Integrand size = 23, antiderivative size = 51 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {a+b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 \sqrt {a+b} d} \] Output:
1/2*arctanh((a+b*sin(d*x+c)^2)/(a+b)^(1/2)/(a+b*sin(d*x+c)^4)^(1/2))/(a+b) ^(1/2)/d
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {a+b-b \cos ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}\right )}{2 \sqrt {a+b} d} \] Input:
Integrate[Tan[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]
Output:
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)
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3708, 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.
\(\displaystyle \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)}{\sqrt {a+b \sin (c+d x)^4}}dx\) |
\(\Big \downarrow \) 3708 |
\(\displaystyle \frac {\int \frac {1}{\left (1-\sin ^2(c+d x)\right ) \sqrt {b \sin ^4(c+d x)+a}}d\sin ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle -\frac {\int \frac {1}{-\sin ^4(c+d x)+a+b}d\frac {-b \sin ^2(c+d x)-a}{\sqrt {b \sin ^4(c+d x)+a}}}{2 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {-a-b \sin ^2(c+d x)}{\sqrt {a+b} \sqrt {a+b \sin ^4(c+d x)}}\right )}{2 d \sqrt {a+b}}\) |
Input:
Int[Tan[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]
Output:
-1/2*ArcTanh[(-a - b*Sin[c + d*x]^2)/(Sqrt[a + b]*Sqrt[a + b*Sin[c + d*x]^ 4])]/(Sqrt[a + b]*d)
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])
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]
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]
Time = 1.71 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.41
method | result | size |
default | \(\frac {\ln \left (\frac {2 a +2 b -2 b \cos \left (d x +c \right )^{2}+2 \sqrt {a +b}\, \sqrt {a +b -2 b \cos \left (d x +c \right )^{2}+b \cos \left (d x +c \right )^{4}}}{\cos \left (d x +c \right )^{2}}\right )}{2 d \sqrt {a +b}}\) | \(72\) |
Input:
int(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (43) = 86\).
Time = 0.16 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.71 \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\left [\frac {\log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {a + b} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}}{\cos \left (d x + c\right )^{4}}\right )}{4 \, \sqrt {a + b} d}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} {\left (b \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right )}{2 \, {\left (a + b\right )} d}\right ] \] Input:
integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")
Output:
[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)]
\[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\tan {\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \] Input:
integrate(tan(d*x+c)/(a+b*sin(d*x+c)**4)**(1/2),x)
Output:
Integral(tan(c + d*x)/sqrt(a + b*sin(c + d*x)**4), x)
\[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \] Input:
integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")
Output:
integrate(tan(d*x + c)/sqrt(b*sin(d*x + c)^4 + a), x)
\[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \] Input:
integrate(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")
Output:
sage0*x
Timed out. \[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\mathrm {tan}\left (c+d\,x\right )}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \] Input:
int(tan(c + d*x)/(a + b*sin(c + d*x)^4)^(1/2),x)
Output:
int(tan(c + d*x)/(a + b*sin(c + d*x)^4)^(1/2), x)
\[ \int \frac {\tan (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {\sqrt {\sin \left (d x +c \right )^{4} b +a}\, \tan \left (d x +c \right )}{\sin \left (d x +c \right )^{4} b +a}d x \] Input:
int(tan(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x)
Output:
int((sqrt(sin(c + d*x)**4*b + a)*tan(c + d*x))/(sin(c + d*x)**4*b + a),x)