3.345 \(\int \frac {\cos ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=79 \[ \frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3190, 388, 217, 206} \[ \frac {(a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 388

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

Rule 3190

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

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Mathematica [A]  time = 0.11, size = 79, normalized size = 1.00 \[ \frac {-\frac {(-a-2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{2 b^{3/2}}-\frac {\sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{2 b}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 461, normalized size = 5.84 \[ \left [\frac {{\left (a + 2 \, b\right )} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{16 \, b^{2} f}, -\frac {{\left (a + 2 \, b\right )} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \sin \left (f x + e\right )}{8 \, b^{2} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((a + 2*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)*sqr
t(b)*sin(f*x + e)) - 8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*sin(f*x + e))/(b^2*f), -1/8*((a + 2*b)*sqrt(-b)*arcta
n(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)*b*sin(f*x + e))/(b^2*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)1/f*(-1/2/b*sin(f*x+exp(1))*sqrt(b*sin(f*x+exp(1))^2+a)+2*(-a-
2*b)/4/b/sqrt(b)*ln(abs(sqrt(b*sin(f*x+exp(1))^2+a)-sqrt(b)*sin(f*x+exp(1)))))

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maple [A]  time = 1.49, size = 98, normalized size = 1.24 \[ -\frac {\sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{2 b f}+\frac {a \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 f \,b^{\frac {3}{2}}}+\frac {\ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{f \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 69, normalized size = 0.87 \[ \frac {\frac {a \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )}{b}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^3/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (e+f\,x\right )}^3}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)**3/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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