3.2.64 \(\int \frac {\sin (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [164]

Optimal. Leaf size=73 \[ -\frac {\cos (e+f x)}{3 (a+b) f \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}-\frac {2 \cos (e+f x)}{3 (a+b)^2 f \sqrt {a+b-b \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3265, 198, 197} \begin {gather*} -\frac {2 \cos (e+f x)}{3 f (a+b)^2 \sqrt {a-b \cos ^2(e+f x)+b}}-\frac {\cos (e+f x)}{3 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]/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-1/3*Cos[e + f*x]/((a + b)*f*(a + b - b*Cos[e + f*x]^2)^(3/2)) - (2*Cos[e + f*x])/(3*(a + b)^2*f*Sqrt[a + b -
b*Cos[e + f*x]^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

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

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Mathematica [A]
time = 0.13, size = 60, normalized size = 0.82 \begin {gather*} \frac {2 \sqrt {2} \cos (e+f x) (-3 a-2 b+b \cos (2 (e+f x)))}{3 (a+b)^2 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 6.44, size = 55, normalized size = 0.75

method	result	size
default	\(-\frac {\left (2 b \left (\sin ^{2}\left (f x +e \right )\right )+3 a +b \right ) \cos \left (f x +e \right )}{3 \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a^{2}+2 a b +b^{2}\right ) f}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 67, normalized size = 0.92 \begin {gather*} -\frac {\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 )}}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*cos(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)))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (65) = 130\).
time = 0.49, size = 134, normalized size = 1.84 \begin {gather*} \frac {{\left (2 \, b \cos \left (f x + e\right )^{3} - 3 \, {\left (a + b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, {\left ({\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*b*cos(f*x + e)^3 - 3*(a + b)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b)/((a^2*b^2 + 2*a*b^3 + b^4)*f
*cos(f*x + e)^4 - 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^
3 + b^4)*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)/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]
time = 0.62, size = 137, normalized size = 1.88 \begin {gather*} \frac {{\left (\frac {2 \, b^{2} f^{2} \cos \left (f x + e\right )^{2}}{a^{2} b f^{2} + 2 \, a b^{2} f^{2} + b^{3} f^{2}} - \frac {3 \, {\left (a b f^{2} + b^{2} f^{2}\right )}}{a^{2} b f^{2} + 2 \, a b^{2} f^{2} + b^{3} f^{2}}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{3 \, {\left (b \cos \left (f x + e\right )^{2} - a - b\right )}^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 20.80, size = 159, normalized size = 2.18 \begin {gather*} \frac {4\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )\,\sqrt {a+b\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,\left (b-6\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-4\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}{3\,f\,{\left (a+b\right )}^2\,{\left (b-4\,a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-2\,b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(e*1i + f*x*1i)*(exp(e*2i + f*x*2i) + 1)*(a + b*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2
)^2)^(1/2)*(b - 6*a*exp(e*2i + f*x*2i) - 4*b*exp(e*2i + f*x*2i) + b*exp(e*4i + f*x*4i)))/(3*f*(a + b)^2*(b - 4
*a*exp(e*2i + f*x*2i) - 2*b*exp(e*2i + f*x*2i) + b*exp(e*4i + f*x*4i))^2)

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