\(\int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [79]

Optimal result

Integrand size = 21, antiderivative size = 49 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 x}{a^3}-\frac {3 \cos (c+d x)}{a^3 d}-\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \]

[Out]

-3*x/a^3-3*cos(d*x+c)/a^3/d-2*cos(d*x+c)^3/a/d/(a+a*sin(d*x+c))^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2759, 2761, 8} \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \cos (c+d x)}{a^3 d}-\frac {3 x}{a^3}-\frac {2 \cos ^3(c+d x)}{a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

(-3*x)/a^3 - (3*Cos[c + d*x])/(a^3*d) - (2*Cos[c + d*x]^3)/(a*d*(a + a*Sin[c + d*x])^2)

Rule 8

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

Rule 2759

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

Rule 2761

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}-\frac {3 \int \frac {\cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{a^2} \\ & = -\frac {3 \cos (c+d x)}{a^3 d}-\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2}-\frac {3 \int 1 \, dx}{a^3} \\ & = -\frac {3 x}{a^3}-\frac {3 \cos (c+d x)}{a^3 d}-\frac {2 \cos ^3(c+d x)}{a d (a+a \sin (c+d x))^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\cos ^5(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{5 \sqrt {2} a^3 d (1+\sin (c+d x))^{5/2}} \]

[In]

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

[Out]

-1/5*(Cos[c + d*x]^5*Hypergeometric2F1[3/2, 5/2, 7/2, (1 - Sin[c + d*x])/2])/(Sqrt[2]*a^3*d*(1 + Sin[c + d*x])
^(5/2))

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a^{3} d}\)	\(54\)
default	\(\frac {-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{a^{3} d}\)	\(54\)
parallelrisch	\(\frac {-6 \cos \left (d x +c \right ) d x +10 \cos \left (d x +c \right )+8 \sin \left (d x +c \right )-\cos \left (2 d x +2 c \right )-9}{2 a^{3} d \cos \left (d x +c \right )}\)	\(56\)
risch	\(-\frac {3 x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}-\frac {8}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\)	\(64\)
norman	\(\frac {-\frac {3 x \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {90 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {396 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {186 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {444 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {412 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {298 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {34 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {153 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {213 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {324 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {90 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {10}{a d}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {42 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {42 x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {90 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {252 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {15 x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {8 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {42 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {106 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {153 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {210 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {252 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {213 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(493\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3 \, d x + {\left (3 \, d x + 5\right )} \cos \left (d x + c\right ) + \cos \left (d x + c\right )^{2} + {\left (3 \, d x + \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{a^{3} d \cos \left (d x + c\right ) + a^{3} d \sin \left (d x + c\right ) + a^{3} d} \]

[In]

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

[Out]

-(3*d*x + (3*d*x + 5)*cos(d*x + c) + cos(d*x + c)^2 + (3*d*x + cos(d*x + c) - 4)*sin(d*x + c) + 4)/(a^3*d*cos(
d*x + c) + a^3*d*sin(d*x + c) + a^3*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (46) = 92\).

Time = 13.89 (sec) , antiderivative size = 478, normalized size of antiderivative = 9.76 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {3 d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {3 d x}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {8 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} - \frac {10}{a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{4}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-3*d*x*tan(c/2 + d*x/2)**3/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3*d*tan(c/
2 + d*x/2) + a**3*d) - 3*d*x*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a*
*3*d*tan(c/2 + d*x/2) + a**3*d) - 3*d*x*tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)
**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 3*d*x/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**
3*d*tan(c/2 + d*x/2) + a**3*d) - 8*tan(c/2 + d*x/2)**2/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**
2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 2*tan(c/2 + d*x/2)/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/
2)**2 + a**3*d*tan(c/2 + d*x/2) + a**3*d) - 10/(a**3*d*tan(c/2 + d*x/2)**3 + a**3*d*tan(c/2 + d*x/2)**2 + a**3
*d*tan(c/2 + d*x/2) + a**3*d), Ne(d, 0)), (x*cos(c)**4/(a*sin(c) + a)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (49) = 98\).

Time = 0.27 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 5}{a^{3} + \frac {a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}} + \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{d} \]

[In]

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

[Out]

-2*((sin(d*x + c)/(cos(d*x + c) + 1) + 4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5)/(a^3 + a^3*sin(d*x + c)/(cos
(d*x + c) + 1) + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3) + 3*arctan
(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {3 \, {\left (d x + c\right )}}{a^{3}} + \frac {2 \, {\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} a^{3}}}{d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {3\,x}{a^3}-\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+10}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

- (3*x)/a^3 - (2*tan(c/2 + (d*x)/2) + 8*tan(c/2 + (d*x)/2)^2 + 10)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 +
(d*x)/2)^2 + 1))