3.457 \(\int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=139 \[ \frac {3 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \sqrt {a^2-b^2}}-\frac {3 a x}{b^4}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 139, 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, 2863, 2735, 2660, 618, 204} \[ \frac {3 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^4 d \sqrt {a^2-b^2}}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}-\frac {3 a x}{b^4}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x)/b^4 + (3*(2*a^2 - b^2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b^4*Sqrt[a^2 - b^2]*d) - Co
s[c + d*x]^3/(2*b*d*(a + b*Sin[c + d*x])^2) - (3*Cos[c + d*x]*(2*a + b*Sin[c + d*x]))/(2*b^3*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 2863

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 + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x)}{(a+b \sin (c+d x))^3} \, dx &=-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \int \frac {\cos ^2(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx}{2 b}\\ &=-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}+\frac {3 \int \frac {-b-2 a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{2 b^3}\\ &=-\frac {3 a x}{b^4}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}+\frac {\left (3 \left (2 a^2-b^2\right )\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{2 b^4}\\ &=-\frac {3 a x}{b^4}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}+\frac {\left (3 \left (2 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 a x}{b^4}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}-\frac {\left (6 \left (2 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 a x}{b^4}+\frac {3 \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^4 \sqrt {a^2-b^2} d}-\frac {\cos ^3(c+d x)}{2 b d (a+b \sin (c+d x))^2}-\frac {3 \cos (c+d x) (2 a+b \sin (c+d x))}{2 b^3 d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 6.28, size = 2641, normalized size = 19.00 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^3*(-1/2*(b*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)
)^(5/2))/(((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[c + d*x])^2) - (-((a*b^3*(-(b
/(a - b)) - (b*Sin[c + d*x])/(a - b))^(5/2)*(b/(a + b) - (b*Sin[c + d*x])/(a + b))^(5/2))/((a^2 - b^2)*((a*b)/
(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))*(a + b*Sin[c + d*x]))) - ((16*Sqrt[2]*a*b^4*(-(b/(a - b))
- (b*Sin[c + d*x])/(a - b))^(5/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b
*Sin[c + d*x])/(a - b)))/(2*b))^(5/2)*((5*(1/(2*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)
)^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1)))/8 - (15*b^3*(((a - b)*(-(b/(a -
b)) - (b*Sin[c + d*x])/(a - b)))/b - ((a - b)^2*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^2)/(3*b^2) - (Sqrt[2
]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/
(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b]*Sqrt[1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2
*b)])))/(32*(a - b)^3*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^3*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x
])/(a - b)))/(2*b))^2)))/(5*(a - b)*(a + b)^3*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b]) + (((4
*a^2*b^5)/((a - b)^2*(a + b)^2) + (b^5*(2*a^2 - 3*b^2))/((a - b)^2*(a + b)^2))*((4*Sqrt[2]*(-(b/(a - b)) - (b*
Sin[c + d*x])/(a - b))^(3/2)*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c
+ d*x])/(a - b)))/(2*b))^(5/2)*((3/(4*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1
+ ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(-1))/2 + (3*b^2*(((a - b)*(-(b/(a - b)) - (b*Si
n[c + d*x])/(a - b)))/b - (Sqrt[2]*Sqrt[a - b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a -
b)])/(Sqrt[2]*Sqrt[b])]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[b]*Sqrt[1 + ((a - b)*(-(b/(a - b)
) - (b*Sin[c + d*x])/(a - b)))/(2*b)])))/(8*(a - b)^2*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b))^2*(1 + ((a - b
)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2)))/(3*(a + b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x]
)/(a + b)))/b]) - ((-((a*b)/(a - b)) + b^2/(a - b))*(-(((-((a*b)/(a - b)) + b^2/(a - b))*(-(((-((a*b)/(a + b))
- b^2/(a + b))*((2*Sqrt[a - b]*ArcTanh[(Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[a +
b]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)])])/(b*Sqrt[a + b]) - (2*Sqrt[-((a*b)/(a + b)) - b^2/(a + b)]*Arc
Tanh[(Sqrt[-((a*b)/(a + b)) - b^2/(a + b)]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[-((a*b)/(a - b
)) + b^2/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)])])/(b*Sqrt[-((a*b)/(a - b)) + b^2/(a - b)])))/b)
+ (2*Sqrt[2]*(a - b)*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a + b)]*
(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(3/2)*((Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-(b/
(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])])/(Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a - b)) - (b*Sin[c +
d*x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(3/2)) + 1/(2*(1 + ((a - b)*(-(
b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b)))))/(b*(a + b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a +
b)))/b])))/b) + (4*Sqrt[2]*Sqrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)]*Sqrt[b/(a + b) - (b*Sin[c + d*x])/(a
+ b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(5/2)*((3*Sqrt[b]*ArcSinh[(Sqrt[a - b]*S
qrt[-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)])/(Sqrt[2]*Sqrt[b])])/(4*Sqrt[2]*Sqrt[a - b]*Sqrt[-(b/(a - b)) - (
b*Sin[c + d*x])/(a - b)]*(1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^(5/2)) + (3/(2*(1 + (
(a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/(a - b)))/(2*b))^2) + (1 + ((a - b)*(-(b/(a - b)) - (b*Sin[c + d*x])/
(a - b)))/(2*b))^(-1))/4))/((a + b)*Sqrt[((a + b)*(b/(a + b) - (b*Sin[c + d*x])/(a + b)))/b])))/b))/b)/(((a*b)
/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/(a + b))))/(2*((a*b)/(a - b) - b^2/(a - b))*((a*b)/(a + b) + b^2/
(a + b)))))/(d*(1 - (a + b*Sin[c + d*x])/(a - b))^(3/2)*(1 - (a + b*Sin[c + d*x])/(a + b))^(3/2))

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fricas [B]  time = 0.78, size = 716, normalized size = 5.15 \[ \left [-\frac {12 \, {\left (a^{3} b^{2} - a b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 4 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (a^{5} - a b^{4}\right )} d x + 3 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4} - {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\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 (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right ) - 6 \, {\left (4 \, {\left (a^{4} b - a^{2} b^{3}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{4} - b^{8}\right )} d\right )}}, -\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} d x \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{5} - a b^{4}\right )} d x - 3 \, {\left (2 \, a^{4} + a^{2} b^{2} - b^{4} - {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )\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^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right ) - 3 \, {\left (4 \, {\left (a^{4} b - a^{2} b^{3}\right )} d x + 3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{3} b^{5} - a b^{7}\right )} d \sin \left (d x + c\right ) - {\left (a^{4} b^{4} - b^{8}\right )} 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))^3,x, algorithm="fricas")

[Out]

[-1/4*(12*(a^3*b^2 - a*b^4)*d*x*cos(d*x + c)^2 + 4*(a^2*b^3 - b^5)*cos(d*x + c)^3 - 12*(a^5 - a*b^4)*d*x + 3*(
2*a^4 + a^2*b^2 - b^4 - (2*a^2*b^2 - b^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*sin(d*x + c))*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*(2*a^4*b - a^2*b^3 - b^5
)*cos(d*x + c) - 6*(4*(a^4*b - a^2*b^3)*d*x + 3*(a^3*b^2 - a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^6 - b^8)
*d*cos(d*x + c)^2 - 2*(a^3*b^5 - a*b^7)*d*sin(d*x + c) - (a^4*b^4 - b^8)*d), -1/2*(6*(a^3*b^2 - a*b^4)*d*x*cos
(d*x + c)^2 + 2*(a^2*b^3 - b^5)*cos(d*x + c)^3 - 6*(a^5 - a*b^4)*d*x - 3*(2*a^4 + a^2*b^2 - b^4 - (2*a^2*b^2 -
b^4)*cos(d*x + c)^2 + 2*(2*a^3*b - a*b^3)*sin(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^
2 - b^2)*cos(d*x + c))) - 3*(2*a^4*b - a^2*b^3 - b^5)*cos(d*x + c) - 3*(4*(a^4*b - a^2*b^3)*d*x + 3*(a^3*b^2 -
a*b^4)*cos(d*x + c))*sin(d*x + c))/((a^2*b^6 - b^8)*d*cos(d*x + c)^2 - 2*(a^3*b^5 - a*b^7)*d*sin(d*x + c) - (
a^4*b^4 - b^8)*d)]

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giac [B]  time = 0.42, size = 272, normalized size = 1.96 \[ -\frac {\frac {3 \, {\left (d x + c\right )} a}{b^{4}} - \frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (2 \, a^{2} - b^{2}\right )}}{\sqrt {a^{2} - b^{2}} b^{4}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b^{3}} + \frac {3 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 9 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{4} + a^{2} b^{2}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}^{2} a^{2} b^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*(d*x + c)*a/b^4 - 3*(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)))*(2*a^2 - b^2)/(sqrt(a^2 - b^2)*b^4) + 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*b^3) + (3*a^3*b*tan(1/2*d*x +
1/2*c)^3 + 2*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*a^4*tan(1/2*d*x + 1/2*c)^2 + 9*a^2*b^2*tan(1/2*d*x + 1/2*c)^2 +
2*b^4*tan(1/2*d*x + 1/2*c)^2 + 13*a^3*b*tan(1/2*d*x + 1/2*c) + 2*a*b^3*tan(1/2*d*x + 1/2*c) + 4*a^4 + a^2*b^2
)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2*a^2*b^3))/d

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maple [B]  time = 0.31, size = 560, normalized size = 4.03 \[ -\frac {2}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \,b^{4}}-\frac {3 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )^{2}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )^{2} a}-\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{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 )^{2}}-\frac {9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{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 )^{2}}-\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\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 )^{2} a^{2}}-\frac {13 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 )^{2}}-\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 )^{2} a}-\frac {4 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 )^{2}}-\frac {1}{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 )^{2}}+\frac {6 \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^{4} \sqrt {a^{2}-b^{2}}}-\frac {3 \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^{2} \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))^3,x)

[Out]

-2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)-6/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a-3/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1
/2*d*x+1/2*c)*b+a)^2*a*tan(1/2*d*x+1/2*c)^3-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a*tan(1/2*
d*x+1/2*c)^3-4/d/b^3/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2*tan(1/2*d*x+1/2*c)^2-9/d/b/(tan(1
/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*tan(1/2*d*x+1/2*c)^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)^2/a^2*tan(1/2*d*x+1/2*c)^2-13/d/b^2/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*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)^2/a*tan(1/2*d*x+1/2*c)-4/d/b^3/(tan(1/2*d*x+
1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a^2-1/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2+6/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^2-3/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))

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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))^3,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 = 7.55, size = 1360, normalized size = 9.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a + b*sin(c + d*x))^3,x)

[Out]

- ((6*a^2 + b^2)/b^3 + (2*tan(c/2 + (d*x)/2)^2*(6*a^4 + b^4 + 9*a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d*x)/2)*(21*
a^2 + 2*b^2))/(a*b^2) + (4*tan(c/2 + (d*x)/2)^3*(6*a^2 + b^2))/(a*b^2) + (tan(c/2 + (d*x)/2)^4*(6*a^4 + 2*b^4
+ 9*a^2*b^2))/(a^2*b^3) + (tan(c/2 + (d*x)/2)^5*(3*a^2 + 2*b^2))/(a*b^2))/(d*(tan(c/2 + (d*x)/2)^2*(3*a^2 + 4*
b^2) + tan(c/2 + (d*x)/2)^4*(3*a^2 + 4*b^2) + a^2*tan(c/2 + (d*x)/2)^6 + a^2 + 8*a*b*tan(c/2 + (d*x)/2)^3 + 4*
a*b*tan(c/2 + (d*x)/2)^5 + 4*a*b*tan(c/2 + (d*x)/2))) - (3*a*x)/b^4 - (atan((((-(a + b)*(a - b))^(1/2)*(2*a^2
- b^2)*((288*a^4)/b^5 - (8*tan(c/2 + (d*x)/2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5*b^3))/b^9 + (3*(-(a + b)*(a - b)
)^(1/2)*(2*a^2 - b^2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 - 48*a^2 + (3*(-(a + b)*(a - b))^(1
/2)*(2*a^2 - b^2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(2*(b^6 - a^2*b^4))))/(2
*(b^6 - a^2*b^4)))*3i)/(2*(b^6 - a^2*b^4)) + ((-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*((288*a^4)/b^5 - (8*tan(c
/2 + (d*x)/2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5*b^3))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(48*a^2 -
(8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(32*a^2*b^3 +
(8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(2*(b^6 - a^2*b^4))))/(2*(b^6 - a^2*b^4)))*3i)/(2*(b^6 -
a^2*b^4)))/((16*(54*a^4 - 27*a^2*b^2))/b^8 + (16*tan(c/2 + (d*x)/2)*(216*a^5 - 108*a^3*b^2))/b^9 - (3*(-(a +
b)*(a - b))^(1/2)*(2*a^2 - b^2)*((288*a^4)/b^5 - (8*tan(c/2 + (d*x)/2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5*b^3))/b
^9 + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*((8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 - 48*a^2 +
(3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))
/(2*(b^6 - a^2*b^4))))/(2*(b^6 - a^2*b^4))))/(2*(b^6 - a^2*b^4)) + (3*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*(
(288*a^4)/b^5 - (8*tan(c/2 + (d*x)/2)*(9*a*b^7 - 108*a^3*b^5 + 72*a^5*b^3))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*
(2*a^2 - b^2)*(48*a^2 - (8*tan(c/2 + (d*x)/2)*(12*a*b^10 - 24*a^3*b^8))/b^9 + (3*(-(a + b)*(a - b))^(1/2)*(2*a
^2 - b^2)*(32*a^2*b^3 + (8*tan(c/2 + (d*x)/2)*(12*a*b^13 - 8*a^3*b^11))/b^9))/(2*(b^6 - a^2*b^4))))/(2*(b^6 -
a^2*b^4))))/(2*(b^6 - a^2*b^4))))*(-(a + b)*(a - b))^(1/2)*(2*a^2 - b^2)*3i)/(d*(b^6 - a^2*b^4))

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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))**3,x)

[Out]

Timed out

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