\(\int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx\) [342]

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\frac {b^{3/4} \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^2 d}+\frac {(a-5 b) \text {arctanh}(\sin (c+d x))}{2 (a-b)^2 d}+\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^2 d}+\frac {\sec (c+d x) \tan (c+d x)}{2 (a-b) d} \] Output:

1/2*b^(3/4)*arctan(b^(1/4)*sin(d*x+c)/a^(1/4))/a^(3/4)/(a^(1/2)+b^(1/2))^2 
/d+1/2*(a-5*b)*arctanh(sin(d*x+c))/(a-b)^2/d+1/2*b^(3/4)*arctanh(b^(1/4)*s 
in(d*x+c)/a^(1/4))/a^(3/4)/(a^(1/2)-b^(1/2))^2/d+1/2*sec(d*x+c)*tan(d*x+c) 
/(a-b)/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.23 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.69 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {-\frac {2 (a-5 b) \text {arctanh}(\sin (c+d x))}{(a-b)^2}+\frac {b^{3/4} \log \left (\sqrt [4]{a}-\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^2}-\frac {i b^{3/4} \log \left (\sqrt [4]{a}-i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^2}+\frac {i b^{3/4} \log \left (\sqrt [4]{a}+i \sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^2}-\frac {b^{3/4} \log \left (\sqrt [4]{a}+\sqrt [4]{b} \sin (c+d x)\right )}{a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^2}+\frac {1}{(a-b) (-1+\sin (c+d x))}+\frac {1}{(a-b) (1+\sin (c+d x))}}{4 d} \] Input:

Integrate[Sec[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]
 

Output:

-1/4*((-2*(a - 5*b)*ArcTanh[Sin[c + d*x]])/(a - b)^2 + (b^(3/4)*Log[a^(1/4 
) - b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] - Sqrt[b])^2) - (I*b^(3/4)*Lo 
g[a^(1/4) - I*b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] + Sqrt[b])^2) + (I* 
b^(3/4)*Log[a^(1/4) + I*b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] + Sqrt[b] 
)^2) - (b^(3/4)*Log[a^(1/4) + b^(1/4)*Sin[c + d*x]])/(a^(3/4)*(Sqrt[a] - S 
qrt[b])^2) + 1/((a - b)*(-1 + Sin[c + d*x])) + 1/((a - b)*(1 + Sin[c + d*x 
])))/d
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3702, 1485, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^3 \left (a-b \sin (c+d x)^4\right )}dx\)

\(\Big \downarrow \) 3702

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

\(\Big \downarrow \) 1485

\(\displaystyle \frac {\int \left (\frac {5 b-a}{2 (a-b)^2 \left (\sin ^2(c+d x)-1\right )}+\frac {b \left (2 b \sin ^2(c+d x)+a+b\right )}{(a-b)^2 \left (a-b \sin ^4(c+d x)\right )}+\frac {1}{4 (a-b) (\sin (c+d x)-1)^2}+\frac {1}{4 (a-b) (\sin (c+d x)+1)^2}\right )d\sin (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^{3/4} \arctan \left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}+\sqrt {b}\right )^2}+\frac {b^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{b} \sin (c+d x)}{\sqrt [4]{a}}\right )}{2 a^{3/4} \left (\sqrt {a}-\sqrt {b}\right )^2}+\frac {(a-5 b) \text {arctanh}(\sin (c+d x))}{2 (a-b)^2}+\frac {1}{4 (a-b) (1-\sin (c+d x))}-\frac {1}{4 (a-b) (\sin (c+d x)+1)}}{d}\)

Input:

Int[Sec[c + d*x]^3/(a - b*Sin[c + d*x]^4),x]
 

Output:

((b^(3/4)*ArcTan[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] + Sq 
rt[b])^2) + ((a - 5*b)*ArcTanh[Sin[c + d*x]])/(2*(a - b)^2) + (b^(3/4)*Arc 
Tanh[(b^(1/4)*Sin[c + d*x])/a^(1/4)])/(2*a^(3/4)*(Sqrt[a] - Sqrt[b])^2) + 
1/(4*(a - b)*(1 - Sin[c + d*x])) - 1/(4*(a - b)*(1 + Sin[c + d*x])))/d
 

Defintions of rubi rules used

rule 1485

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> Int[Expa 
ndIntegrand[(d + e*x^2)^q/(a + c*x^4), x], x] /; FreeQ[{a, c, d, e}, x] && 
NeQ[c*d^2 + 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 3702

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x 
_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Si 
mp[ff/f   Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p, x], x, 
 Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 
1)/2] && (EqQ[n, 4] || GtQ[m, 0] || IGtQ[p, 0] || IntegersQ[m, p])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(240\) vs. \(2(119)=238\).

Time = 2.41 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.60

method	result	size
derivativedivides	\(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a +5 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a -b \right )^{2}}-\frac {b \left (\frac {\left (-a -b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -5 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\)	\(241\)
default	\(\frac {-\frac {1}{\left (4 a -4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\left (-a +5 b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{4 \left (a -b \right )^{2}}-\frac {b \left (\frac {\left (-a -b \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 a}+\frac {2 \arctan \left (\frac {\sin \left (d x +c \right )}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {\sin \left (d x +c \right )+\left (\frac {a}{b}\right )^{\frac {1}{4}}}{\sin \left (d x +c \right )-\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a -b \right )^{2}}-\frac {1}{\left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -5 b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2}}}{d}\)	\(241\)
risch	\(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left (a -b \right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 d \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a}{2 d \left (a^{2}-2 a b +b^{2}\right )}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 d \left (a^{2}-2 a b +b^{2}\right )}+8 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (1048576 a^{7} d^{4}-4194304 a^{6} b \,d^{4}+6291456 a^{5} b^{2} d^{4}-4194304 a^{4} b^{3} d^{4}+1048576 a^{3} b^{4} d^{4}\right ) \textit {\_Z}^{4}+\left (-8192 a^{3} b^{2} d^{2}-8192 d^{2} b^{3} a^{2}\right ) \textit {\_Z}^{2}-b^{3}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\left (\left (-\frac {131072 i d^{3} a^{7}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}+\frac {524288 i d^{3} a^{6} b}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}-\frac {786432 i d^{3} a^{5} b^{2}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}+\frac {524288 i d^{3} a^{4} b^{3}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}-\frac {131072 i d^{3} a^{3} b^{4}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}\right ) \textit {\_R}^{3}+\left (\frac {64 i a^{4} b d}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}+\frac {960 i d \,a^{3} b^{2}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}+\frac {960 i d \,a^{2} b^{3}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}+\frac {64 i a \,b^{4} d}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}\right ) \textit {\_R} \right ) {\mathrm e}^{i \left (d x +c \right )}-\frac {a^{2} b^{2}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}-\frac {6 a \,b^{3}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}-\frac {b^{4}}{a^{2} b^{2}+6 a \,b^{3}+b^{4}}\right )\right )\)	\(641\)

Input:

int(sec(d*x+c)^3/(a-b*sin(d*x+c)^4),x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/(4*a-4*b)/(sin(d*x+c)-1)+1/4/(a-b)^2*(-a+5*b)*ln(sin(d*x+c)-1)-b/( 
a-b)^2*(1/4*(-a-b)*(1/b*a)^(1/4)/a*(ln((sin(d*x+c)+(1/b*a)^(1/4))/(sin(d*x 
+c)-(1/b*a)^(1/4)))+2*arctan(sin(d*x+c)/(1/b*a)^(1/4)))+1/2/(1/b*a)^(1/4)* 
(2*arctan(sin(d*x+c)/(1/b*a)^(1/4))-ln((sin(d*x+c)+(1/b*a)^(1/4))/(sin(d*x 
+c)-(1/b*a)^(1/4)))))-1/(4*a-4*b)/(1+sin(d*x+c))+1/4*(a-5*b)/(a-b)^2*ln(1+ 
sin(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2529 vs. \(2 (119) = 238\).

Time = 0.53 (sec) , antiderivative size = 2529, normalized size of antiderivative = 16.75 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*((a^2 - 2*a*b + b^2)*d*sqrt(((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + 
 a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^ 
11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5 
*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) + 4*a*b^2 + 4*b^3)/((a^5 - 4*a^4*b + 6*a 
^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*cos(d*x + c)^2*log(1/2*(a^2*b^2 + 6*a*b^ 
3 + b^4)*sin(d*x + c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4*b^3 + a^ 
3*b^4)*d^3*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^1 
1 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5* 
b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - (a^4*b + 7*a^3*b^2 + 7*a^2*b^3 + a*b^4) 
*d)*sqrt(((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^ 
3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b^7)/((a^11 - 8*a^10*b + 28*a^9*b 
^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 + 28*a^5*b^6 - 8*a^4*b^7 + a^3*b 
^8)*d^4)) + 4*a*b^2 + 4*b^3)/((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b 
^4)*d^2))) - (a^2 - 2*a*b + b^2)*d*sqrt(-((a^5 - 4*a^4*b + 6*a^3*b^2 - 4*a 
^2*b^3 + a*b^4)*d^2*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 + b 
^7)/((a^11 - 8*a^10*b + 28*a^9*b^2 - 56*a^8*b^3 + 70*a^7*b^4 - 56*a^6*b^5 
+ 28*a^5*b^6 - 8*a^4*b^7 + a^3*b^8)*d^4)) - 4*a*b^2 - 4*b^3)/((a^5 - 4*a^4 
*b + 6*a^3*b^2 - 4*a^2*b^3 + a*b^4)*d^2))*cos(d*x + c)^2*log(1/2*(a^2*b^2 
+ 6*a*b^3 + b^4)*sin(d*x + c) + 1/2*(2*(a^7 - 4*a^6*b + 6*a^5*b^2 - 4*a^4* 
b^3 + a^3*b^4)*d^3*sqrt((a^4*b^3 + 12*a^3*b^4 + 38*a^2*b^5 + 12*a*b^6 +...
 

Sympy [F]

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

integrate(sec(d*x+c)**3/(a-b*sin(d*x+c)**4),x)
 

Output:

Integral(sec(c + d*x)**3/(a - b*sin(c + d*x)**4), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (119) = 238\).

Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {\frac {b {\left (\frac {2 \, {\left (b {\left (2 \, \sqrt {a} - \sqrt {b}\right )} - a \sqrt {b}\right )} \arctan \left (\frac {\sqrt {b} \sin \left (d x + c\right )}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {{\left (b {\left (2 \, \sqrt {a} + \sqrt {b}\right )} + a \sqrt {b}\right )} \log \left (\frac {\sqrt {b} \sin \left (d x + c\right ) - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} \sin \left (d x + c\right ) + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}\right )}}{a^{2} - 2 \, a b + b^{2}} - \frac {{\left (a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {{\left (a - 5 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {2 \, \sin \left (d x + c\right )}{{\left (a - b\right )} \sin \left (d x + c\right )^{2} - a + b}}{4 \, d} \] Input:

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

Output:

-1/4*(b*(2*(b*(2*sqrt(a) - sqrt(b)) - a*sqrt(b))*arctan(sqrt(b)*sin(d*x + 
c)/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + (b*(2* 
sqrt(a) + sqrt(b)) + a*sqrt(b))*log((sqrt(b)*sin(d*x + c) - sqrt(sqrt(a)*s 
qrt(b)))/(sqrt(b)*sin(d*x + c) + sqrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqr 
t(a)*sqrt(b))*sqrt(b)))/(a^2 - 2*a*b + b^2) - (a - 5*b)*log(sin(d*x + c) + 
 1)/(a^2 - 2*a*b + b^2) + (a - 5*b)*log(sin(d*x + c) - 1)/(a^2 - 2*a*b + b 
^2) + 2*sin(d*x + c)/((a - b)*sin(d*x + c)^2 - a + b))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (119) = 238\).

Time = 0.39 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.28 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=-\frac {{\left (a - 5 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {{\left (a - 5 \, b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{4 \, {\left (a^{2} d - 2 \, a b d + b^{2} d\right )}} + \frac {{\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{3} b - 2 \, \sqrt {2} a^{2} b^{2} + \sqrt {2} a b^{3}\right )} d} + \frac {{\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} + 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sin \left (d x + c\right )\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, {\left (\sqrt {2} a^{3} b - 2 \, \sqrt {2} a^{2} b^{2} + \sqrt {2} a b^{3}\right )} d} + \frac {{\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a^{3} b - 2 \, \sqrt {2} a^{2} b^{2} + \sqrt {2} a b^{3}\right )} d} - \frac {{\left (\left (-a b^{3}\right )^{\frac {1}{4}} {\left (a b + b^{2}\right )} - 2 \, \left (-a b^{3}\right )^{\frac {3}{4}}\right )} \log \left (\sin \left (d x + c\right )^{2} - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}} \sin \left (d x + c\right ) + \sqrt {-\frac {a}{b}}\right )}{4 \, {\left (\sqrt {2} a^{3} b - 2 \, \sqrt {2} a^{2} b^{2} + \sqrt {2} a b^{3}\right )} d} - \frac {\sin \left (d x + c\right )}{2 \, {\left (a d - b d\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}} \] Input:

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

Output:

-1/4*(a - 5*b)*log(abs(-sin(d*x + c) + 1))/(a^2*d - 2*a*b*d + b^2*d) + 1/4 
*(a - 5*b)*log(abs(-sin(d*x + c) - 1))/(a^2*d - 2*a*b*d + b^2*d) + 1/2*((- 
a*b^3)^(1/4)*(a*b + b^2) + 2*(-a*b^3)^(3/4))*arctan(1/2*sqrt(2)*(sqrt(2)*( 
-a/b)^(1/4) + 2*sin(d*x + c))/(-a/b)^(1/4))/((sqrt(2)*a^3*b - 2*sqrt(2)*a^ 
2*b^2 + sqrt(2)*a*b^3)*d) + 1/2*((-a*b^3)^(1/4)*(a*b + b^2) + 2*(-a*b^3)^( 
3/4))*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a/b)^(1/4) - 2*sin(d*x + c))/(-a/b)^( 
1/4))/((sqrt(2)*a^3*b - 2*sqrt(2)*a^2*b^2 + sqrt(2)*a*b^3)*d) + 1/4*((-a*b 
^3)^(1/4)*(a*b + b^2) - 2*(-a*b^3)^(3/4))*log(sin(d*x + c)^2 + sqrt(2)*(-a 
/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/((sqrt(2)*a^3*b - 2*sqrt(2)*a^2*b^2 + 
 sqrt(2)*a*b^3)*d) - 1/4*((-a*b^3)^(1/4)*(a*b + b^2) - 2*(-a*b^3)^(3/4))*l 
og(sin(d*x + c)^2 - sqrt(2)*(-a/b)^(1/4)*sin(d*x + c) + sqrt(-a/b))/((sqrt 
(2)*a^3*b - 2*sqrt(2)*a^2*b^2 + sqrt(2)*a*b^3)*d) - 1/2*sin(d*x + c)/((a*d 
 - b*d)*(sin(d*x + c)^2 - 1))
 

Mupad [B] (verification not implemented)

Time = 39.09 (sec) , antiderivative size = 7758, normalized size of antiderivative = 51.38 \[ \int \frac {\sec ^3(c+d x)}{a-b \sin ^4(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((((128*a*b^11 + 256*a^2*b^10 - 3456*a^3*b^9 + 8960*a^4*b^8 - 1088 
0*a^5*b^7 + 6912*a^6*b^6 - 2176*a^7*b^5 + 256*a^8*b^4)/(2*(a^4 - 4*a^3*b - 
 4*a*b^3 + b^4 + 6*a^2*b^2)) - (sin(c + d*x)*(-(a^2*(a^3*b^3)^(1/2) + b^2* 
(a^3*b^3)^(1/2) - 4*a^2*b^3 - 4*a^3*b^2 + 6*a*b*(a^3*b^3)^(1/2))/(16*(a^7 
- 4*a^6*b + a^3*b^4 - 4*a^4*b^3 + 6*a^5*b^2)))^(1/2)*(512*a^2*b^11 - 2560* 
a^3*b^10 + 4608*a^4*b^9 - 2560*a^5*b^8 - 2560*a^6*b^7 + 4608*a^7*b^6 - 256 
0*a^8*b^5 + 512*a^9*b^4))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2))*(-( 
a^2*(a^3*b^3)^(1/2) + b^2*(a^3*b^3)^(1/2) - 4*a^2*b^3 - 4*a^3*b^2 + 6*a*b* 
(a^3*b^3)^(1/2))/(16*(a^7 - 4*a^6*b + a^3*b^4 - 4*a^4*b^3 + 6*a^5*b^2)))^( 
1/2) + (sin(c + d*x)*(48*a*b^10 - 16*b^11 + 1024*a^2*b^9 - 2208*a^3*b^8 + 
1264*a^4*b^7 - 144*a^5*b^6 + 32*a^6*b^5))/(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 
 6*a^2*b^2))*(-(a^2*(a^3*b^3)^(1/2) + b^2*(a^3*b^3)^(1/2) - 4*a^2*b^3 - 4* 
a^3*b^2 + 6*a*b*(a^3*b^3)^(1/2))/(16*(a^7 - 4*a^6*b + a^3*b^4 - 4*a^4*b^3 
+ 6*a^5*b^2)))^(1/2) - (200*a*b^9 + 480*a^2*b^8 - 784*a^3*b^7 + 96*a^4*b^6 
 + 8*a^5*b^5)/(2*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)))*(-(a^2*(a^3 
*b^3)^(1/2) + b^2*(a^3*b^3)^(1/2) - 4*a^2*b^3 - 4*a^3*b^2 + 6*a*b*(a^3*b^3 
)^(1/2))/(16*(a^7 - 4*a^6*b + a^3*b^4 - 4*a^4*b^3 + 6*a^5*b^2)))^(1/2) + ( 
sin(c + d*x)*(11*a*b^8 + 27*b^9 - 7*a^2*b^7 + a^3*b^6))/(a^4 - 4*a^3*b - 4 
*a*b^3 + b^4 + 6*a^2*b^2))*(-(a^2*(a^3*b^3)^(1/2) + b^2*(a^3*b^3)^(1/2) - 
4*a^2*b^3 - 4*a^3*b^2 + 6*a*b*(a^3*b^3)^(1/2))/(16*(a^7 - 4*a^6*b + a^3...
 

Reduce [F]

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

int(sec(d*x+c)^3/(a-b*sin(d*x+c)^4),x)
 

Output:

 - int(sec(c + d*x)**3/(sin(c + d*x)**4*b - a),x)