3.48 \(\int \frac {\sin ^4(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=191 \[ -\frac {3 \sqrt {b} \sqrt {a+b} (a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^4 f}-\frac {3 b (3 a+4 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {(5 a+6 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {3 x \left (a^2+8 a b+8 b^2\right )}{8 a^4}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4132, 470, 527, 522, 203, 205} \[ \frac {3 x \left (a^2+8 a b+8 b^2\right )}{8 a^4}-\frac {3 \sqrt {b} \sqrt {a+b} (a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^4 f}-\frac {3 b (3 a+4 b) \tan (e+f x)}{8 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}-\frac {(5 a+6 b) \sin (e+f x) \cos (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a^2 + 8*a*b + 8*b^2)*x)/(8*a^4) - (3*Sqrt[b]*Sqrt[a + b]*(a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a +
b]])/(2*a^4*f) - ((5*a + 6*b)*Cos[e + f*x]*Sin[e + f*x])/(8*a^2*f*(a + b + b*Tan[e + f*x]^2)) + (Cos[e + f*x]^
3*Sin[e + f*x])/(4*a*f*(a + b + b*Tan[e + f*x]^2)) - (3*b*(3*a + 4*b)*Tan[e + f*x])/(8*a^3*f*(a + b + b*Tan[e
+ f*x]^2))

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 205

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

Rule 470

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

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rubi steps

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

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Mathematica [C]  time = 13.24, size = 1105, normalized size = 5.79 \[ -\frac {(\cos (2 e+2 f x) a+a+2 b)^2 \left (16 x+\frac {\left (-a^3+6 b a^2+24 b^2 a+16 b^3\right ) \tan ^{-1}\left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\left (a^2+8 b a+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (\cos (2 (e+f x)) a+a+2 b) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right ) \sec ^4(e+f x)}{256 a^2 \left (b \sec ^2(e+f x)+a\right )^2}+\frac {3 (\cos (2 e+2 f x) a+a+2 b)^2 \left (\frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (\cos (2 (e+f x)) a+a+2 b)}\right ) \sec ^4(e+f x)}{1024 b^{3/2} f \left (b \sec ^2(e+f x)+a\right )^2}+\frac {(\cos (2 e+2 f x) a+a+2 b)^2 \left (\frac {\sec (2 e) \left (\sin (2 e) a^5-\sin (2 f x) a^5+80 b f x \cos (4 e+2 f x) a^4+34 b \sin (2 e) a^4-62 b \sin (2 f x) a^4-12 b \sin (2 (e+2 f x)) a^4-30 b \sin (4 e+2 f x) a^4-12 b \sin (6 e+4 f x) a^4+2 b \sin (4 e+6 f x) a^4+2 b \sin (8 e+6 f x) a^4+464 b^2 f x \cos (4 e+2 f x) a^3+224 b^2 \sin (2 e) a^3-318 b^2 \sin (2 f x) a^3-36 b^2 \sin (2 (e+2 f x)) a^3-158 b^2 \sin (4 e+2 f x) a^3-36 b^2 \sin (6 e+4 f x) a^3+2 b^2 \sin (4 e+6 f x) a^3+2 b^2 \sin (8 e+6 f x) a^3+768 b^3 f x \cos (4 e+2 f x) a^2+576 b^3 \sin (2 e) a^2-512 b^3 \sin (2 f x) a^2-24 b^3 \sin (2 (e+2 f x)) a^2-256 b^3 \sin (4 e+2 f x) a^2-24 b^3 \sin (6 e+4 f x) a^2+16 b \left (5 a^3+29 b a^2+48 b^2 a+24 b^3\right ) f x \cos (2 f x) a+384 b^4 f x \cos (4 e+2 f x) a+640 b^4 \sin (2 e) a-256 b^4 \sin (2 f x) a-128 b^4 \sin (4 e+2 f x) a+32 b \left (5 a^4+39 b a^3+106 b^2 a^2+120 b^3 a+48 b^4\right ) f x \cos (2 e)+256 b^5 \sin (2 e)\right )}{\cos (2 (e+f x)) a+a+2 b}-\frac {\left (a^5-30 b a^4-480 b^2 a^3-1600 b^3 a^2-1920 b^4 a-768 b^5\right ) \tan ^{-1}\left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) \sec ^4(e+f x)}{1024 a^4 b (a+b) f \left (b \sec ^2(e+f x)+a\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/256*((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(
Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I
*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b
^2)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e]
+ Sin[e]))))/(a^2*(a + b*Sec[e + f*x]^2)^2) + (3*(a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*x]^4*(((a + 2*b)*
ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2*(e + f*x)])/((a + b)*(a + 2*b + a
*Cos[2*(e + f*x)]))))/(1024*b^(3/2)*f*(a + b*Sec[e + f*x]^2)^2) + ((a + 2*b + a*Cos[2*e + 2*f*x])^2*Sec[e + f*
x]^4*(-(((a^5 - 30*a^4*b - 480*a^3*b^2 - 1600*a^2*b^3 - 1920*a*b^4 - 768*b^5)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*S
in[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e]
- I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])) + (Sec[2*e]*(32*b*(5*a^4 + 39*a^3*b + 106*a^2*b^2
+ 120*a*b^3 + 48*b^4)*f*x*Cos[2*e] + 16*a*b*(5*a^3 + 29*a^2*b + 48*a*b^2 + 24*b^3)*f*x*Cos[2*f*x] + 80*a^4*b*f
*x*Cos[4*e + 2*f*x] + 464*a^3*b^2*f*x*Cos[4*e + 2*f*x] + 768*a^2*b^3*f*x*Cos[4*e + 2*f*x] + 384*a*b^4*f*x*Cos[
4*e + 2*f*x] + a^5*Sin[2*e] + 34*a^4*b*Sin[2*e] + 224*a^3*b^2*Sin[2*e] + 576*a^2*b^3*Sin[2*e] + 640*a*b^4*Sin[
2*e] + 256*b^5*Sin[2*e] - a^5*Sin[2*f*x] - 62*a^4*b*Sin[2*f*x] - 318*a^3*b^2*Sin[2*f*x] - 512*a^2*b^3*Sin[2*f*
x] - 256*a*b^4*Sin[2*f*x] - 12*a^4*b*Sin[2*(e + 2*f*x)] - 36*a^3*b^2*Sin[2*(e + 2*f*x)] - 24*a^2*b^3*Sin[2*(e
+ 2*f*x)] - 30*a^4*b*Sin[4*e + 2*f*x] - 158*a^3*b^2*Sin[4*e + 2*f*x] - 256*a^2*b^3*Sin[4*e + 2*f*x] - 128*a*b^
4*Sin[4*e + 2*f*x] - 12*a^4*b*Sin[6*e + 4*f*x] - 36*a^3*b^2*Sin[6*e + 4*f*x] - 24*a^2*b^3*Sin[6*e + 4*f*x] + 2
*a^4*b*Sin[4*e + 6*f*x] + 2*a^3*b^2*Sin[4*e + 6*f*x] + 2*a^4*b*Sin[8*e + 6*f*x] + 2*a^3*b^2*Sin[8*e + 6*f*x]))
/(a + 2*b + a*Cos[2*(e + f*x)])))/(1024*a^4*b*(a + b)*f*(a + b*Sec[e + f*x]^2)^2)

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fricas [A]  time = 0.62, size = 522, normalized size = 2.73 \[ \left [\frac {3 \, {\left (a^{3} + 8 \, a^{2} b + 8 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} f x + 3 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + {\left (2 \, a^{3} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}, \frac {3 \, {\left (a^{3} + 8 \, a^{2} b + 8 \, a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 8 \, a b^{2} + 8 \, b^{3}\right )} f x + 6 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + {\left (2 \, a^{3} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} + 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(a^3 + 8*a^2*b + 8*a*b^2)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 8*a*b^2 + 8*b^3)*f*x + 3*((a^2 + 2*a*b)*cos(
f*x + e)^2 + a*b + 2*b^2)*sqrt(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2)*cos(f
*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x +
e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)) + (2*a^3*cos(f*x + e)^5 - (5*a^3 + 6*a^2*b)*cos(f*x + e)^3 - 3*(3*a^2*b +
4*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f), 1/8*(3*(a^3 + 8*a^2*b + 8*a*b^2)*f*x*co
s(f*x + e)^2 + 3*(a^2*b + 8*a*b^2 + 8*b^3)*f*x + 6*((a^2 + 2*a*b)*cos(f*x + e)^2 + a*b + 2*b^2)*sqrt(a*b + b^2
)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) + (2*a^3*cos(f*x + e)
^5 - (5*a^3 + 6*a^2*b)*cos(f*x + e)^3 - 3*(3*a^2*b + 4*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^5*f*cos(f*x + e)^
2 + a^4*b*f)]

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giac [A]  time = 0.35, size = 204, normalized size = 1.07 \[ \frac {\frac {3 \, {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}} - \frac {12 \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{4}} - \frac {4 \, {\left (a b \tan \left (f x + e\right ) + b^{2} \tan \left (f x + e\right )\right )}}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a^{3}} - \frac {5 \, a \tan \left (f x + e\right )^{3} + 8 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 8 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2} a^{3}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*(a^2 + 8*a*b + 8*b^2)*(f*x + e)/a^4 - 12*(a^2*b + 3*a*b^2 + 2*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b)
+ arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*a^4) - 4*(a*b*tan(f*x + e) + b^2*tan(f*x + e))/((b
*tan(f*x + e)^2 + a + b)*a^3) - (5*a*tan(f*x + e)^3 + 8*b*tan(f*x + e)^3 + 3*a*tan(f*x + e) + 8*b*tan(f*x + e)
)/((tan(f*x + e)^2 + 1)^2*a^3))/f

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maple [A]  time = 0.89, size = 323, normalized size = 1.69 \[ -\frac {b \tan \left (f x +e \right )}{2 a^{2} f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{2} \sqrt {\left (a +b \right ) b}}-\frac {9 b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f \,a^{3} \sqrt {\left (a +b \right ) b}}-\frac {b^{2} \tan \left (f x +e \right )}{2 f \,a^{3} \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b^{3} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{4} \sqrt {\left (a +b \right ) b}}-\frac {\left (\tan ^{3}\left (f x +e \right )\right ) b}{f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {3 \tan \left (f x +e \right )}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {\tan \left (f x +e \right ) b}{f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}+\frac {3 \arctan \left (\tan \left (f x +e \right )\right ) b}{f \,a^{3}}+\frac {3 \arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{f \,a^{4}}+\frac {3 \arctan \left (\tan \left (f x +e \right )\right )}{8 f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*tan(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)-3/2/f*b/a^2/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-
9/2/f*b^2/a^3/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/2/f*b^2/a^3*tan(f*x+e)/(a+b+b*tan(f*x+e)^
2)-3/f*b^3/a^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))-1/f/a^3/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3*b-
5/8/f/a^2/(tan(f*x+e)^2+1)^2*tan(f*x+e)^3-3/8/f/a^2/(tan(f*x+e)^2+1)^2*tan(f*x+e)-1/f/a^3/(tan(f*x+e)^2+1)^2*t
an(f*x+e)*b+3/f/a^3*arctan(tan(f*x+e))*b+3/f/a^4*arctan(tan(f*x+e))*b^2+3/8/f/a^2*arctan(tan(f*x+e))

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maxima [A]  time = 0.48, size = 205, normalized size = 1.07 \[ -\frac {\frac {3 \, {\left (3 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{2} + 24 \, a b + 24 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} + 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{3} b \tan \left (f x + e\right )^{6} + {\left (a^{4} + 3 \, a^{3} b\right )} \tan \left (f x + e\right )^{4} + a^{4} + a^{3} b + {\left (2 \, a^{4} + 3 \, a^{3} b\right )} \tan \left (f x + e\right )^{2}} - \frac {3 \, {\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )}}{a^{4}} + \frac {12 \, {\left (a^{2} b + 3 \, a b^{2} + 2 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{4}}}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((3*(3*a*b + 4*b^2)*tan(f*x + e)^5 + (5*a^2 + 24*a*b + 24*b^2)*tan(f*x + e)^3 + 3*(a^2 + 5*a*b + 4*b^2)*t
an(f*x + e))/(a^3*b*tan(f*x + e)^6 + (a^4 + 3*a^3*b)*tan(f*x + e)^4 + a^4 + a^3*b + (2*a^4 + 3*a^3*b)*tan(f*x
+ e)^2) - 3*(a^2 + 8*a*b + 8*b^2)*(f*x + e)/a^4 + 12*(a^2*b + 3*a*b^2 + 2*b^3)*arctan(b*tan(f*x + e)/sqrt((a +
b)*b))/(sqrt((a + b)*b)*a^4))/f

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mupad [B]  time = 5.77, size = 435, normalized size = 2.28 \[ \frac {3\,\mathrm {atanh}\left (\frac {27\,b^3\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^2-a\,b}}{64\,\left (\frac {27\,a\,b^3}{64}+\frac {81\,b^4}{64}+\frac {27\,b^5}{32\,a}\right )}+\frac {27\,b^4\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^2-a\,b}}{32\,\left (\frac {27\,a^2\,b^3}{64}+\frac {81\,a\,b^4}{64}+\frac {27\,b^5}{32}\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{2\,a^4\,f}-\frac {\frac {3\,{\mathrm {tan}\left (e+f\,x\right )}^5\,\left (4\,b^2+3\,a\,b\right )}{8\,a^3}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+24\,a\,b+24\,b^2\right )}{8\,a^3}+\frac {3\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^2+5\,a\,b+4\,b^2\right )}{8\,a^3}}{f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^6+\left (a+3\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^4+\left (2\,a+3\,b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {\mathrm {atan}\left (\frac {27\,b^2\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,b^2}{256}+\frac {243\,b^3}{256\,a}+\frac {27\,b^4}{16\,a^2}+\frac {27\,b^5}{32\,a^3}\right )}+\frac {243\,b^3\,\mathrm {tan}\left (e+f\,x\right )}{256\,\left (\frac {27\,a\,b^2}{256}+\frac {243\,b^3}{256}+\frac {27\,b^4}{16\,a}+\frac {27\,b^5}{32\,a^2}\right )}+\frac {27\,b^4\,\mathrm {tan}\left (e+f\,x\right )}{16\,\left (\frac {243\,a\,b^3}{256}+\frac {27\,b^4}{16}+\frac {27\,a^2\,b^2}{256}+\frac {27\,b^5}{32\,a}\right )}+\frac {27\,b^5\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {27\,a^3\,b^2}{256}+\frac {243\,a^2\,b^3}{256}+\frac {27\,a\,b^4}{16}+\frac {27\,b^5}{32}\right )}\right )\,\left (a^2\,1{}\mathrm {i}+a\,b\,8{}\mathrm {i}+b^2\,8{}\mathrm {i}\right )\,3{}\mathrm {i}}{8\,a^4\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*atanh((27*b^3*tan(e + f*x)*(- a*b - b^2)^(1/2))/(64*((27*a*b^3)/64 + (81*b^4)/64 + (27*b^5)/(32*a))) + (27*
b^4*tan(e + f*x)*(- a*b - b^2)^(1/2))/(32*((81*a*b^4)/64 + (27*b^5)/32 + (27*a^2*b^3)/64)))*(a + 2*b)*(-b*(a +
b))^(1/2))/(2*a^4*f) - (atan((27*b^2*tan(e + f*x))/(256*((27*b^2)/256 + (243*b^3)/(256*a) + (27*b^4)/(16*a^2)
+ (27*b^5)/(32*a^3))) + (243*b^3*tan(e + f*x))/(256*((27*a*b^2)/256 + (243*b^3)/256 + (27*b^4)/(16*a) + (27*b
^5)/(32*a^2))) + (27*b^4*tan(e + f*x))/(16*((243*a*b^3)/256 + (27*b^4)/16 + (27*a^2*b^2)/256 + (27*b^5)/(32*a)
)) + (27*b^5*tan(e + f*x))/(32*((27*a*b^4)/16 + (27*b^5)/32 + (243*a^2*b^3)/256 + (27*a^3*b^2)/256)))*(a*b*8i
+ a^2*1i + b^2*8i)*3i)/(8*a^4*f) - ((3*tan(e + f*x)^5*(3*a*b + 4*b^2))/(8*a^3) + (tan(e + f*x)^3*(24*a*b + 5*a
^2 + 24*b^2))/(8*a^3) + (3*tan(e + f*x)*(5*a*b + a^2 + 4*b^2))/(8*a^3))/(f*(a + b + tan(e + f*x)^2*(2*a + 3*b)
+ b*tan(e + f*x)^6 + tan(e + f*x)^4*(a + 3*b)))

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sympy [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)**4/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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