3.43 \(\int \sin ^5(e+f x) (a+b \tan ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=107 \[ -\frac {\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}-\frac {(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac {2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}+\frac {2 b (a-2 b) \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3664, 448} \[ -\frac {\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}-\frac {(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac {2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}+\frac {2 b (a-2 b) \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

-(((a^2 - 6*a*b + 6*b^2)*Cos[e + f*x])/f) + (2*(a - 2*b)*(a - b)*Cos[e + f*x]^3)/(3*f) - ((a - b)^2*Cos[e + f*
x]^5)/(5*f) + (2*(a - 2*b)*b*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^2}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 (a-2 b) b+\frac {(a-b)^2}{x^6}+\frac {2 (a-2 b) (-a+b)}{x^4}+\frac {a^2-6 a b+6 b^2}{x^2}+b^2 x^2\right ) \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\left (a^2-6 a b+6 b^2\right ) \cos (e+f x)}{f}+\frac {2 (a-2 b) (a-b) \cos ^3(e+f x)}{3 f}-\frac {(a-b)^2 \cos ^5(e+f x)}{5 f}+\frac {2 (a-2 b) b \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [A]  time = 0.72, size = 97, normalized size = 0.91 \[ \frac {-30 \left (5 a^2-38 a b+41 b^2\right ) \cos (e+f x)+5 (5 a-13 b) (a-b) \cos (3 (e+f x))-3 (a-b)^2 \cos (5 (e+f x))+480 b (a-2 b) \sec (e+f x)+80 b^2 \sec ^3(e+f x)}{240 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5*(a + b*Tan[e + f*x]^2)^2,x]

[Out]

(-30*(5*a^2 - 38*a*b + 41*b^2)*Cos[e + f*x] + 5*(5*a - 13*b)*(a - b)*Cos[3*(e + f*x)] - 3*(a - b)^2*Cos[5*(e +
f*x)] + 480*(a - 2*b)*b*Sec[e + f*x] + 80*b^2*Sec[e + f*x]^3)/(240*f)

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fricas [A]  time = 0.44, size = 105, normalized size = 0.98 \[ -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{8} - 10 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{6} + 15 \, {\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 30 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 5 \, b^{2}}{15 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^8 - 10*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^6 + 15*(a^2 - 6*a*b + 6*b^
2)*cos(f*x + e)^4 - 30*(a*b - 2*b^2)*cos(f*x + e)^2 - 5*b^2)/(f*cos(f*x + e)^3)

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giac [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(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.74, size = 185, normalized size = 1.73 \[ \frac {-\frac {a^{2} \left (\frac {8}{3}+\sin ^{4}\left (f x +e \right )+\frac {4 \left (\sin ^{2}\left (f x +e \right )\right )}{3}\right ) \cos \left (f x +e \right )}{5}+2 a b \left (\frac {\sin ^{8}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\frac {16}{5}+\sin ^{6}\left (f x +e \right )+\frac {6 \left (\sin ^{4}\left (f x +e \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (f x +e \right )\right )}{5}\right ) \cos \left (f x +e \right )\right )+b^{2} \left (\frac {\sin ^{10}\left (f x +e \right )}{3 \cos \left (f x +e \right )^{3}}-\frac {7 \left (\sin ^{10}\left (f x +e \right )\right )}{3 \cos \left (f x +e \right )}-\frac {7 \left (\frac {128}{35}+\sin ^{8}\left (f x +e \right )+\frac {8 \left (\sin ^{6}\left (f x +e \right )\right )}{7}+\frac {48 \left (\sin ^{4}\left (f x +e \right )\right )}{35}+\frac {64 \left (\sin ^{2}\left (f x +e \right )\right )}{35}\right ) \cos \left (f x +e \right )}{3}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x)

[Out]

1/f*(-1/5*a^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+2*a*b*(sin(f*x+e)^8/cos(f*x+e)+(16/5+sin(f*x+e)^6
+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*cos(f*x+e))+b^2*(1/3*sin(f*x+e)^10/cos(f*x+e)^3-7/3*sin(f*x+e)^10/cos(f*x+
e)-7/3*(128/35+sin(f*x+e)^8+8/7*sin(f*x+e)^6+48/35*sin(f*x+e)^4+64/35*sin(f*x+e)^2)*cos(f*x+e)))

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maxima [A]  time = 0.44, size = 104, normalized size = 0.97 \[ -\frac {3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right ) - \frac {5 \, {\left (6 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

-1/15*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 10*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^3 + 15*(a^2 - 6*a*b + 6*b^
2)*cos(f*x + e) - 5*(6*(a*b - 2*b^2)*cos(f*x + e)^2 + b^2)/cos(f*x + e)^3)/f

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mupad [B]  time = 12.65, size = 183, normalized size = 1.71 \[ \frac {2\,a^2\,{\cos \left (e+f\,x\right )}^3}{3\,f}-\frac {6\,b^2\,\cos \left (e+f\,x\right )}{f}-\frac {a^2\,\cos \left (e+f\,x\right )}{f}-\frac {a^2\,{\cos \left (e+f\,x\right )}^5}{5\,f}-\frac {4\,b^2}{f\,\cos \left (e+f\,x\right )}+\frac {b^2}{3\,f\,{\cos \left (e+f\,x\right )}^3}+\frac {4\,b^2\,{\cos \left (e+f\,x\right )}^3}{3\,f}-\frac {b^2\,{\cos \left (e+f\,x\right )}^5}{5\,f}+\frac {6\,a\,b\,\cos \left (e+f\,x\right )}{f}+\frac {2\,a\,b}{f\,\cos \left (e+f\,x\right )}-\frac {2\,a\,b\,{\cos \left (e+f\,x\right )}^3}{f}+\frac {2\,a\,b\,{\cos \left (e+f\,x\right )}^5}{5\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^2,x)

[Out]

(2*a^2*cos(e + f*x)^3)/(3*f) - (6*b^2*cos(e + f*x))/f - (a^2*cos(e + f*x))/f - (a^2*cos(e + f*x)^5)/(5*f) - (4
*b^2)/(f*cos(e + f*x)) + b^2/(3*f*cos(e + f*x)^3) + (4*b^2*cos(e + f*x)^3)/(3*f) - (b^2*cos(e + f*x)^5)/(5*f)
+ (6*a*b*cos(e + f*x))/f + (2*a*b)/(f*cos(e + f*x)) - (2*a*b*cos(e + f*x)^3)/f + (2*a*b*cos(e + f*x)^5)/(5*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{2} \sin ^{5}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**5*(a+b*tan(f*x+e)**2)**2,x)

[Out]

Integral((a + b*tan(e + f*x)**2)**2*sin(e + f*x)**5, x)

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