3.4.65 \(\int \frac {\cos (e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [365]

Optimal. Leaf size=65 \[ \frac {\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {2 \sin (e+f x)}{3 a^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 65, 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 = {3269, 198, 197} \begin {gather*} \frac {2 \sin (e+f x)}{3 a^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sin (e+f x)}{3 a f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[e + f*x]/(3*a*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*Sin[e + f*x])/(3*a^2*f*Sqrt[a + b*Sin[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 3269

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

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Mathematica [A]
time = 0.04, size = 47, normalized size = 0.72 \begin {gather*} \frac {\sin (e+f x) \left (3 a+2 b \sin ^2(e+f x)\right )}{3 a^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.16, size = 56, normalized size = 0.86

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (f x +e \right )}{3 a \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}}{f}\)	\(56\)
default	\(\frac {\frac {\sin \left (f x +e \right )}{3 a \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}+\frac {2 \sin \left (f x +e \right )}{3 a^{2} \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}}{f}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 59, normalized size = 0.91 \begin {gather*} \frac {\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}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(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))/f

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Fricas [A]
time = 0.53, size = 104, normalized size = 1.60 \begin {gather*} -\frac {{\left (2 \, b \cos \left (f x + e\right )^{2} - 3 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{3 \, {\left (a^{2} b^{2} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} b + a^{2} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]
time = 0.75, size = 48, normalized size = 0.74 \begin {gather*} \frac {{\left (\frac {2 \, b \sin \left (f x + e\right )^{2}}{a^{2}} + \frac {3}{a}\right )} \sin \left (f x + e\right )}{3 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 21.84, size = 164, normalized size = 2.52 \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\,1{}\mathrm {i}-a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,6{}\mathrm {i}-b\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,2{}\mathrm {i}+b\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,1{}\mathrm {i}\right )}{3\,a^2\,f\,{\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(cos(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*1i - a*exp(e*2i + f*x*2i)*6i - b*exp(e*2i + f*x*2i)*2i + b*exp(e*4i + f*x*4i)*1i))/(3*a^2*f*(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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