3.363 \(\int \frac {\cos ^5(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=130 \[ -\frac {(3 a-2 b) (a+b) \sin (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {(a+b) \sin (e+f x) \cos ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3190, 413, 385, 217, 206} \[ -\frac {(3 a-2 b) (a+b) \sin (e+f x)}{3 a^2 b^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {(a+b) \sin (e+f x) \cos ^2(e+f x)}{3 a b f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 3190

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

Rubi steps

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

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Mathematica [A]  time = 0.81, size = 128, normalized size = 0.98 \[ \frac {\frac {2 \sqrt {2} (a+b) \sin (e+f x) \left (-3 a^2+b (2 a-b) \cos (2 (e+f x))+a b+b^2\right )}{a^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {-b} \sin (e+f x)}{\sqrt {2 a-b \cos (2 (e+f x))+b}}\right )}{\sqrt {-b}}}{3 b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.62, size = 799, normalized size = 6.15 \[ \left [\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, {\left (3 \, a^{3} b + 4 \, a^{2} b^{2} - a b^{3} - 2 \, b^{4} - 2 \, {\left (2 \, a^{2} b^{2} + a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{24 \, {\left (a^{2} b^{5} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} b^{2} \cos \left (f x + e\right )^{4} + a^{4} + 2 \, a^{3} b + a^{2} b^{2} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (3 \, a^{3} b + 4 \, a^{2} b^{2} - a b^{3} - 2 \, b^{4} - 2 \, {\left (2 \, a^{2} b^{2} + a b^{3} - b^{4}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{12 \, {\left (a^{2} b^{5} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + a^{2} b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 2 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*sqrt(b)*log(1
28*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^3 + 24*b^4)*cos(f*x + e)^4
+ a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*a*b^3 + 16*b^4)*cos(f*x +
e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b - 24*a*b^2 - 16*b^3 + 2*(
5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)*sin(f*x + e)) - 8*(3*a^3*
b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b^2 + a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*si
n(f*x + e))/(a^2*b^5*f*cos(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^4*b^3 + 2*a^3*b^4 + a^2*b^
5)*f), -1/12*(3*(a^2*b^2*cos(f*x + e)^4 + a^4 + 2*a^3*b + a^2*b^2 - 2*(a^3*b + a^2*b^2)*cos(f*x + e)^2)*sqrt(-
b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2)*sqrt(-b*cos(f*x +
e)^2 + a + b)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3)*cos(f*x + e)^2)*si
n(f*x + e))) + 4*(3*a^3*b + 4*a^2*b^2 - a*b^3 - 2*b^4 - 2*(2*a^2*b^2 + a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(-b*co
s(f*x + e)^2 + a + b)*sin(f*x + e))/(a^2*b^5*f*cos(f*x + e)^4 - 2*(a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^2 + (a^4*
b^3 + 2*a^3*b^4 + a^2*b^5)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 4.00, size = 383, normalized size = 2.95 \[ \frac {3 \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) a^{4} b^{4}+6 \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) a^{3} b^{5}+3 \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) a^{2} b^{6}+3 \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) a^{2} b^{6} \left (\cos ^{4}\left (f x +e \right )\right )-6 \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\right ) a^{2} b^{5} \left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, b^{\frac {11}{2}} \left (2 a^{2}+a b -b^{2}\right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-\sin \left (f x +e \right ) \sqrt {-b \left (\cos ^{2}\left (f x +e \right )\right )+\frac {a \,b^{2}+b^{3}}{b^{2}}}\, b^{\frac {9}{2}} \left (3 a^{3}+4 a^{2} b -a \,b^{2}-2 b^{3}\right )}{3 b^{\frac {13}{2}} a^{2} \left (b^{2} \left (\cos ^{4}\left (f x +e \right )\right )-2 a b \left (\cos ^{2}\left (f x +e \right )\right )-2 b^{2} \left (\cos ^{2}\left (f x +e \right )\right )+a^{2}+2 a b +b^{2}\right ) f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/b^(13/2)/a^2/(b^2*cos(f*x+e)^4-2*a*b*cos(f*x+e)^2-2*b^2*cos(f*x+e)^2+a^2+2*a*b+b^2)*(3*ln(sin(f*x+e)*b^(1/
2)+(a+b-b*cos(f*x+e)^2)^(1/2))*a^4*b^4+6*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*a^3*b^5+3*ln(sin(f*
x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*a^2*b^6+3*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*a^2*b^6*c
os(f*x+e)^4-6*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))*a^2*b^5*(a+b)*cos(f*x+e)^2+2*(-b*cos(f*x+e)^2+
(a*b^2+b^3)/b^2)^(1/2)*b^(11/2)*(2*a^2+a*b-b^2)*sin(f*x+e)*cos(f*x+e)^2-sin(f*x+e)*(-b*cos(f*x+e)^2+(a*b^2+b^3
)/b^2)^(1/2)*b^(9/2)*(3*a^3+4*a^2*b-a*b^2-2*b^3))/f

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maxima [A]  time = 1.36, size = 207, normalized size = 1.59 \[ -\frac {{\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} \sin \left (f x + e\right ) - \frac {3 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a^{2}} - \frac {\sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} a} + \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} b^{2}} - \frac {2 \, \sin \left (f x + e\right )}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, \sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a b}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^5}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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