^3*(3*a^2 - b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)
^5*(3*a^2 - 2*b^2))/(a*b^2))/(d*(a + 2*b*tan(c/2 + (d*x)/2) + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2
)^4 + a*tan(c/2 + (d*x)/2)^6 + 4*b*tan(c/2 + (d*x)/2)^3 + 2*b*tan(c/2 + (d*x)/2)^5)) - (atan((432*a^5*tan(c/2
+ (d*x)/2))/(216*a*b^4 + 432*a^5 - 648*a^3*b^2) - (648*a^3*tan(c/2 + (d*x)/2))/(216*a*b^2 - 648*a^3 + (432*a^5
)/b^2) + (216*a*tan(c/2 + (d*x)/2))/(216*a - (648*a^3)/b^2 + (432*a^5)/b^4))*(a^2*2i - b^2*1i)*3i)/(b^4*d) + (
6*a*atanh((432*a^3*(b^2 - a^2)^(1/2))/(432*a^3*b - (432*a^5)/b - 864*a^4*tan(c/2 + (d*x)/2) + 864*a^2*b^2*tan(
c/2 + (d*x)/2)) + (864*a^2*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(432*a^3 - (432*a^5)/b^2 + 864*a^2*b*tan(c/2
+ (d*x)/2) - (864*a^4*tan(c/2 + (d*x)/2))/b) + (432*a^4*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(432*a^5 - 432*a
^3*b^2 + 864*a^4*b*tan(c/2 + (d*x)/2) - 864*a^2*b^3*tan(c/2 + (d*x)/2)))*(b^2 - a^2)^(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))**2,x)

[Out]

Timed out

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