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

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

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.136832, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3186, 413, 385, 217, 203} \[ \frac{a (3 a+5 b) \cos (e+f x)}{3 b^2 f (a+b)^2 \sqrt{a-b \cos ^2(e+f x)+b}}-\frac{\tan ^{-1}\left (\frac{\sqrt{b} \cos (e+f x)}{\sqrt{a-b \cos ^2(e+f x)+b}}\right )}{b^{5/2} f}+\frac{a \sin ^2(e+f x) \cos (e+f x)}{3 b f (a+b) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}} \]

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 3186

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]

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 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 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 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])

Rubi steps

\begin{align*} \int \frac{\sin ^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-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{\operatorname{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{\operatorname{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{\operatorname{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*}

Mathematica [A]  time = 0.754213, size = 133, normalized size = 0.97 \[ \frac{\frac{2 \sqrt{2} a \cos (e+f x) \left (3 a^2-b (2 a+3 b) \cos (2 (e+f x))+7 a b+3 b^2\right )}{(a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}}-\frac{3 \log \left (\sqrt{2 a-b \cos (2 (e+f x))+b}+\sqrt{2} \sqrt{-b} \cos (e+f x)\right )}{\sqrt{-b}}}{3 b^2 f} \]

Warning: Unable to verify antiderivative.

[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)

________________________________________________________________________________________

Maple [A]  time = 2.746, size = 243, normalized size = 1.8 \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) }\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({\frac{1}{2}\arctan \left ({\sqrt{b} \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}-{\frac{-a+b}{2\,b}} \right ){\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \right ){b}^{-{\frac{5}{2}}}}-{\frac{{a}^{2} \left ( 2\,b \left ( \sin \left ( fx+e \right ) \right ) ^{2}+3\,a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{3\,{b}^{2} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ({a}^{2}+2\,ab+{b}^{2} \right ) }{\frac{1}{\sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}}+2\,{\frac{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{{b}^{2} \left ( a+b \right ) \sqrt{- \left ( -b \left ( \sin \left ( fx+e \right ) \right ) ^{2}-a \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}} \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

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)

[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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 6.0517, size = 2045, normalized size = 14.93 \begin{align*} \text{result too large to display} \end{align*}

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)]

________________________________________________________________________________________

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

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{5}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

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]

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