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

Optimal. Leaf size=108 \[ -\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} f \sqrt {a+b}}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}+\frac {\sin ^5(e+f x)}{5 a f} \]

[Out]

(a^2-a*b+b^2)*sin(f*x+e)/a^3/f-1/3*(2*a-b)*sin(f*x+e)^3/a^2/f+1/5*sin(f*x+e)^5/a/f-b^3*arctanh(sin(f*x+e)*a^(1
/2)/(a+b)^(1/2))/a^(7/2)/f/(a+b)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4147, 390, 208} \[ \frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}-\frac {b^3 \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} f \sqrt {a+b}}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\sin ^5(e+f x)}{5 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^3*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(a^(7/2)*Sqrt[a + b]*f)) + ((a^2 - a*b + b^2)*Sin[e + f*x]
)/(a^3*f) - ((2*a - b)*Sin[e + f*x]^3)/(3*a^2*f) + Sin[e + f*x]^5/(5*a*f)

Rule 208

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

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 4147

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

Rubi steps

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

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Mathematica [A]  time = 0.69, size = 136, normalized size = 1.26 \[ \frac {5 a^{3/2} (5 a-4 b) \sin (3 (e+f x))+3 a^{5/2} \sin (5 (e+f x))+30 \sqrt {a} \left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x)+\frac {120 b^3 \left (\log \left (\sqrt {a+b}-\sqrt {a} \sin (e+f x)\right )-\log \left (\sqrt {a+b}+\sqrt {a} \sin (e+f x)\right )\right )}{\sqrt {a+b}}}{240 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((120*b^3*(Log[Sqrt[a + b] - Sqrt[a]*Sin[e + f*x]] - Log[Sqrt[a + b] + Sqrt[a]*Sin[e + f*x]]))/Sqrt[a + b] + 3
0*Sqrt[a]*(5*a^2 - 6*a*b + 8*b^2)*Sin[e + f*x] + 5*a^(3/2)*(5*a - 4*b)*Sin[3*(e + f*x)] + 3*a^(5/2)*Sin[5*(e +
f*x)])/(240*a^(7/2)*f)

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fricas [A]  time = 0.47, size = 305, normalized size = 2.82 \[ \left [\frac {15 \, \sqrt {a^{2} + a b} b^{3} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{5} + a^{4} b\right )} f}, \frac {15 \, \sqrt {-a^{2} - a b} b^{3} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{5} + a^{4} b\right )} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/30*(15*sqrt(a^2 + a*b)*b^3*log(-(a*cos(f*x + e)^2 + 2*sqrt(a^2 + a*b)*sin(f*x + e) - 2*a - b)/(a*cos(f*x +
e)^2 + b)) + 2*(3*(a^4 + a^3*b)*cos(f*x + e)^4 + 8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3 + (4*a^4 - a^3*b - 5*a
^2*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)*f), 1/15*(15*sqrt(-a^2 - a*b)*b^3*arctan(sqrt(-a^2 - a*b)
*sin(f*x + e)/(a + b)) + (3*(a^4 + a^3*b)*cos(f*x + e)^4 + 8*a^4 - 2*a^3*b + 5*a^2*b^2 + 15*a*b^3 + (4*a^4 - a
^3*b - 5*a^2*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^5 + a^4*b)*f)]

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giac [A]  time = 0.24, size = 136, normalized size = 1.26 \[ \frac {\frac {15 \, b^{3} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} a^{3}} + \frac {3 \, a^{4} \sin \left (f x + e\right )^{5} - 10 \, a^{4} \sin \left (f x + e\right )^{3} + 5 \, a^{3} b \sin \left (f x + e\right )^{3} + 15 \, a^{4} \sin \left (f x + e\right ) - 15 \, a^{3} b \sin \left (f x + e\right ) + 15 \, a^{2} b^{2} \sin \left (f x + e\right )}{a^{5}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*b^3*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/(sqrt(-a^2 - a*b)*a^3) + (3*a^4*sin(f*x + e)^5 - 10*a^4*s
in(f*x + e)^3 + 5*a^3*b*sin(f*x + e)^3 + 15*a^4*sin(f*x + e) - 15*a^3*b*sin(f*x + e) + 15*a^2*b^2*sin(f*x + e)
)/a^5)/f

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maple [A]  time = 1.39, size = 110, normalized size = 1.02 \[ \frac {\frac {\frac {\left (\sin ^{5}\left (f x +e \right )\right ) a^{2}}{5}-\frac {2 \left (\sin ^{3}\left (f x +e \right )\right ) a^{2}}{3}+\frac {\left (\sin ^{3}\left (f x +e \right )\right ) a b}{3}+a^{2} \sin \left (f x +e \right )-\sin \left (f x +e \right ) a b +b^{2} \sin \left (f x +e \right )}{a^{3}}-\frac {b^{3} \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{a^{3} \sqrt {\left (a +b \right ) a}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/a^3*(1/5*sin(f*x+e)^5*a^2-2/3*sin(f*x+e)^3*a^2+1/3*sin(f*x+e)^3*a*b+a^2*sin(f*x+e)-sin(f*x+e)*a*b+b^2*s
in(f*x+e))-b^3/a^3/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2)))

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maxima [A]  time = 0.46, size = 117, normalized size = 1.08 \[ \frac {\frac {15 \, b^{3} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {2 \, {\left (3 \, a^{2} \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - a b + b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{3}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(15*b^3*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/(sqrt((a + b)*a)*a^3)
+ 2*(3*a^2*sin(f*x + e)^5 - 5*(2*a^2 - a*b)*sin(f*x + e)^3 + 15*(a^2 - a*b + b^2)*sin(f*x + e))/a^3)/f

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mupad [B]  time = 0.14, size = 111, normalized size = 1.03 \[ \frac {\sin \left (e+f\,x\right )\,\left (\frac {3}{a}+\frac {\left (a+b\right )\,\left (\frac {a+b}{a^2}-\frac {3}{a}\right )}{a}\right )}{f}+\frac {{\sin \left (e+f\,x\right )}^5}{5\,a\,f}+\frac {{\sin \left (e+f\,x\right )}^3\,\left (\frac {a+b}{3\,a^2}-\frac {1}{a}\right )}{f}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{a^{7/2}\,f\,\sqrt {a+b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sin(e + f*x)*(3/a + ((a + b)*((a + b)/a^2 - 3/a))/a))/f + sin(e + f*x)^5/(5*a*f) + (sin(e + f*x)^3*((a + b)/(
3*a^2) - 1/a))/f - (b^3*atanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2)))/(a^(7/2)*f*(a + b)^(1/2))

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

[Out]

Timed out

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