3.2.62 \(\int \frac {\sin ^5(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [162]

Optimal. Leaf size=137 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{b^{5/2} f}+\frac {a (3 a+5 b) \cos (e+f x)}{3 b^2 (a+b)^2 f \sqrt {a+b-b \cos ^2(e+f x)}}+\frac {a \cos (e+f x) \sin ^2(e+f x)}{3 b (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 137, 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 = {3265, 424, 393, 223, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{b^{5/2} f}+\frac {a (3 a+5 b) \cos (e+f x)}{3 b^2 f (a+b)^2 \sqrt {a-b \cos ^2(e+f x)+b}}+\frac {a \sin ^2(e+f x) \cos (e+f x)}{3 b f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 223

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 393

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 424

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 3265

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

Rubi steps

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

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Mathematica [A]
time = 0.55, size = 133, normalized size = 0.97 \begin {gather*} \frac {\frac {2 \sqrt {2} a \cos (e+f x) \left (3 a^2+7 a b+3 b^2-b (2 a+3 b) \cos (2 (e+f x))\right )}{(a+b)^2 (2 a+b-b \cos (2 (e+f x)))^{3/2}}-\frac {3 \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{\sqrt {-b}}}{3 b^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[2]*a*Cos[e + f*x]*(3*a^2 + 7*a*b + 3*b^2 - b*(2*a + 3*b)*Cos[2*(e + f*x)]))/((a + b)^2*(2*a + b - b*C
os[2*(e + f*x)])^(3/2)) - (3*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b]
)/(3*b^2*f)

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Maple [A]
time = 18.34, size = 243, normalized size = 1.77

method	result	size
default	\(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\frac {\arctan \left (\frac {\sqrt {b}\, \left (\sin ^{2}\left (f x +e \right )-\frac {-a +b}{2 b}\right )}{\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{2 b^{\frac {5}{2}}}-\frac {a^{2} \left (2 b \left (\sin ^{2}\left (f x +e \right )\right )+3 a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 b^{2} \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 a \left (\cos ^{2}\left (f x +e \right )\right )}{b^{2} \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(243\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (130) = 260\).
time = 0.49, size = 301, normalized size = 2.20 \begin {gather*} -\frac {{\left (\frac {3 \, \cos \left (f x + e\right )^{2}}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} - \frac {2 \, a}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b^{2}} - \frac {2}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b}\right )} \cos \left (f x + e\right ) + \frac {3 \, \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {5}{2}}} + \frac {2 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}^{2}} + \frac {\cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} {\left (a + b\right )}} - \frac {3 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b^{2}} + \frac {2 \, a \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b^{2}} - \frac {2 \, \cos \left (f x + e\right )}{{\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} b} + \frac {4 \, \cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} b}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((3*cos(f*x + e)^2/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b) - 2*a/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b^2) -
2/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b))*cos(f*x + e) + 3*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(5/2) + 2*c
os(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)^2) + cos(f*x + e)/((-b*cos(f*x + e)^2 + a + b)^(3/2)*(a +
b)) - 3*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*b^2) + 2*a*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b
)*(a + b)*b^2) - 2*cos(f*x + e)/((-b*cos(f*x + e)^2 + a + b)^(3/2)*b) + 4*cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2
+ a + b)*(a + b)*b))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (123) = 246\).
time = 0.87, size = 885, normalized size = 6.46 \begin {gather*} \left [-\frac {3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\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} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right ) + 8 \, {\left (2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{24 \, {\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 4 \, a^{3} b^{4} + 6 \, a^{2} b^{5} + 4 \, a b^{6} + b^{7}\right )} f\right )}}, \frac {3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\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 + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + 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 )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) - 4 \, {\left (2 \, {\left (2 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 2 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, {\left ({\left (a^{2} b^{5} + 2 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b^{4} + 3 \, a^{2} b^{5} + 3 \, a b^{6} + b^{7}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{3} + 4 \, a^{3} b^{4} + 6 \, a^{2} b^{5} + 4 \, a b^{6} + b^{7}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*((a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3
*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)
^6 + 160*(a^2*b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*
a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 + 8*(16*b^3*cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2
*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a
+ b)*sqrt(-b)) + 8*(2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 - 3*(a^3*b + 3*a^2*b^2 + 2*a*b^3)*cos(f*x + e))*sqr
t(-b*cos(f*x + e)^2 + a + b))/((a^2*b^5 + 2*a*b^6 + b^7)*f*cos(f*x + e)^4 - 2*(a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 +
b^7)*f*cos(f*x + e)^2 + (a^4*b^3 + 4*a^3*b^4 + 6*a^2*b^5 + 4*a*b^6 + b^7)*f), 1/12*(3*((a^2*b^2 + 2*a*b^3 + b
^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x
+ e)^2)*sqrt(b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*
cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a*b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b^
3)*cos(f*x + e))) - 4*(2*(2*a^2*b^2 + 3*a*b^3)*cos(f*x + e)^3 - 3*(a^3*b + 3*a^2*b^2 + 2*a*b^3)*cos(f*x + e))*
sqrt(-b*cos(f*x + e)^2 + a + b))/((a^2*b^5 + 2*a*b^6 + b^7)*f*cos(f*x + e)^4 - 2*(a^3*b^4 + 3*a^2*b^5 + 3*a*b^
6 + b^7)*f*cos(f*x + e)^2 + (a^4*b^3 + 4*a^3*b^4 + 6*a^2*b^5 + 4*a*b^6 + b^7)*f)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (130) = 260\).
time = 0.78, size = 338, normalized size = 2.47 \begin {gather*} -\frac {\frac {{\left ({\left (\frac {{\left (3 \, a^{3} b^{8} + 5 \, a^{2} b^{9}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}} + \frac {3 \, {\left (a^{3} b^{8} + 7 \, a^{2} b^{9} + 8 \, a b^{10}\right )}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, {\left (a^{3} b^{8} + 7 \, a^{2} b^{9} + 8 \, a b^{10}\right )}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}\right )} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \frac {3 \, a^{3} b^{8} + 5 \, a^{2} b^{9}}{a^{2} b^{10} + 2 \, a b^{11} + b^{12}}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {6 \, \arctan \left (-\frac {\sqrt {a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 4 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a} + \sqrt {a}}{2 \, \sqrt {b}}\right )}{b^{\frac {5}{2}}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(((((3*a^3*b^8 + 5*a^2*b^9)*tan(1/2*f*x + 1/2*e)^2/(a^2*b^10 + 2*a*b^11 + b^12) + 3*(a^3*b^8 + 7*a^2*b^9
+ 8*a*b^10)/(a^2*b^10 + 2*a*b^11 + b^12))*tan(1/2*f*x + 1/2*e)^2 - 3*(a^3*b^8 + 7*a^2*b^9 + 8*a*b^10)/(a^2*b^1
0 + 2*a*b^11 + b^12))*tan(1/2*f*x + 1/2*e)^2 - (3*a^3*b^8 + 5*a^2*b^9)/(a^2*b^10 + 2*a*b^11 + b^12))/(a*tan(1/
2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a)^(3/2) - 6*arctan(-1/2*(sqrt(a)
*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e
)^2 + a) + sqrt(a))/sqrt(b))/b^(5/2))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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