\(\int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [347]

Optimal result

Integrand size = 24, antiderivative size = 142 \[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{3/2} d}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{3/2} d}+\frac {\tan (c+d x)}{(a-b) d} \] Output:

-1/2*b^(1/2)*arctan((a^(1/2)-b^(1/2))^(1/2)*tan(d*x+c)/a^(1/4))/a^(3/4)/(a 
^(1/2)-b^(1/2))^(3/2)/d+1/2*b^(1/2)*arctan((a^(1/2)+b^(1/2))^(1/2)*tan(d*x 
+c)/a^(1/4))/a^(3/4)/(a^(1/2)+b^(1/2))^(3/2)/d+tan(d*x+c)/(a-b)/d
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.23 \[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {\frac {\left (\sqrt {a} \sqrt {b}-b\right ) \arctan \left (\frac {\left (\sqrt {a}+\sqrt {b}\right ) \tan (c+d x)}{\sqrt {a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {a+\sqrt {a} \sqrt {b}}}+\frac {\left (\sqrt {a} \sqrt {b}+b\right ) \text {arctanh}\left (\frac {\left (\sqrt {a}-\sqrt {b}\right ) \tan (c+d x)}{\sqrt {-a+\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {-a+\sqrt {a} \sqrt {b}}}+2 \tan (c+d x)}{2 (a-b) d} \] Input:

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

Output:

(((Sqrt[a]*Sqrt[b] - b)*ArcTan[((Sqrt[a] + Sqrt[b])*Tan[c + d*x])/Sqrt[a + 
 Sqrt[a]*Sqrt[b]]])/(Sqrt[a]*Sqrt[a + Sqrt[a]*Sqrt[b]]) + ((Sqrt[a]*Sqrt[b 
] + b)*ArcTanh[((Sqrt[a] - Sqrt[b])*Tan[c + d*x])/Sqrt[-a + Sqrt[a]*Sqrt[b 
]]])/(Sqrt[a]*Sqrt[-a + Sqrt[a]*Sqrt[b]]) + 2*Tan[c + d*x])/(2*(a - b)*d)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3703, 1484, 2009}

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.

\(\displaystyle \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3703

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

\(\Big \downarrow \) 1484

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}-\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^{3/2}}+\frac {\sqrt {b} \arctan \left (\frac {\sqrt {\sqrt {a}+\sqrt {b}} \tan (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^{3/2}}+\frac {\tan (c+d x)}{a-b}}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 1484

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symb 
ol] :> Int[ExpandIntegrand[(d + e*x^2)^q/(a + b*x^2 + c*x^4), x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 
 0] && IntegerQ[q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3703

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Su 
bst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(m/2 + 2*p + 
 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2 
] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.64 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\frac {\tan \left (d x +c \right )}{a -b}-b \left (\frac {\left (a +b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(166\)
default	\(\frac {\frac {\tan \left (d x +c \right )}{a -b}-b \left (\frac {\left (a +b +2 \sqrt {a b}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}+a \right ) \left (a -b \right )}}+\frac {\left (-a -b +2 \sqrt {a b}\right ) \operatorname {arctanh}\left (\frac {\left (-a +b \right ) \tan \left (d x +c \right )}{\sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{2 \sqrt {a b}\, \left (a -b \right ) \sqrt {\left (\sqrt {a b}-a \right ) \left (a -b \right )}}\right )}{d}\)	\(166\)
risch	\(\frac {2 i}{d \left (a -b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (65536 a^{6} d^{4}-196608 a^{5} b \,d^{4}+196608 a^{4} b^{2} d^{4}-65536 a^{3} b^{3} d^{4}\right ) \textit {\_Z}^{4}+\left (512 a^{3} b \,d^{2}+1536 a^{2} b^{2} d^{2}\right ) \textit {\_Z}^{2}+b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\frac {16384 i d^{3} a^{6}}{3 b^{2} a +b^{3}}-\frac {49152 i a^{5} b \,d^{3}}{3 b^{2} a +b^{3}}+\frac {49152 i a^{4} b^{2} d^{3}}{3 b^{2} a +b^{3}}-\frac {16384 i a^{3} b^{3} d^{3}}{3 b^{2} a +b^{3}}\right ) \textit {\_R}^{3}+\left (-\frac {512 d^{2} a^{5}}{3 b^{2} a +b^{3}}+\frac {1536 a^{4} b \,d^{2}}{3 b^{2} a +b^{3}}-\frac {1536 a^{3} b^{2} d^{2}}{3 b^{2} a +b^{3}}+\frac {512 a^{2} b^{3} d^{2}}{3 b^{2} a +b^{3}}\right ) \textit {\_R}^{2}+\left (\frac {160 i d \,a^{3} b}{3 b^{2} a +b^{3}}+\frac {320 i a^{2} b^{2} d}{3 b^{2} a +b^{3}}+\frac {32 i a \,b^{3} d}{3 b^{2} a +b^{3}}\right ) \textit {\_R} -\frac {2 a^{2} b}{3 b^{2} a +b^{3}}-\frac {9 b^{2} a}{3 b^{2} a +b^{3}}-\frac {b^{3}}{3 b^{2} a +b^{3}}\right )\right )\)	\(426\)

Input:

int(sec(d*x+c)^2/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/(a-b)*tan(d*x+c)-b*(1/2*(a+b+2*(a*b)^(1/2))/(a*b)^(1/2)/(a-b)/(((a* 
b)^(1/2)+a)*(a-b))^(1/2)*arctan((a-b)*tan(d*x+c)/(((a*b)^(1/2)+a)*(a-b))^( 
1/2))+1/2*(-a-b+2*(a*b)^(1/2))/(a*b)^(1/2)/(a-b)/(((a*b)^(1/2)-a)*(a-b))^( 
1/2)*arctanh((-a+b)*tan(d*x+c)/(((a*b)^(1/2)-a)*(a-b))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2589 vs. \(2 (102) = 204\).

Time = 0.40 (sec) , antiderivative size = 2589, normalized size of antiderivative = 18.23 \[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*((a - b)*d*sqrt(((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b 
^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 
 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - 
 a*b^3)*d^2))*cos(d*x + c)*log(3/4*a*b^2 + 1/4*b^3 - 1/4*(3*a*b^2 + b^3)*c 
os(d*x + c)^2 + 1/2*(2*(a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*a 
^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5 
*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))*cos(d*x + c)*sin(d*x + c) + (3*a^3*b + 4 
*a^2*b^2 + a*b^3)*d*cos(d*x + c)*sin(d*x + c))*sqrt(((a^4 - 3*a^3*b + 3*a^ 
2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a 
^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4)) - a*b - 3*b^ 
2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2)) - 1/4*(2*(a^5 - 3*a^4*b + 3* 
a^3*b^2 - a^2*b^3)*d^2*cos(d*x + c)^2 - (a^5 - 3*a^4*b + 3*a^3*b^2 - a^2*b 
^3)*d^2)*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^9 - 6*a^8*b + 15*a^7*b^2 - 2 
0*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)*d^4))) - (a - b)*d*sqrt(((a^ 
4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/((a^ 
9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^3*b^6)* 
d^4)) - a*b - 3*b^2)/((a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*d^2))*cos(d*x + 
c)*log(3/4*a*b^2 + 1/4*b^3 - 1/4*(3*a*b^2 + b^3)*cos(d*x + c)^2 - 1/2*(2*( 
a^6 - 3*a^5*b + 3*a^4*b^2 - a^3*b^3)*d^3*sqrt((9*a^2*b^3 + 6*a*b^4 + b^5)/ 
((a^9 - 6*a^8*b + 15*a^7*b^2 - 20*a^6*b^3 + 15*a^5*b^4 - 6*a^4*b^5 + a^...
 

Sympy [F]

\[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a - b \sin ^{4}{\left (c + d x \right )}}\, dx \] Input:

integrate(sec(d*x+c)**2/(a-b*sin(d*x+c)**4),x)
 

Output:

Integral(sec(c + d*x)**2/(a - b*sin(c + d*x)**4), x)
 

Maxima [F]

\[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\int { -\frac {\sec \left (d x + c\right )^{2}}{b \sin \left (d x + c\right )^{4} - a} \,d x } \] Input:

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

Output:

(((a - b)*d*cos(2*d*x + 2*c)^2 + (a - b)*d*sin(2*d*x + 2*c)^2 + 2*(a - b)* 
d*cos(2*d*x + 2*c) + (a - b)*d)*integrate(4*(4*b^2*cos(6*d*x + 6*c)^2 + 4* 
b^2*cos(2*d*x + 2*c)^2 + 4*b^2*sin(6*d*x + 6*c)^2 + 4*b^2*sin(2*d*x + 2*c) 
^2 - 12*(8*a*b - 3*b^2)*cos(4*d*x + 4*c)^2 - b^2*cos(2*d*x + 2*c) - 12*(8* 
a*b - 3*b^2)*sin(4*d*x + 4*c)^2 + 2*(8*a*b - 15*b^2)*sin(4*d*x + 4*c)*sin( 
2*d*x + 2*c) - (b^2*cos(6*d*x + 6*c) - 6*b^2*cos(4*d*x + 4*c) + b^2*cos(2* 
d*x + 2*c))*cos(8*d*x + 8*c) + (8*b^2*cos(2*d*x + 2*c) - b^2 + 2*(8*a*b - 
15*b^2)*cos(4*d*x + 4*c))*cos(6*d*x + 6*c) + 2*(3*b^2 + (8*a*b - 15*b^2)*c 
os(2*d*x + 2*c))*cos(4*d*x + 4*c) - (b^2*sin(6*d*x + 6*c) - 6*b^2*sin(4*d* 
x + 4*c) + b^2*sin(2*d*x + 2*c))*sin(8*d*x + 8*c) + 2*(4*b^2*sin(2*d*x + 2 
*c) + (8*a*b - 15*b^2)*sin(4*d*x + 4*c))*sin(6*d*x + 6*c))/(a*b^2 - b^3 + 
(a*b^2 - b^3)*cos(8*d*x + 8*c)^2 + 16*(a*b^2 - b^3)*cos(6*d*x + 6*c)^2 + 4 
*(64*a^3 - 112*a^2*b + 57*a*b^2 - 9*b^3)*cos(4*d*x + 4*c)^2 + 16*(a*b^2 - 
b^3)*cos(2*d*x + 2*c)^2 + (a*b^2 - b^3)*sin(8*d*x + 8*c)^2 + 16*(a*b^2 - b 
^3)*sin(6*d*x + 6*c)^2 + 4*(64*a^3 - 112*a^2*b + 57*a*b^2 - 9*b^3)*sin(4*d 
*x + 4*c)^2 + 16*(8*a^2*b - 11*a*b^2 + 3*b^3)*sin(4*d*x + 4*c)*sin(2*d*x + 
 2*c) + 16*(a*b^2 - b^3)*sin(2*d*x + 2*c)^2 + 2*(a*b^2 - b^3 - 4*(a*b^2 - 
b^3)*cos(6*d*x + 6*c) - 2*(8*a^2*b - 11*a*b^2 + 3*b^3)*cos(4*d*x + 4*c) - 
4*(a*b^2 - b^3)*cos(2*d*x + 2*c))*cos(8*d*x + 8*c) - 8*(a*b^2 - b^3 - 2*(8 
*a^2*b - 11*a*b^2 + 3*b^3)*cos(4*d*x + 4*c) - 4*(a*b^2 - b^3)*cos(2*d*x...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1211 vs. \(2 (102) = 204\).

Time = 0.78 (sec) , antiderivative size = 1211, normalized size of antiderivative = 8.53 \[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*((3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^5 - 9*sqrt(a^2 - 
a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a^4*b + 2*sqrt(a^2 - a*b - sqrt(a*b)*(a 
 - b))*sqrt(a*b)*a^3*b^2 + 10*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b 
)*a^2*b^3 - 5*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^4 - sqrt(a 
^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^5 - 2*(3*sqrt(a^2 - a*b - sqrt(a 
*b)*(a - b))*sqrt(a*b)*a^2*b - 6*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt( 
a*b)*a*b^2 - sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*sqrt(a*b)*b^3)*(a - b)^2 
+ (3*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^4*b - 12*sqrt(a^2 - a*b - sqrt( 
a*b)*(a - b))*a^3*b^2 + 14*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a^2*b^3 - 4 
*sqrt(a^2 - a*b - sqrt(a*b)*(a - b))*a*b^4 - sqrt(a^2 - a*b - sqrt(a*b)*(a 
 - b))*b^5)*abs(-a + b))*(pi*floor((d*x + c)/pi + 1/2) + arctan(tan(d*x + 
c)/sqrt((a^2 - a*b + sqrt((a^2 - a*b)^2 - (a^2 - a*b)*(a^2 - 2*a*b + b^2)) 
)/(a^2 - 2*a*b + b^2))))/(3*a^8 - 21*a^7*b + 59*a^6*b^2 - 85*a^5*b^3 + 65* 
a^4*b^4 - 23*a^3*b^5 + a^2*b^6 + a*b^7) - (3*sqrt(a^2 - a*b + sqrt(a*b)*(a 
 - b))*sqrt(a*b)*a^5 - 9*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^4 
*b + 2*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^3*b^2 + 10*sqrt(a^2 
 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b^3 - 5*sqrt(a^2 - a*b + sqrt(a* 
b)*(a - b))*sqrt(a*b)*a*b^4 - sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b 
)*b^5 - 2*(3*sqrt(a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a^2*b - 6*sqrt( 
a^2 - a*b + sqrt(a*b)*(a - b))*sqrt(a*b)*a*b^2 - sqrt(a^2 - a*b + sqrt(...
 

Mupad [B] (verification not implemented)

Time = 37.39 (sec) , antiderivative size = 2832, normalized size of antiderivative = 19.94 \[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

tan(c + d*x)/(d*(a - b)) + (atan(((((2*(8*a*b^4 - 16*a^2*b^3 + 8*a^3*b^2)) 
/(a - b) - (4*tan(c + d*x)*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3 
*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))^(1/2)*(16*a^5* 
b - 16*a^2*b^4 + 48*a^3*b^3 - 48*a^4*b^2))/(a - b))*((3*a*(a^3*b^3)^(1/2) 
+ b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 + a^3*b^3 - 3* 
a^4*b^2)))^(1/2) - (4*tan(c + d*x)*(6*a*b^3 + b^4 + a^2*b^2))/(a - b))*((3 
*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - 
 a^6 + a^3*b^3 - 3*a^4*b^2)))^(1/2)*1i - (((2*(8*a*b^4 - 16*a^2*b^3 + 8*a^ 
3*b^2))/(a - b) + (4*tan(c + d*x)*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2 
) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))^(1/2)*( 
16*a^5*b - 16*a^2*b^4 + 48*a^3*b^3 - 48*a^4*b^2))/(a - b))*((3*a*(a^3*b^3) 
^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 + a^3*b 
^3 - 3*a^4*b^2)))^(1/2) + (4*tan(c + d*x)*(6*a*b^3 + b^4 + a^2*b^2))/(a - 
b))*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3* 
a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))^(1/2)*1i)/((((2*(8*a*b^4 - 16*a^2*b^3 
 + 8*a^3*b^2))/(a - b) - (4*tan(c + d*x)*((3*a*(a^3*b^3)^(1/2) + b*(a^3*b^ 
3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 + a^3*b^3 - 3*a^4*b^2)))^ 
(1/2)*(16*a^5*b - 16*a^2*b^4 + 48*a^3*b^3 - 48*a^4*b^2))/(a - b))*((3*a*(a 
^3*b^3)^(1/2) + b*(a^3*b^3)^(1/2) + a^3*b + 3*a^2*b^2)/(16*(3*a^5*b - a^6 
+ a^3*b^3 - 3*a^4*b^2)))^(1/2) - (4*tan(c + d*x)*(6*a*b^3 + b^4 + a^2*b...
 

Reduce [F]

\[ \int \frac {\sec ^2(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\left (\int \frac {\sec \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{4} b -a}d x \right ) \] Input:

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

Output:

 - int(sec(c + d*x)**2/(sin(c + d*x)**4*b - a),x)