3.557 \(\int \frac{\tan (c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=51 \[ \frac{\tanh ^{-1}\left (\frac{a+b \sin ^2(c+d x)}{\sqrt{a+b} \sqrt{a+b \sin ^4(c+d x)}}\right )}{2 d \sqrt{a+b}} \]

[Out]

ArcTanh[(a + b*Sin[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b*Sin[c + d*x]^4])]/(2*Sqrt[a + b]*d)

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Rubi [A]  time = 0.0555082, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3229, 725, 206} \[ \frac{\tanh ^{-1}\left (\frac{a+b \sin ^2(c+d x)}{\sqrt{a+b} \sqrt{a+b \sin ^4(c+d x)}}\right )}{2 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

ArcTanh[(a + b*Sin[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b*Sin[c + d*x]^4])]/(2*Sqrt[a + b]*d)

Rule 3229

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff^(n/2)*x^(n/2))^p
)/(1 - ff*x)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] &
& IntegerQ[n/2]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

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])

Rubi steps

\begin{align*} \int \frac{\tan (c+d x)}{\sqrt{a+b \sin ^4(c+d x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\frac{-a-b \sin ^2(c+d x)}{\sqrt{a+b \sin ^4(c+d x)}}\right )}{2 d}\\ &=\frac{\tanh ^{-1}\left (\frac{a+b \sin ^2(c+d x)}{\sqrt{a+b} \sqrt{a+b \sin ^4(c+d x)}}\right )}{2 \sqrt{a+b} d}\\ \end{align*}

Mathematica [A]  time = 0.0906018, size = 65, normalized size = 1.27 \[ \frac{\tanh ^{-1}\left (\frac{a-b \cos ^2(c+d x)+b}{\sqrt{a+b} \sqrt{a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )}{2 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

ArcTanh[(a + b - b*Cos[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])]/(2*Sqrt[
a + b]*d)

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Maple [A]  time = 0.203, size = 72, normalized size = 1.4 \begin{align*}{\frac{1}{2\,d}\ln \left ({\frac{1}{ \left ( \cos \left ( dx+c \right ) \right ) ^{2}} \left ( 2\,a+2\,b-2\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2\,\sqrt{a+b}\sqrt{a+b-2\,b \left ( \cos \left ( dx+c \right ) \right ) ^{2}+b \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \right ) } \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

1/2/d/(a+b)^(1/2)*ln((2*a+2*b-2*b*cos(d*x+c)^2+2*(a+b)^(1/2)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2))/cos(
d*x+c)^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)/sqrt(b*sin(d*x + c)^4 + a), x)

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Fricas [B]  time = 2.66844, size = 589, normalized size = 11.55 \begin{align*} \left [\frac{\log \left (\frac{{\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{a + b} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{\cos \left (d x + c\right )^{4}}\right )}{4 \, \sqrt{a + b} d}, \frac{\sqrt{-a - b} \arctan \left (\frac{\sqrt{b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}{\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt{-a - b}}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{2 \,{\left (a + b\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

[1/4*log(((a*b + 2*b^2)*cos(d*x + c)^4 - 4*(a*b + b^2)*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x
+ c)^2 + a + b)*(b*cos(d*x + c)^2 - a - b)*sqrt(a + b) + 2*a^2 + 4*a*b + 2*b^2)/cos(d*x + c)^4)/(sqrt(a + b)*d
), 1/2*sqrt(-a - b)*arctan(sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*(b*cos(d*x + c)^2 - a - b)*sqrt
(-a - b)/((a*b + b^2)*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2))/((a + b)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (c + d x \right )}}{\sqrt{a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(tan(c + d*x)/sqrt(a + b*sin(c + d*x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)/sqrt(b*sin(d*x + c)^4 + a), x)