\(\int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [487]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 (a+b)^{3/2} d}+\frac {\sec ^2(c+d x) \sqrt {a+b \sin ^4(c+d x)}}{2 (a+b) d} \] Input:

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

Output:

-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)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3708, 588, 488, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3708

\(\displaystyle \frac {\int \frac {\sin ^2(c+d x)}{\left (1-\sin ^2(c+d x)\right )^2 \sqrt {b \sin ^4(c+d x)+a}}d\sin ^2(c+d x)}{2 d}\)

\(\Big \downarrow \) 588

\(\displaystyle \frac {\frac {\sqrt {a+b \sin ^4(c+d x)}}{(a+b) \left (1-\sin ^2(c+d x)\right )}-\frac {a \int \frac {1}{\left (1-\sin ^2(c+d x)\right ) \sqrt {b \sin ^4(c+d x)+a}}d\sin ^2(c+d x)}{a+b}}{2 d}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {\frac {a \int \frac {1}{-\sin ^4(c+d x)+a+b}d\frac {-b \sin ^2(c+d x)-a}{\sqrt {b \sin ^4(c+d x)+a}}}{a+b}+\frac {\sqrt {a+b \sin ^4(c+d x)}}{(a+b) \left (1-\sin ^2(c+d x)\right )}}{2 d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {a \text {arctanh}\left (\frac {-a-b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{(a+b)^{3/2}}+\frac {\sqrt {a+b \sin ^4(c+d x)}}{(a+b) \left (1-\sin ^2(c+d x)\right )}}{2 d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 488

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

rule 588

Int[(x_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[c*(c + d*x)^(n + 1)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + a*d^2))) 
, x] + Simp[a*(d/(b*c^2 + a*d^2))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x] 
, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[Simplify[n + 2*p + 3], 0] && Ne 
Q[b*c^2 + a*d^2, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ 
)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[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]
 

Maple [F]

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

Input:

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

Output:

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 = 0.18 (sec) , antiderivative size = 361, normalized size of antiderivative = 4.06 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\left [\frac {\sqrt {a + b} a \cos \left (d x + c\right )^{2} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a + b} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{\cos \left (d x + c\right )^{4}}\right ) + 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}, -\frac {a \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) \cos \left (d x + c\right )^{2} - \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (a + b\right )}}{2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{2}}\right ] \] Input:

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

Output:

[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)/co 
s(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(s 
qrt(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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

sage0*x
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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