3.556 \(\int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\)

Optimal. Leaf size=89 \[ \frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 d (a+b)}-\frac {a \tanh ^{-1}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 d (a+b)^{3/2}} \]

[Out]

-1/2*a*arctanh((a+b*sin(d*x+c)^2)/(a+b)^(1/2)/(a+b*sin(d*x+c)^4)^(1/2))/(a+b)^(3/2)/d+1/2*sec(d*x+c)^2*(a+b*si
n(d*x+c)^4)^(1/2)/(a+b)/d

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Rubi [A]  time = 0.12, antiderivative size = 89, 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 = {3229, 807, 725, 206} \[ \frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 d (a+b)}-\frac {a \tanh ^{-1}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 d (a+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-(a*ArcTanh[(a + b*Sin[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b*Sin[c + d*x]^4])])/(2*(a + b)^(3/2)*d) + (Sec[c + d
*x]^2*Sqrt[a + b*Sin[c + d*x]^4])/(2*(a + b)*d)

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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

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]

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x}{(1-x)^2 \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{(1-x) \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{2 (a+b) d}\\ &=\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d}+\frac {a \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 (a+b) d}\\ &=-\frac {a \tanh ^{-1}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 (a+b)^{3/2} d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 85, normalized size = 0.96 \[ -\frac {\frac {a \tanh ^{-1}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{(a+b)^{3/2}}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{a+b}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/2*((a*ArcTanh[(a + b*Sin[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b*Sin[c + d*x]^4])])/(a + b)^(3/2) - (Sec[c + d*
x]^2*Sqrt[a + b*Sin[c + d*x]^4])/(a + b))/d

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fricas [B]  time = 0.64, size = 361, normalized size = 4.06 \[ \left [\frac {\sqrt {a + b} a \cos \left (d x + c\right )^{2} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a + b} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{\cos \left (d x + c\right )^{4}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {a \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{3}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^3/sqrt(b*sin(d*x + c)^4 + a), x)

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maple [F]  time = 1.52, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

int(tan(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (d x + c\right )^{3}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^3/sqrt(b*sin(d*x + c)^4 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(tan(c + d*x)^3/(a + b*sin(c + d*x)^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(tan(c + d*x)**3/sqrt(a + b*sin(c + d*x)**4), x)

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