\(\int \frac {\cos ^3(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [354]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{3/2} f}+\frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3269, 393, 223, 212} \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{3/2} f} \]

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[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 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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1-x^2}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{b f} \\ & = \frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\text {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 f} \\ & = -\frac {\text {arctanh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{b^{3/2} f}+\frac {(a+b) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\sqrt {b} (a+b) \sin (e+f x)-a^{3/2} \text {arcsinh}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a}}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{a b^{3/2} f \sqrt {a+b \sin ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (f x +e \right )}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\sin \left (f x +e \right )}{b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{b^{\frac {3}{2}}}}{f}\)	\(85\)
default	\(\frac {\frac {\sin \left (f x +e \right )}{a \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}+\frac {\sin \left (f x +e \right )}{b \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}-\frac {\ln \left (\sqrt {b}\, \sin \left (f x +e \right )+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{b^{\frac {3}{2}}}}{f}\)	\(85\)

[In]

int(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 225 vs. \(2 (67) = 134\).

Time = 0.42 (sec) , antiderivative size = 559, normalized size of antiderivative = 7.45 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [\frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\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 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left (a b^{3} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + a b^{3}\right )} f\right )}}, \frac {{\left (a b \cos \left (f x + e\right )^{2} - a^{2} - a b\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 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a b + b^{2}\right )} \sin \left (f x + e\right )}{4 \, {\left (a b^{3} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + a b^{3}\right )} f\right )}}\right ] \]

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*((a*b*cos(f*x + e)^2 - a^2 - a*b)*sqrt(b)*log(128*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*sqrt(-b*cos(f*x + e)^2 + a + b)*(a*b + b^2)*sin(f*x + e))/(a*b^3*
f*cos(f*x + e)^2 - (a^2*b^2 + a*b^3)*f), 1/4*((a*b*cos(f*x + e)^2 - a^2 - a*b)*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)*sin(f*x + e))) - 4*sqrt(-b*
cos(f*x + e)^2 + a + b)*(a*b + b^2)*sin(f*x + e))/(a*b^3*f*cos(f*x + e)^2 - (a^2*b^2 + a*b^3)*f)]

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**3/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} a} - \frac {\sin \left (f x + e\right )}{\sqrt {b \sin \left (f x + e\right )^{2} + a} b}}{f} \]

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\cos \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(f*x + e)^3/(b*sin(f*x + e)^2 + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

[In]

int(cos(e + f*x)^3/(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(cos(e + f*x)^3/(a + b*sin(e + f*x)^2)^(3/2), x)