\(\int \cos ^4(e+f x) (a+b \sin ^2(e+f x)) \, dx\) [286]

Optimal result

Integrand size = 21, antiderivative size = 83 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {1}{16} (6 a+b) x+\frac {(6 a+b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(6 a+b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f} \]

[Out]

1/16*(6*a+b)*x+1/16*(6*a+b)*cos(f*x+e)*sin(f*x+e)/f+1/24*(6*a+b)*cos(f*x+e)^3*sin(f*x+e)/f-1/6*b*cos(f*x+e)^5*
sin(f*x+e)/f

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3270, 393, 205, 209} \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {(6 a+b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac {(6 a+b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {1}{16} x (6 a+b)-\frac {b \sin (e+f x) \cos ^5(e+f x)}{6 f} \]

[In]

Int[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2),x]

[Out]

((6*a + b)*x)/16 + ((6*a + b)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + ((6*a + b)*Cos[e + f*x]^3*Sin[e + f*x])/(24*
f) - (b*Cos[e + f*x]^5*Sin[e + f*x])/(6*f)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 3270

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a+(a+b) x^2}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {(6 a+b) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f} \\ & = \frac {(6 a+b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {(6 a+b) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = \frac {(6 a+b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(6 a+b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f}+\frac {(6 a+b) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f} \\ & = \frac {1}{16} (6 a+b) x+\frac {(6 a+b) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {(6 a+b) \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {b \cos ^5(e+f x) \sin (e+f x)}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {72 a e+72 a f x+12 b f x+3 (16 a+b) \sin (2 (e+f x))+(6 a-3 b) \sin (4 (e+f x))-b \sin (6 (e+f x))}{192 f} \]

[In]

Integrate[Cos[e + f*x]^4*(a + b*Sin[e + f*x]^2),x]

[Out]

(72*a*e + 72*a*f*x + 12*b*f*x + 3*(16*a + b)*Sin[2*(e + f*x)] + (6*a - 3*b)*Sin[4*(e + f*x)] - b*Sin[6*(e + f*
x)])/(192*f)

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {\left (48 a +3 b \right ) \sin \left (2 f x +2 e \right )+\left (6 a -3 b \right ) \sin \left (4 f x +4 e \right )-\sin \left (6 f x +6 e \right ) b +72 f \left (a +\frac {b}{6}\right ) x}{192 f}\)	\(62\)
risch	\(\frac {3 a x}{8}+\frac {b x}{16}-\frac {\sin \left (6 f x +6 e \right ) b}{192 f}+\frac {\sin \left (4 f x +4 e \right ) a}{32 f}-\frac {\sin \left (4 f x +4 e \right ) b}{64 f}+\frac {\sin \left (2 f x +2 e \right ) a}{4 f}+\frac {\sin \left (2 f x +2 e \right ) b}{64 f}\)	\(85\)
derivativedivides	\(\frac {b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+a \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(92\)
default	\(\frac {b \left (-\frac {\left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+a \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}\)	\(92\)
norman	\(\frac {\left (\frac {3 a}{8}+\frac {b}{16}\right ) x +\left (\frac {3 a}{8}+\frac {b}{16}\right ) x \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {9 a}{4}+\frac {3 b}{8}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {9 a}{4}+\frac {3 b}{8}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {15 a}{2}+\frac {5 b}{4}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {45 a}{8}+\frac {15 b}{16}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (\frac {45 a}{8}+\frac {15 b}{16}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\frac {\left (2 a -13 b \right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {\left (2 a -13 b \right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}+\frac {\left (10 a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}-\frac {\left (10 a -b \right ) \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}+\frac {\left (42 a +47 b \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}-\frac {\left (42 a +47 b \right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{6}}\)	\(283\)

[In]

int(cos(f*x+e)^4*(a+b*sin(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/192*((48*a+3*b)*sin(2*f*x+2*e)+(6*a-3*b)*sin(4*f*x+4*e)-sin(6*f*x+6*e)*b+72*f*(a+1/6*b)*x)/f

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (6 \, a + b\right )} f x - {\left (8 \, b \cos \left (f x + e\right )^{5} - 2 \, {\left (6 \, a + b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (6 \, a + b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \]

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="fricas")

[Out]

1/48*(3*(6*a + b)*f*x - (8*b*cos(f*x + e)^5 - 2*(6*a + b)*cos(f*x + e)^3 - 3*(6*a + b)*cos(f*x + e))*sin(f*x +
e))/f

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (76) = 152\).

Time = 0.40 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.01 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\begin {cases} \frac {3 a x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 a x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 a \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 a \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {b x \sin ^{6}{\left (e + f x \right )}}{16} + \frac {3 b x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{16} + \frac {3 b x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{16} + \frac {b x \cos ^{6}{\left (e + f x \right )}}{16} + \frac {b \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{16 f} + \frac {b \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{6 f} - \frac {b \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{16 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin ^{2}{\left (e \right )}\right ) \cos ^{4}{\left (e \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(f*x+e)**4*(a+b*sin(f*x+e)**2),x)

[Out]

Piecewise((3*a*x*sin(e + f*x)**4/8 + 3*a*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 3*a*x*cos(e + f*x)**4/8 + 3*a*s
in(e + f*x)**3*cos(e + f*x)/(8*f) + 5*a*sin(e + f*x)*cos(e + f*x)**3/(8*f) + b*x*sin(e + f*x)**6/16 + 3*b*x*si
n(e + f*x)**4*cos(e + f*x)**2/16 + 3*b*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + b*x*cos(e + f*x)**6/16 + b*sin(e
+ f*x)**5*cos(e + f*x)/(16*f) + b*sin(e + f*x)**3*cos(e + f*x)**3/(6*f) - b*sin(e + f*x)*cos(e + f*x)**5/(16*
f), Ne(f, 0)), (x*(a + b*sin(e)**2)*cos(e)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {3 \, {\left (f x + e\right )} {\left (6 \, a + b\right )} + \frac {3 \, {\left (6 \, a + b\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (6 \, a + b\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (10 \, a - b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="maxima")

[Out]

1/48*(3*(f*x + e)*(6*a + b) + (3*(6*a + b)*tan(f*x + e)^5 + 8*(6*a + b)*tan(f*x + e)^3 + 3*(10*a - b)*tan(f*x
+ e))/(tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1))/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.77 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=\frac {1}{16} \, {\left (6 \, a + b\right )} x - \frac {b \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {{\left (2 \, a - b\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {{\left (16 \, a + b\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \]

[In]

integrate(cos(f*x+e)^4*(a+b*sin(f*x+e)^2),x, algorithm="giac")

[Out]

1/16*(6*a + b)*x - 1/192*b*sin(6*f*x + 6*e)/f + 1/64*(2*a - b)*sin(4*f*x + 4*e)/f + 1/64*(16*a + b)*sin(2*f*x
+ 2*e)/f

Mupad [B] (verification not implemented)

Time = 14.56 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right ) \, dx=x\,\left (\frac {3\,a}{8}+\frac {b}{16}\right )+\frac {\left (\frac {3\,a}{8}+\frac {b}{16}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (a+\frac {b}{6}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a}{8}-\frac {b}{16}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

[In]

int(cos(e + f*x)^4*(a + b*sin(e + f*x)^2),x)

[Out]

x*((3*a)/8 + b/16) + (tan(e + f*x)^5*((3*a)/8 + b/16) + tan(e + f*x)*((5*a)/8 - b/16) + tan(e + f*x)^3*(a + b/
6))/(f*(3*tan(e + f*x)^2 + 3*tan(e + f*x)^4 + tan(e + f*x)^6 + 1))