\(\int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [556]

Optimal result

Integrand size = 25, antiderivative size = 89 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 (a+b)^{3/2} d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d} \]

[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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, 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 = {3308, 821, 739, 212} \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 d (a+b)}-\frac {a \text {arctanh}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 d (a+b)^{3/2}} \]

[In]

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

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 739

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 821

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 3308

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*} \text {integral}& = \frac {\text {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 \text {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 \text {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 \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 (a+b)^{3/2} d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d} \]

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

Maple [F]

\[\int \frac {\tan ^{3}\left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}d x\]

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 177 vs. \(2 (77) = 154\).

Time = 0.41 (sec) , antiderivative size = 361, normalized size of antiderivative = 4.06 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\left [\frac {\sqrt {a + b} a \cos \left (d x + c\right )^{2} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a + b} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{\cos \left (d x + c\right )^{4}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {a \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{3}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{3}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

[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)