3.41 \(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=161 \[ \frac {\sqrt {b} (a+b) (3 a+7 b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{9/2} f}-\frac {(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac {(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (a \cos ^2(e+f x)+b\right )}-\frac {\cos ^5(e+f x)}{5 a^2 f} \]

[Out]

-1/2*(a+b)*(3*a+7*b)*cos(f*x+e)/a^4/f+1/6*(a+b)*(3*a+7*b)*cos(f*x+e)^3/a^3/b/f-1/5*cos(f*x+e)^5/a^2/f-1/2*(a+b
)^2*cos(f*x+e)^5/a^2/b/f/(b+a*cos(f*x+e)^2)+1/2*(a+b)*(3*a+7*b)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^(
9/2)/f

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Rubi [A]  time = 0.18, antiderivative size = 161, 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 = {4133, 463, 459, 302, 205} \[ -\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (a \cos ^2(e+f x)+b\right )}+\frac {(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac {(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac {\sqrt {b} (a+b) (3 a+7 b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{9/2} f}-\frac {\cos ^5(e+f x)}{5 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*(a + b)*(3*a + 7*b)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/(2*a^(9/2)*f) - ((a + b)*(3*a + 7*b)*Cos[
e + f*x])/(2*a^4*f) + ((a + b)*(3*a + 7*b)*Cos[e + f*x]^3)/(6*a^3*b*f) - Cos[e + f*x]^5/(5*a^2*f) - ((a + b)^2
*Cos[e + f*x]^5)/(2*a^2*b*f*(b + a*Cos[e + f*x]^2))

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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 459

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

Rule 463

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

Rule 4133

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

Rubi steps

\begin {align*} \int \frac {\sin ^5(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 )^2}{\left (b+a x^2\right )^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (b+a \cos ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {x^4 \left (-2 a^2+5 (a+b)^2-2 a b x^2\right )}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{2 a^2 b f}\\ &=-\frac {\cos ^5(e+f x)}{5 a^2 f}-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (b+a \cos ^2(e+f x)\right )}+\frac {((a+b) (3 a+7 b)) \operatorname {Subst}\left (\int \frac {x^4}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{2 a^2 b f}\\ &=-\frac {\cos ^5(e+f x)}{5 a^2 f}-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (b+a \cos ^2(e+f x)\right )}+\frac {((a+b) (3 a+7 b)) \operatorname {Subst}\left (\int \left (-\frac {b}{a^2}+\frac {x^2}{a}+\frac {b^2}{a^2 \left (b+a x^2\right )}\right ) \, dx,x,\cos (e+f x)\right )}{2 a^2 b f}\\ &=-\frac {(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac {(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac {\cos ^5(e+f x)}{5 a^2 f}-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (b+a \cos ^2(e+f x)\right )}+\frac {(b (a+b) (3 a+7 b)) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{2 a^4 f}\\ &=\frac {\sqrt {b} (a+b) (3 a+7 b) \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{2 a^{9/2} f}-\frac {(a+b) (3 a+7 b) \cos (e+f x)}{2 a^4 f}+\frac {(a+b) (3 a+7 b) \cos ^3(e+f x)}{6 a^3 b f}-\frac {\cos ^5(e+f x)}{5 a^2 f}-\frac {(a+b)^2 \cos ^5(e+f x)}{2 a^2 b f \left (b+a \cos ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [C]  time = 6.30, size = 454, normalized size = 2.82 \[ \frac {-\frac {45 a^4 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {45 a^4 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {a}}{\sqrt {b}}\right )}{b^{3/2}}+\frac {15 \left (3 a^4+384 a^2 b^2+1280 a b^3+896 b^4\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}+\frac {15 \left (3 a^4+384 a^2 b^2+1280 a b^3+896 b^4\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )}{b^{3/2}}-\frac {16 \sqrt {a} \cos (e+f x) \left (3 a^3 \cos (6 (e+f x))+150 a^3+a \left (125 a^2+688 a b+560 b^2\right ) \cos (2 (e+f x))-2 a^2 (11 a+14 b) \cos (4 (e+f x))+1436 a^2 b+2960 a b^2+1680 b^3\right )}{a \cos (2 (e+f x))+a+2 b}}{3840 a^{9/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*b^4)*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2
])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]])/b^
(3/2) + (15*(3*a^4 + 384*a^2*b^2 + 1280*a*b^3 + 896*b^4)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Si
n[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[
b]])/b^(3/2) - (45*a^4*ArcTan[(Sqrt[a] - Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (45*a^4*ArcTan[(Sqr
t[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]])/b^(3/2) - (16*Sqrt[a]*Cos[e + f*x]*(150*a^3 + 1436*a^2*b + 2960
*a*b^2 + 1680*b^3 + a*(125*a^2 + 688*a*b + 560*b^2)*Cos[2*(e + f*x)] - 2*a^2*(11*a + 14*b)*Cos[4*(e + f*x)] +
3*a^3*Cos[6*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f*x)]))/(3840*a^(9/2)*f)

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fricas [A]  time = 0.70, size = 405, normalized size = 2.52 \[ \left [-\frac {12 \, a^{3} \cos \left (f x + e\right )^{7} - 4 \, {\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 20 \, {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{60 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}, -\frac {6 \, a^{3} \cos \left (f x + e\right )^{7} - 2 \, {\left (10 \, a^{3} + 7 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3} + {\left (3 \, a^{3} + 10 \, a^{2} b + 7 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \cos \left (f x + e\right )}{30 \, {\left (a^{5} f \cos \left (f x + e\right )^{2} + a^{4} b f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/60*(12*a^3*cos(f*x + e)^7 - 4*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 + 20*(3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x
+ e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^3 + (3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^2)*sqrt(-b/a)*log(-(a*cos(
f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 30*(3*a^2*b + 10*a*b^2 + 7*b^3)*cos(f*
x + e))/(a^5*f*cos(f*x + e)^2 + a^4*b*f), -1/30*(6*a^3*cos(f*x + e)^7 - 2*(10*a^3 + 7*a^2*b)*cos(f*x + e)^5 +
10*(3*a^3 + 10*a^2*b + 7*a*b^2)*cos(f*x + e)^3 - 15*(3*a^2*b + 10*a*b^2 + 7*b^3 + (3*a^3 + 10*a^2*b + 7*a*b^2)
*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b) + 15*(3*a^2*b + 10*a*b^2 + 7*b^3)*cos(f*x + e))/
(a^5*f*cos(f*x + e)^2 + a^4*b*f)]

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giac [B]  time = 1.41, size = 545, normalized size = 3.39 \[ -\frac {\frac {15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b}{\sqrt {a b} \cos \left (f x + e\right ) + \sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {30 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3} + \frac {a^{2} b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {b^{3} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )}}{{\left (a + b + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} a^{4}} - \frac {4 \, {\left (8 \, a^{2} + 50 \, a b + 45 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {220 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {180 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {320 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {270 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {180 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {180 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {30 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {45 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a^{4} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/30*(15*(3*a^2*b + 10*a*b^2 + 7*b^3)*arctan(-(a*cos(f*x + e) - b)/(sqrt(a*b)*cos(f*x + e) + sqrt(a*b)))/(sqr
t(a*b)*a^4) + 30*(a^2*b + 2*a*b^2 + b^3 + a^2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - b^3*(cos(f*x + e) - 1)
/(cos(f*x + e) + 1))/((a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e
) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*a^4) - 4*(
8*a^2 + 50*a*b + 45*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 220*a*b*(cos(f*x + e) - 1)/(cos(f*x +
e) + 1) - 180*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 +
320*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 270*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 180*a*
b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 180*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 30*a*b*(cos(
f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/(a^4*((cos(f*x + e) -
1)/(cos(f*x + e) + 1) - 1)^5))/f

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maple [A]  time = 0.95, size = 276, normalized size = 1.71 \[ -\frac {\cos ^{5}\left (f x +e \right )}{5 a^{2} f}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{3 a^{2} f}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right ) b}{3 f \,a^{3}}-\frac {\cos \left (f x +e \right )}{a^{2} f}-\frac {4 \cos \left (f x +e \right ) b}{f \,a^{3}}-\frac {3 \cos \left (f x +e \right ) b^{2}}{f \,a^{4}}-\frac {b \cos \left (f x +e \right )}{2 f \,a^{2} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{2} \cos \left (f x +e \right )}{f \,a^{3} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}-\frac {b^{3} \cos \left (f x +e \right )}{2 f \,a^{4} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {3 b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \,a^{2} \sqrt {a b}}+\frac {5 b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \,a^{3} \sqrt {a b}}+\frac {7 b^{3} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{2 f \,a^{4} \sqrt {a b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*cos(f*x+e)^5/a^2/f+2/3*cos(f*x+e)^3/a^2/f+2/3/f/a^3*cos(f*x+e)^3*b-cos(f*x+e)/a^2/f-4/f/a^3*cos(f*x+e)*b-
3/f/a^4*cos(f*x+e)*b^2-1/2/f*b/a^2*cos(f*x+e)/(b+a*cos(f*x+e)^2)-1/f*b^2/a^3*cos(f*x+e)/(b+a*cos(f*x+e)^2)-1/2
/f*b^3/a^4*cos(f*x+e)/(b+a*cos(f*x+e)^2)+3/2/f*b/a^2/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+5/f*b^2/a^3/
(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))+7/2/f*b^3/a^4/(a*b)^(1/2)*arctan(a*cos(f*x+e)/(a*b)^(1/2))

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maxima [A]  time = 0.53, size = 148, normalized size = 0.92 \[ -\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )}{a^{5} \cos \left (f x + e\right )^{2} + a^{4} b} - \frac {15 \, {\left (3 \, a^{2} b + 10 \, a b^{2} + 7 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{4}} + \frac {2 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )\right )}}{a^{4}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(15*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)/(a^5*cos(f*x + e)^2 + a^4*b) - 15*(3*a^2*b + 10*a*b^2 + 7*b^3)*
arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^4) + 2*(3*a^2*cos(f*x + e)^5 - 10*(a^2 + a*b)*cos(f*x + e)^3 + 1
5*(a^2 + 4*a*b + 3*b^2)*cos(f*x + e))/a^4)/f

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mupad [B]  time = 0.16, size = 195, normalized size = 1.21 \[ \frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {2\,b}{3\,a^3}+\frac {2}{3\,a^2}\right )}{f}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a^2\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a^2}-\frac {b^2}{a^4}+\frac {2\,b\,\left (\frac {2\,b}{a^3}+\frac {2}{a^2}\right )}{a}\right )}{f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {a^2\,b}{2}+a\,b^2+\frac {b^3}{2}\right )}{f\,\left (a^5\,{\cos \left (e+f\,x\right )}^2+b\,a^4\right )}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (a+b\right )\,\left (3\,a+7\,b\right )}{3\,a^2\,b+10\,a\,b^2+7\,b^3}\right )\,\left (a+b\right )\,\left (3\,a+7\,b\right )}{2\,a^{9/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(e + f*x)^3*((2*b)/(3*a^3) + 2/(3*a^2)))/f - cos(e + f*x)^5/(5*a^2*f) - (cos(e + f*x)*(1/a^2 - b^2/a^4 + (
2*b*((2*b)/a^3 + 2/a^2))/a))/f - (cos(e + f*x)*(a*b^2 + (a^2*b)/2 + b^3/2))/(f*(a^4*b + a^5*cos(e + f*x)^2)) +
(b^(1/2)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(a + b)*(3*a + 7*b))/(10*a*b^2 + 3*a^2*b + 7*b^3))*(a + b)*(3*a +
7*b))/(2*a^(9/2)*f)

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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)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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