\(\int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx\) [265]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {f x}{4 a d}+\frac {f \cos (c+d x)}{a d^2}+\frac {(e+f x) \sin (c+d x)}{a d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 a d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d} \]

[Out]

1/4*f*x/a/d+f*cos(d*x+c)/a/d^2+(f*x+e)*sin(d*x+c)/a/d-1/4*f*cos(d*x+c)*sin(d*x+c)/a/d^2-1/2*(f*x+e)*sin(d*x+c)
^2/a/d

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4619, 3377, 2718, 4489, 2715, 8} \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {f \cos (c+d x)}{a d^2}-\frac {f \sin (c+d x) \cos (c+d x)}{4 a d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d}+\frac {(e+f x) \sin (c+d x)}{a d}+\frac {f x}{4 a d} \]

[In]

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

[Out]

(f*x)/(4*a*d) + (f*Cos[c + d*x])/(a*d^2) + ((e + f*x)*Sin[c + d*x])/(a*d) - (f*Cos[c + d*x]*Sin[c + d*x])/(4*a
*d^2) - ((e + f*x)*Sin[c + d*x]^2)/(2*a*d)

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4489

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[(c + d
*x)^m*(Sin[a + b*x]^(n + 1)/(b*(n + 1))), x] - Dist[d*(m/(b*(n + 1))), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 4619

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol
] :> Dist[1/a, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Dist[1/b, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*S
in[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x) \cos (c+d x) \, dx}{a}-\frac {\int (e+f x) \cos (c+d x) \sin (c+d x) \, dx}{a} \\ & = \frac {(e+f x) \sin (c+d x)}{a d}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d}+\frac {f \int \sin ^2(c+d x) \, dx}{2 a d}-\frac {f \int \sin (c+d x) \, dx}{a d} \\ & = \frac {f \cos (c+d x)}{a d^2}+\frac {(e+f x) \sin (c+d x)}{a d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 a d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d}+\frac {f \int 1 \, dx}{4 a d} \\ & = \frac {f x}{4 a d}+\frac {f \cos (c+d x)}{a d^2}+\frac {(e+f x) \sin (c+d x)}{a d}-\frac {f \cos (c+d x) \sin (c+d x)}{4 a d^2}-\frac {(e+f x) \sin ^2(c+d x)}{2 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.57 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-f \cos (c+d x) (-4+\sin (c+d x))+d (e+f x) (\cos (2 (c+d x))+4 \sin (c+d x))}{4 a d^2} \]

[In]

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

[Out]

(-(f*Cos[c + d*x]*(-4 + Sin[c + d*x])) + d*(e + f*x)*(Cos[2*(c + d*x)] + 4*Sin[c + d*x]))/(4*a*d^2)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(\frac {2 d \left (f x +e \right ) \cos \left (2 d x +2 c \right )-f \sin \left (2 d x +2 c \right )+8 d \left (f x +e \right ) \sin \left (d x +c \right )-2 d e +8 f \cos \left (d x +c \right )-8 f}{8 a \,d^{2}}\)	\(69\)
risch	\(\frac {f \cos \left (d x +c \right )}{a \,d^{2}}+\frac {\left (f x +e \right ) \sin \left (d x +c \right )}{a d}+\frac {\left (f x +e \right ) \cos \left (2 d x +2 c \right )}{4 a d}-\frac {f \sin \left (2 d x +2 c \right )}{8 a \,d^{2}}\)	\(74\)
derivativedivides	\(\frac {-\frac {f c \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\left (\cos ^{2}\left (d x +c \right )\right ) d e}{2}-f \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )-\sin \left (d x +c \right ) c f +d e \sin \left (d x +c \right )+f \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2} a}\)	\(113\)
default	\(\frac {-\frac {f c \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\left (\cos ^{2}\left (d x +c \right )\right ) d e}{2}-f \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )-\sin \left (d x +c \right ) c f +d e \sin \left (d x +c \right )+f \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )}{d^{2} a}\)	\(113\)
norman	\(\frac {\frac {2 f}{a \,d^{2}}+\frac {\left (2 d e +2 f \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{2}}+\frac {\left (2 d e +4 f \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,d^{2}}+\frac {f x}{4 a d}+\frac {5 f \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {7 f \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {\left (4 d e +f \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a \,d^{2}}+\frac {\left (4 d e +3 f \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a \,d^{2}}+\frac {9 f x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {3 f x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {11 f x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {11 f x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {3 f x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {9 f x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}+\frac {f x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(337\)

[In]

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

[Out]

1/8*(2*d*(f*x+e)*cos(2*d*x+2*c)-f*sin(2*d*x+2*c)+8*d*(f*x+e)*sin(d*x+c)-2*d*e+8*f*cos(d*x+c)-8*f)/a/d^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.74 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {d f x - 2 \, {\left (d f x + d e\right )} \cos \left (d x + c\right )^{2} - 4 \, f \cos \left (d x + c\right ) - {\left (4 \, d f x + 4 \, d e - f \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, a d^{2}} \]

[In]

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

[Out]

-1/4*(d*f*x - 2*(d*f*x + d*e)*cos(d*x + c)^2 - 4*f*cos(d*x + c) - (4*d*f*x + 4*d*e - f*cos(d*x + c))*sin(d*x +
c))/(a*d^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 724 vs. \(2 (78) = 156\).

Time = 2.49 (sec) , antiderivative size = 724, normalized size of antiderivative = 7.96 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {8 d e \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} - \frac {8 d e \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {8 d e \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {d f x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {8 d f x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} - \frac {6 d f x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {8 d f x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {d f x}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {2 f \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {8 f \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} - \frac {2 f \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} + \frac {8 f}{4 a d^{2} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8 a d^{2} \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 4 a d^{2}} & \text {for}\: d \neq 0 \\\frac {\left (e x + \frac {f x^{2}}{2}\right ) \cos ^{3}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((8*d*e*tan(c/2 + d*x/2)**3/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2)
- 8*d*e*tan(c/2 + d*x/2)**2/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*e*t
an(c/2 + d*x/2)/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + d*f*x*tan(c/2 + d*x
/2)**4/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*f*x*tan(c/2 + d*x/2)**3/
(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) - 6*d*f*x*tan(c/2 + d*x/2)**2/(4*a*d*
*2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*d*f*x*tan(c/2 + d*x/2)/(4*a*d**2*tan(c/2
+ d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + d*f*x/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c
/2 + d*x/2)**2 + 4*a*d**2) + 2*f*tan(c/2 + d*x/2)**3/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)
**2 + 4*a*d**2) + 8*f*tan(c/2 + d*x/2)**2/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d
**2) - 2*f*tan(c/2 + d*x/2)/(4*a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2) + 8*f/(4*
a*d**2*tan(c/2 + d*x/2)**4 + 8*a*d**2*tan(c/2 + d*x/2)**2 + 4*a*d**2), Ne(d, 0)), ((e*x + f*x**2/2)*cos(c)**3/
(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.25 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} e}{a} - \frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c f}{a d} - \frac {{\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} f}{a d}}{8 \, d} \]

[In]

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

[Out]

-1/8*(4*(sin(d*x + c)^2 - 2*sin(d*x + c))*e/a - 4*(sin(d*x + c)^2 - 2*sin(d*x + c))*c*f/(a*d) - (2*(d*x + c)*c
os(2*d*x + 2*c) + 8*(d*x + c)*sin(d*x + c) + 8*cos(d*x + c) - sin(2*d*x + 2*c))*f/(a*d))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 947 vs. \(2 (85) = 170\).

Time = 0.35 (sec) , antiderivative size = 947, normalized size of antiderivative = 10.41 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(d*f*x*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*d*f*x*tan(1/2*d*x)^4*tan(1/2*c)^3 - 8*d*f*x*tan(1/2*d*x)^3*tan(1/2*
c)^4 + d*e*tan(1/2*d*x)^4*tan(1/2*c)^4 - 6*d*f*x*tan(1/2*d*x)^4*tan(1/2*c)^2 - 16*d*f*x*tan(1/2*d*x)^3*tan(1/2
*c)^3 - 8*d*e*tan(1/2*d*x)^4*tan(1/2*c)^3 - 6*d*f*x*tan(1/2*d*x)^2*tan(1/2*c)^4 - 8*d*e*tan(1/2*d*x)^3*tan(1/2
*c)^4 + 4*f*tan(1/2*d*x)^4*tan(1/2*c)^4 - 8*d*f*x*tan(1/2*d*x)^4*tan(1/2*c) - 6*d*e*tan(1/2*d*x)^4*tan(1/2*c)^
2 - 16*d*e*tan(1/2*d*x)^3*tan(1/2*c)^3 + 2*f*tan(1/2*d*x)^4*tan(1/2*c)^3 - 8*d*f*x*tan(1/2*d*x)*tan(1/2*c)^4 -
6*d*e*tan(1/2*d*x)^2*tan(1/2*c)^4 + 2*f*tan(1/2*d*x)^3*tan(1/2*c)^4 + d*f*x*tan(1/2*d*x)^4 + 16*d*f*x*tan(1/2
*d*x)^3*tan(1/2*c) - 8*d*e*tan(1/2*d*x)^4*tan(1/2*c) + 36*d*f*x*tan(1/2*d*x)^2*tan(1/2*c)^2 + 16*d*f*x*tan(1/2
*d*x)*tan(1/2*c)^3 - 16*f*tan(1/2*d*x)^3*tan(1/2*c)^3 + d*f*x*tan(1/2*c)^4 - 8*d*e*tan(1/2*d*x)*tan(1/2*c)^4 +
8*d*f*x*tan(1/2*d*x)^3 + d*e*tan(1/2*d*x)^4 + 16*d*e*tan(1/2*d*x)^3*tan(1/2*c) - 2*f*tan(1/2*d*x)^4*tan(1/2*c
) + 36*d*e*tan(1/2*d*x)^2*tan(1/2*c)^2 - 12*f*tan(1/2*d*x)^3*tan(1/2*c)^2 + 8*d*f*x*tan(1/2*c)^3 + 16*d*e*tan(
1/2*d*x)*tan(1/2*c)^3 - 12*f*tan(1/2*d*x)^2*tan(1/2*c)^3 + d*e*tan(1/2*c)^4 - 2*f*tan(1/2*d*x)*tan(1/2*c)^4 -
6*d*f*x*tan(1/2*d*x)^2 + 8*d*e*tan(1/2*d*x)^3 - 4*f*tan(1/2*d*x)^4 - 16*d*f*x*tan(1/2*d*x)*tan(1/2*c) - 16*f*t
an(1/2*d*x)^3*tan(1/2*c) - 6*d*f*x*tan(1/2*c)^2 + 8*d*e*tan(1/2*c)^3 - 16*f*tan(1/2*d*x)*tan(1/2*c)^3 - 4*f*ta
n(1/2*c)^4 + 8*d*f*x*tan(1/2*d*x) - 6*d*e*tan(1/2*d*x)^2 + 2*f*tan(1/2*d*x)^3 + 8*d*f*x*tan(1/2*c) - 16*d*e*ta
n(1/2*d*x)*tan(1/2*c) + 12*f*tan(1/2*d*x)^2*tan(1/2*c) - 6*d*e*tan(1/2*c)^2 + 12*f*tan(1/2*d*x)*tan(1/2*c)^2 +
2*f*tan(1/2*c)^3 + d*f*x + 8*d*e*tan(1/2*d*x) + 8*d*e*tan(1/2*c) - 16*f*tan(1/2*d*x)*tan(1/2*c) + d*e - 2*f*t
an(1/2*d*x) - 2*f*tan(1/2*c) + 4*f)/(a*d^2*tan(1/2*d*x)^4*tan(1/2*c)^4 + 2*a*d^2*tan(1/2*d*x)^4*tan(1/2*c)^2 +
2*a*d^2*tan(1/2*d*x)^2*tan(1/2*c)^4 + a*d^2*tan(1/2*d*x)^4 + 4*a*d^2*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*d^2*tan(
1/2*c)^4 + 2*a*d^2*tan(1/2*d*x)^2 + 2*a*d^2*tan(1/2*c)^2 + a*d^2)

Mupad [B] (verification not implemented)

Time = 2.60 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.92 \[ \int \frac {(e+f x) \cos ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {f\,\sin \left (2\,c+2\,d\,x\right )}{2}+8\,f\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,d\,e\,\sin \left (c+d\,x\right )+2\,d\,e\,{\sin \left (c+d\,x\right )}^2-4\,d\,f\,x\,\sin \left (c+d\,x\right )+d\,f\,x\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )}{4\,a\,d^2} \]

[In]

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

[Out]

-((f*sin(2*c + 2*d*x))/2 + 8*f*sin(c/2 + (d*x)/2)^2 - 4*d*e*sin(c + d*x) + 2*d*e*sin(c + d*x)^2 - 4*d*f*x*sin(
c + d*x) + d*f*x*(2*sin(c + d*x)^2 - 1))/(4*a*d^2)