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

Optimal. Leaf size=156 \[ \frac {\left (48 a^2+16 a b+3 b^2\right ) \sin (e+f x) \cos ^3(e+f x)}{192 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{128 f}+\frac {1}{128} x \left (48 a^2+16 a b+3 b^2\right )-\frac {b (10 a+3 b) \sin (e+f x) \cos ^5(e+f x)}{48 f}-\frac {b \sin (e+f x) \cos ^7(e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 f} \]

[Out]

1/128*(48*a^2+16*a*b+3*b^2)*x+1/128*(48*a^2+16*a*b+3*b^2)*cos(f*x+e)*sin(f*x+e)/f+1/192*(48*a^2+16*a*b+3*b^2)*
cos(f*x+e)^3*sin(f*x+e)/f-1/48*b*(10*a+3*b)*cos(f*x+e)^5*sin(f*x+e)/f-1/8*b*cos(f*x+e)^7*sin(f*x+e)*(a+(a+b)*t
an(f*x+e)^2)/f

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Rubi [A]  time = 0.15, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3191, 413, 385, 199, 203} \[ \frac {\left (48 a^2+16 a b+3 b^2\right ) \sin (e+f x) \cos ^3(e+f x)}{192 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \sin (e+f x) \cos (e+f x)}{128 f}+\frac {1}{128} x \left (48 a^2+16 a b+3 b^2\right )-\frac {b (10 a+3 b) \sin (e+f x) \cos ^5(e+f x)}{48 f}-\frac {b \sin (e+f x) \cos ^7(e+f x) \left ((a+b) \tan ^2(e+f x)+a\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((48*a^2 + 16*a*b + 3*b^2)*x)/128 + ((48*a^2 + 16*a*b + 3*b^2)*Cos[e + f*x]*Sin[e + f*x])/(128*f) + ((48*a^2 +
16*a*b + 3*b^2)*Cos[e + f*x]^3*Sin[e + f*x])/(192*f) - (b*(10*a + 3*b)*Cos[e + f*x]^5*Sin[e + f*x])/(48*f) -
(b*Cos[e + f*x]^7*Sin[e + f*x]*(a + (a + b)*Tan[e + f*x]^2))/(8*f)

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

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

Rule 385

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 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 3191

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*} \int \cos ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+(a+b) x^2\right )^2}{\left (1+x^2\right )^5} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f}+\frac {\operatorname {Subst}\left (\int \frac {a (8 a+b)+(a+b) (8 a+3 b) x^2}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac {b (10 a+3 b) \cos ^5(e+f x) \sin (e+f x)}{48 f}-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{48 f}\\ &=\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos ^3(e+f x) \sin (e+f x)}{192 f}-\frac {b (10 a+3 b) \cos ^5(e+f x) \sin (e+f x)}{48 f}-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{64 f}\\ &=\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{128 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos ^3(e+f x) \sin (e+f x)}{192 f}-\frac {b (10 a+3 b) \cos ^5(e+f x) \sin (e+f x)}{48 f}-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{128 f}\\ &=\frac {1}{128} \left (48 a^2+16 a b+3 b^2\right ) x+\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos (e+f x) \sin (e+f x)}{128 f}+\frac {\left (48 a^2+16 a b+3 b^2\right ) \cos ^3(e+f x) \sin (e+f x)}{192 f}-\frac {b (10 a+3 b) \cos ^5(e+f x) \sin (e+f x)}{48 f}-\frac {b \cos ^7(e+f x) \sin (e+f x) \left (a+(a+b) \tan ^2(e+f x)\right )}{8 f}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 96, normalized size = 0.62 \[ \frac {24 \left (48 a^2+16 a b+3 b^2\right ) (e+f x)+24 \left (4 a^2-4 a b-b^2\right ) \sin (4 (e+f x))-32 a b \sin (6 (e+f x))+96 a (8 a+b) \sin (2 (e+f x))+3 b^2 \sin (8 (e+f x))}{3072 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*(48*a^2 + 16*a*b + 3*b^2)*(e + f*x) + 96*a*(8*a + b)*Sin[2*(e + f*x)] + 24*(4*a^2 - 4*a*b - b^2)*Sin[4*(e
+ f*x)] - 32*a*b*Sin[6*(e + f*x)] + 3*b^2*Sin[8*(e + f*x)])/(3072*f)

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fricas [A]  time = 0.42, size = 114, normalized size = 0.73 \[ \frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} f x + {\left (48 \, b^{2} \cos \left (f x + e\right )^{7} - 8 \, {\left (16 \, a b + 9 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{384 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*(48*a^2 + 16*a*b + 3*b^2)*f*x + (48*b^2*cos(f*x + e)^7 - 8*(16*a*b + 9*b^2)*cos(f*x + e)^5 + 2*(48*a^
2 + 16*a*b + 3*b^2)*cos(f*x + e)^3 + 3*(48*a^2 + 16*a*b + 3*b^2)*cos(f*x + e))*sin(f*x + e))/f

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giac [A]  time = 0.24, size = 108, normalized size = 0.69 \[ \frac {1}{128} \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} x + \frac {b^{2} \sin \left (8 \, f x + 8 \, e\right )}{1024 \, f} - \frac {a b \sin \left (6 \, f x + 6 \, e\right )}{96 \, f} + \frac {{\left (4 \, a^{2} - 4 \, a b - b^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{128 \, f} + \frac {{\left (8 \, a^{2} + a b\right )} \sin \left (2 \, f x + 2 \, e\right )}{32 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(48*a^2 + 16*a*b + 3*b^2)*x + 1/1024*b^2*sin(8*f*x + 8*e)/f - 1/96*a*b*sin(6*f*x + 6*e)/f + 1/128*(4*a^2
- 4*a*b - b^2)*sin(4*f*x + 4*e)/f + 1/32*(8*a^2 + a*b)*sin(2*f*x + 2*e)/f

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maple [A]  time = 0.51, size = 167, normalized size = 1.07 \[ \frac {b^{2} \left (-\frac {\left (\sin ^{3}\left (f x +e \right )\right ) \left (\cos ^{5}\left (f x +e \right )\right )}{8}-\frac {\sin \left (f x +e \right ) \left (\cos ^{5}\left (f x +e \right )\right )}{16}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{64}+\frac {3 f x}{128}+\frac {3 e}{128}\right )+2 a b \left (-\frac {\sin \left (f x +e \right ) \left (\cos ^{5}\left (f x +e \right )\right )}{6}+\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{24}+\frac {f x}{16}+\frac {e}{16}\right )+a^{2} \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(b^2*(-1/8*sin(f*x+e)^3*cos(f*x+e)^5-1/16*sin(f*x+e)*cos(f*x+e)^5+1/64*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f
*x+e)+3/128*f*x+3/128*e)+2*a*b*(-1/6*sin(f*x+e)*cos(f*x+e)^5+1/24*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+1/1
6*f*x+1/16*e)+a^2*(1/4*(cos(f*x+e)^3+3/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e))

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maxima [A]  time = 0.50, size = 169, normalized size = 1.08 \[ \frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} {\left (f x + e\right )} + \frac {3 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{7} + 11 \, {\left (48 \, a^{2} + 16 \, a b + 3 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + {\left (624 \, a^{2} + 80 \, a b - 33 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (80 \, a^{2} - 16 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{8} + 4 \, \tan \left (f x + e\right )^{6} + 6 \, \tan \left (f x + e\right )^{4} + 4 \, \tan \left (f x + e\right )^{2} + 1}}{384 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*(48*a^2 + 16*a*b + 3*b^2)*(f*x + e) + (3*(48*a^2 + 16*a*b + 3*b^2)*tan(f*x + e)^7 + 11*(48*a^2 + 16*a
*b + 3*b^2)*tan(f*x + e)^5 + (624*a^2 + 80*a*b - 33*b^2)*tan(f*x + e)^3 + 3*(80*a^2 - 16*a*b - 3*b^2)*tan(f*x
+ e))/(tan(f*x + e)^8 + 4*tan(f*x + e)^6 + 6*tan(f*x + e)^4 + 4*tan(f*x + e)^2 + 1))/f

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mupad [B]  time = 15.49, size = 160, normalized size = 1.03 \[ x\,\left (\frac {3\,a^2}{8}+\frac {a\,b}{8}+\frac {3\,b^2}{128}\right )+\frac {\left (\frac {3\,a^2}{8}+\frac {a\,b}{8}+\frac {3\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^7+\left (\frac {11\,a^2}{8}+\frac {11\,a\,b}{24}+\frac {11\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {13\,a^2}{8}+\frac {5\,a\,b}{24}-\frac {11\,b^2}{128}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {5\,a^2}{8}-\frac {a\,b}{8}-\frac {3\,b^2}{128}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^8+4\,{\mathrm {tan}\left (e+f\,x\right )}^6+6\,{\mathrm {tan}\left (e+f\,x\right )}^4+4\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((a*b)/8 + (3*a^2)/8 + (3*b^2)/128) + (tan(e + f*x)^7*((a*b)/8 + (3*a^2)/8 + (3*b^2)/128) - tan(e + f*x)*((a
*b)/8 - (5*a^2)/8 + (3*b^2)/128) + tan(e + f*x)^3*((5*a*b)/24 + (13*a^2)/8 - (11*b^2)/128) + tan(e + f*x)^5*((
11*a*b)/24 + (11*a^2)/8 + (11*b^2)/128))/(f*(4*tan(e + f*x)^2 + 6*tan(e + f*x)^4 + 4*tan(e + f*x)^6 + tan(e +
f*x)^8 + 1))

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sympy [A]  time = 14.62, size = 481, normalized size = 3.08 \[ \begin {cases} \frac {3 a^{2} x \sin ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {3 a^{2} x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {3 a^{2} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {5 a^{2} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} + \frac {a b x \sin ^{6}{\left (e + f x \right )}}{8} + \frac {3 a b x \sin ^{4}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{8} + \frac {3 a b x \sin ^{2}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{8} + \frac {a b x \cos ^{6}{\left (e + f x \right )}}{8} + \frac {a b \sin ^{5}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {a b \sin ^{3}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {a b \sin {\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{8 f} + \frac {3 b^{2} x \sin ^{8}{\left (e + f x \right )}}{128} + \frac {3 b^{2} x \sin ^{6}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{32} + \frac {9 b^{2} x \sin ^{4}{\left (e + f x \right )} \cos ^{4}{\left (e + f x \right )}}{64} + \frac {3 b^{2} x \sin ^{2}{\left (e + f x \right )} \cos ^{6}{\left (e + f x \right )}}{32} + \frac {3 b^{2} x \cos ^{8}{\left (e + f x \right )}}{128} + \frac {3 b^{2} \sin ^{7}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{128 f} + \frac {11 b^{2} \sin ^{5}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{128 f} - \frac {11 b^{2} \sin ^{3}{\left (e + f x \right )} \cos ^{5}{\left (e + f x \right )}}{128 f} - \frac {3 b^{2} \sin {\left (e + f x \right )} \cos ^{7}{\left (e + f x \right )}}{128 f} & \text {for}\: f \neq 0 \\x \left (a + b \sin ^{2}{\relax (e )}\right )^{2} \cos ^{4}{\relax (e )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((3*a**2*x*sin(e + f*x)**4/8 + 3*a**2*x*sin(e + f*x)**2*cos(e + f*x)**2/4 + 3*a**2*x*cos(e + f*x)**4/
8 + 3*a**2*sin(e + f*x)**3*cos(e + f*x)/(8*f) + 5*a**2*sin(e + f*x)*cos(e + f*x)**3/(8*f) + a*b*x*sin(e + f*x)
**6/8 + 3*a*b*x*sin(e + f*x)**4*cos(e + f*x)**2/8 + 3*a*b*x*sin(e + f*x)**2*cos(e + f*x)**4/8 + a*b*x*cos(e +
f*x)**6/8 + a*b*sin(e + f*x)**5*cos(e + f*x)/(8*f) + a*b*sin(e + f*x)**3*cos(e + f*x)**3/(3*f) - a*b*sin(e + f
*x)*cos(e + f*x)**5/(8*f) + 3*b**2*x*sin(e + f*x)**8/128 + 3*b**2*x*sin(e + f*x)**6*cos(e + f*x)**2/32 + 9*b**
2*x*sin(e + f*x)**4*cos(e + f*x)**4/64 + 3*b**2*x*sin(e + f*x)**2*cos(e + f*x)**6/32 + 3*b**2*x*cos(e + f*x)**
8/128 + 3*b**2*sin(e + f*x)**7*cos(e + f*x)/(128*f) + 11*b**2*sin(e + f*x)**5*cos(e + f*x)**3/(128*f) - 11*b**
2*sin(e + f*x)**3*cos(e + f*x)**5/(128*f) - 3*b**2*sin(e + f*x)*cos(e + f*x)**7/(128*f), Ne(f, 0)), (x*(a + b*
sin(e)**2)**2*cos(e)**4, True))

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