3.4.46 \(\int \frac {\cos (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [346]

Optimal. Leaf size=38 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 38, 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, 223, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{\sqrt {b} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.12, size = 34, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{f \sqrt {b}}\)	\(34\)
default	\(\frac {\ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{f \sqrt {b}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 = 0.29, size = 22, normalized size = 0.58 \begin {gather*} \frac {\operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b} 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)^(1/2),x, algorithm="maxima")

[Out]

arcsinh(b*sin(f*x + e)/sqrt(a*b))/(sqrt(b)*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (32) = 64\).
time = 0.46, size = 394, normalized size = 10.37 \begin {gather*} \left [\frac {\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} f}, -\frac {\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 \, b 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)^(1/2),x, algorithm="fricas")

[Out]

[1/8*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))/(
sqrt(b)*f), -1/4*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)))/(b*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (e + f x \right )}}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \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)**(1/2),x)

[Out]

Integral(cos(e + f*x)/sqrt(a + b*sin(e + f*x)**2), x)

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Giac [A]
time = 0.68, size = 65, normalized size = 1.71 \begin {gather*} -\frac {\frac {a \log \left ({\left | -\sqrt {b} \sin \left (f x + e\right ) + \sqrt {b \sin \left (f x + e\right )^{2} + a} \right |}\right )}{\sqrt {b}} - \sqrt {b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right )}{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)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 14.74, size = 33, normalized size = 0.87 \begin {gather*} \frac {\ln \left (\sqrt {b}\,\sin \left (e+f\,x\right )+\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}\right )}{\sqrt {b}\,f} \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)^(1/2),x)

[Out]

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

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