\(\int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx\) [1359]

Optimal result

Integrand size = 27, antiderivative size = 190 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a (3 a+b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {a (3 a-b) \log (1+\sin (c+d x))}{16 (a-b)^3 d}-\frac {a^4 b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x))}{4 \left (a^2-b^2\right ) d}+\frac {\sec ^2(c+d x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \] Output:

-1/16*a*(3*a+b)*ln(1-sin(d*x+c))/(a+b)^3/d+1/16*a*(3*a-b)*ln(1+sin(d*x+c)) 
/(a-b)^3/d-a^4*b*ln(a+b*sin(d*x+c))/(a^2-b^2)^3/d-1/4*sec(d*x+c)^4*(b-a*si 
n(d*x+c))/(a^2-b^2)/d+1/8*sec(d*x+c)^2*(4*b*(2*a^2-b^2)-a*(5*a^2-b^2)*sin( 
d*x+c))/(a^2-b^2)^2/d
 

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-\frac {a (3 a+b) \log (1-\sin (c+d x))}{(a+b)^3}+\frac {a (3 a-b) \log (1+\sin (c+d x))}{(a-b)^3}-\frac {16 a^4 b \log (a+b \sin (c+d x))}{(a-b)^3 (a+b)^3}+\frac {1}{(a+b) (-1+\sin (c+d x))^2}+\frac {5 a+3 b}{(a+b)^2 (-1+\sin (c+d x))}-\frac {1}{(a-b) (1+\sin (c+d x))^2}+\frac {5 a-3 b}{(a-b)^2 (1+\sin (c+d x))}}{16 d} \] Input:

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

Output:

(-((a*(3*a + b)*Log[1 - Sin[c + d*x]])/(a + b)^3) + (a*(3*a - b)*Log[1 + S 
in[c + d*x]])/(a - b)^3 - (16*a^4*b*Log[a + b*Sin[c + d*x]])/((a - b)^3*(a 
 + b)^3) + 1/((a + b)*(-1 + Sin[c + d*x])^2) + (5*a + 3*b)/((a + b)^2*(-1 
+ Sin[c + d*x])) - 1/((a - b)*(1 + Sin[c + d*x])^2) + (5*a - 3*b)/((a - b) 
^2*(1 + Sin[c + d*x])))/(16*d)
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.43, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3316, 27, 601, 2178, 25, 27, 657, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^5 (a+b \sin (c+d x))}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^5 \int \frac {\sin ^4(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \int \frac {b^4 \sin ^4(c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 601

\(\displaystyle \frac {b \left (-\frac {\int \frac {-\frac {3 a \sin (c+d x) b^5}{a^2-b^2}+4 \sin ^2(c+d x) b^4+\frac {a^2 b^4}{a^2-b^2}}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 2178

\(\displaystyle \frac {b \left (-\frac {\frac {\int -\frac {a b^4 \left (a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b \left (-\frac {-\frac {\int \frac {a b^4 \left (a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{\left (a^2-b^2\right )^2 (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 b^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \left (-\frac {-\frac {a b^2 \int \frac {a \left (3 a^2+b^2\right )-b \left (5 a^2-b^2\right ) \sin (c+d x)}{(a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )}d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {b \left (-\frac {-\frac {a b^2 \int \left (-\frac {8 a^3}{(a-b) (a+b) (a+b \sin (c+d x))}+\frac {(a-b)^2 (3 a+b)}{2 b (a+b) (b-b \sin (c+d x))}+\frac {(3 a-b) (a+b)^2}{2 (a-b) b (\sin (c+d x) b+b)}\right )d(b \sin (c+d x))}{2 \left (a^2-b^2\right )^2}-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (-\frac {b^2 \left (\frac {b^2}{a^2-b^2}-\frac {a b \sin (c+d x)}{a^2-b^2}\right )}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {-\frac {b^2 \left (4 b^2 \left (2 a^2-b^2\right )-a b \left (5 a^2-b^2\right ) \sin (c+d x)\right )}{2 \left (a^2-b^2\right )^2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {a b^2 \left (-\frac {8 a^3 \log (a+b \sin (c+d x))}{a^2-b^2}-\frac {(a-b)^2 (3 a+b) \log (b-b \sin (c+d x))}{2 b (a+b)}+\frac {(3 a-b) (a+b)^2 \log (b \sin (c+d x)+b)}{2 b (a-b)}\right )}{2 \left (a^2-b^2\right )^2}}{4 b^2}\right )}{d}\)

Input:

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

Output:

(b*(-1/4*(b^2*(b^2/(a^2 - b^2) - (a*b*Sin[c + d*x])/(a^2 - b^2)))/(b^2 - b 
^2*Sin[c + d*x]^2)^2 - (-1/2*(a*b^2*(-1/2*((a - b)^2*(3*a + b)*Log[b - b*S 
in[c + d*x]])/(b*(a + b)) - (8*a^3*Log[a + b*Sin[c + d*x]])/(a^2 - b^2) + 
((3*a - b)*(a + b)^2*Log[b + b*Sin[c + d*x]])/(2*(a - b)*b)))/(a^2 - b^2)^ 
2 - (b^2*(4*b^2*(2*a^2 - b^2) - a*b*(5*a^2 - b^2)*Sin[c + d*x]))/(2*(a^2 - 
 b^2)^2*(b^2 - b^2*Sin[c + d*x]^2)))/(4*b^2)))/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 
2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
 

rule 2009

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

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 3042

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

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-5 a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {a \left (3 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-5 a -3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (3 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{4} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(175\)
default	\(\frac {-\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-5 a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {a \left (3 a -b \right ) \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {-5 a -3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {a \left (3 a +b \right ) \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}-\frac {a^{4} b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\)	\(175\)
risch	\(\frac {i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {3 i a^{2} c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a b x}{8 a^{3}+24 a^{2} b +24 a \,b^{2}+8 b^{3}}-\frac {3 i a^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {2 i a^{4} b c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {2 i a^{4} b x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {3 i a^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i a^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {i a b c}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a b x}{8 a^{3}-24 a^{2} b +24 a \,b^{2}-8 b^{3}}-\frac {-5 i a^{3} {\mathrm e}^{7 i \left (d x +c \right )}+i a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+3 i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}-7 i a \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+8 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+7 i a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}+5 i a^{3} {\mathrm e}^{i \left (d x +c \right )}-i a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-16 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2} b +8 \,{\mathrm e}^{2 i \left (d x +c \right )} b^{3}}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2}}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2}}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {a^{4} b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\)	\(782\)

Input:

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

Output:

1/d*(-1/2/(8*a-8*b)/(1+sin(d*x+c))^2-1/16*(-5*a+3*b)/(a-b)^2/(1+sin(d*x+c) 
)+1/16*a*(3*a-b)/(a-b)^3*ln(1+sin(d*x+c))+1/2/(8*a+8*b)/(sin(d*x+c)-1)^2-1 
/16*(-5*a-3*b)/(a+b)^2/(sin(d*x+c)-1)-1/16*a*(3*a+b)/(a+b)^3*ln(sin(d*x+c) 
-1)-a^4*b/(a+b)^3/(a-b)^3*ln(a+b*sin(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.37 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {16 \, a^{4} b \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (3 \, a^{5} + 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{5} - 8 \, a^{4} b + 6 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{4} b - 8 \, a^{2} b^{3} + 4 \, b^{5} - 8 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{5} - 4 \, a^{3} b^{2} + 2 \, a b^{4} - {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

-1/16*(16*a^4*b*cos(d*x + c)^4*log(b*sin(d*x + c) + a) - (3*a^5 + 8*a^4*b 
+ 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) + (3*a^5 - 8*a^4 
*b + 6*a^3*b^2 - a*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 4*a^4*b - 
8*a^2*b^3 + 4*b^5 - 8*(2*a^4*b - 3*a^2*b^3 + b^5)*cos(d*x + c)^2 - 2*(2*a^ 
5 - 4*a^3*b^2 + 2*a*b^4 - (5*a^5 - 6*a^3*b^2 + a*b^4)*cos(d*x + c)^2)*sin( 
d*x + c))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.45 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {16 \, a^{4} b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} - \frac {{\left (3 \, a^{2} - a b\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} + \frac {{\left (3 \, a^{2} + a b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {2 \, {\left ({\left (5 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{3} + 6 \, a^{2} b - 2 \, b^{3} - 4 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{2} - {\left (3 \, a^{3} + a b^{2}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \] Input:

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

Output:

-1/16*(16*a^4*b*log(b*sin(d*x + c) + a)/(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 
) - (3*a^2 - a*b)*log(sin(d*x + c) + 1)/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) + 
(3*a^2 + a*b)*log(sin(d*x + c) - 1)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - 2*(( 
5*a^3 - a*b^2)*sin(d*x + c)^3 + 6*a^2*b - 2*b^3 - 4*(2*a^2*b - b^3)*sin(d* 
x + c)^2 - (3*a^3 + a*b^2)*sin(d*x + c))/((a^4 - 2*a^2*b^2 + b^4)*sin(d*x 
+ c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*sin(d*x + c)^2) 
)/d
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.56 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {a^{4} b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d - 3 \, a^{4} b^{3} d + 3 \, a^{2} b^{5} d - b^{7} d} - \frac {{\left (3 \, a^{2} + a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, {\left (a^{3} d + 3 \, a^{2} b d + 3 \, a b^{2} d + b^{3} d\right )}} + \frac {{\left (3 \, a^{2} - a b\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, {\left (a^{3} d - 3 \, a^{2} b d + 3 \, a b^{2} d - b^{3} d\right )}} + \frac {6 \, a^{4} b - 8 \, a^{2} b^{3} + 2 \, b^{5} + {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )^{3} - 4 \, {\left (2 \, a^{4} b - 3 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - {\left (3 \, a^{5} - 2 \, a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{8 \, {\left (a + b\right )}^{3} {\left (a - b\right )}^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:

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

Output:

-a^4*b^2*log(abs(b*sin(d*x + c) + a))/(a^6*b*d - 3*a^4*b^3*d + 3*a^2*b^5*d 
 - b^7*d) - 1/16*(3*a^2 + a*b)*log(abs(-sin(d*x + c) + 1))/(a^3*d + 3*a^2* 
b*d + 3*a*b^2*d + b^3*d) + 1/16*(3*a^2 - a*b)*log(abs(-sin(d*x + c) - 1))/ 
(a^3*d - 3*a^2*b*d + 3*a*b^2*d - b^3*d) + 1/8*(6*a^4*b - 8*a^2*b^3 + 2*b^5 
 + (5*a^5 - 6*a^3*b^2 + a*b^4)*sin(d*x + c)^3 - 4*(2*a^4*b - 3*a^2*b^3 + b 
^5)*sin(d*x + c)^2 - (3*a^5 - 2*a^3*b^2 - a*b^4)*sin(d*x + c))/((a + b)^3* 
(a - b)^3*d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 19.98 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.67 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {b^2}{4\,{\left (a-b\right )}^3}+\frac {5\,b}{8\,{\left (a-b\right )}^2}+\frac {3}{8\,\left (a-b\right )}\right )}{d}-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (3\,a^3+a\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a\,b^2-11\,a^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (2\,a^2\,b-b^3\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^4-2\,a^2\,b^2+b^4}-\frac {2\,a^2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{a^4-2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2+b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3}{8\,\left (a+b\right )}-\frac {5\,b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}-\frac {a^4\,b\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )} \] Input:

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

Output:

(log(tan(c/2 + (d*x)/2) + 1)*(b^2/(4*(a - b)^3) + (5*b)/(8*(a - b)^2) + 3/ 
(8*(a - b))))/d - ((tan(c/2 + (d*x)/2)^7*(a*b^2 + 3*a^3))/(4*(a^4 + b^4 - 
2*a^2*b^2)) + (tan(c/2 + (d*x)/2)^3*(7*a*b^2 - 11*a^3))/(4*(a^4 + b^4 - 2* 
a^2*b^2)) + (tan(c/2 + (d*x)/2)^5*(7*a*b^2 - 11*a^3))/(4*(a^4 + b^4 - 2*a^ 
2*b^2)) + (4*tan(c/2 + (d*x)/2)^4*(2*a^2*b - b^3))/(a^4 + b^4 - 2*a^2*b^2) 
 - (2*a^2*b*tan(c/2 + (d*x)/2)^2)/(a^4 + b^4 - 2*a^2*b^2) - (2*a^2*b*tan(c 
/2 + (d*x)/2)^6)/(a^4 + b^4 - 2*a^2*b^2) + (a*tan(c/2 + (d*x)/2)*(3*a^2 + 
b^2))/(4*(a^4 + b^4 - 2*a^2*b^2)))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 
+ (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (log( 
tan(c/2 + (d*x)/2) - 1)*(3/(8*(a + b)) - (5*b)/(8*(a + b)^2) + b^2/(4*(a + 
 b)^3)))/d - (a^4*b*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^ 
2))/(d*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 995, normalized size of antiderivative = 5.24 \[ \int \frac {\sec (c+d x) \tan ^4(c+d x)}{a+b \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**5 + 8*log(tan((c + d*x) 
/2) - 1)*sin(c + d*x)**4*a**4*b - 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**4*a**3*b**2 + log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**4 + 6*log(t 
an((c + d*x)/2) - 1)*sin(c + d*x)**2*a**5 - 16*log(tan((c + d*x)/2) - 1)*s 
in(c + d*x)**2*a**4*b + 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3* 
b**2 - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**4 - 3*log(tan((c + 
 d*x)/2) - 1)*a**5 + 8*log(tan((c + d*x)/2) - 1)*a**4*b - 6*log(tan((c + d 
*x)/2) - 1)*a**3*b**2 + log(tan((c + d*x)/2) - 1)*a*b**4 + 3*log(tan((c + 
d*x)/2) + 1)*sin(c + d*x)**4*a**5 + 8*log(tan((c + d*x)/2) + 1)*sin(c + d* 
x)**4*a**4*b + 6*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3*b**2 - log 
(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**4 - 6*log(tan((c + d*x)/2) + 1 
)*sin(c + d*x)**2*a**5 - 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 
*b - 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b**2 + 2*log(tan((c 
 + d*x)/2) + 1)*sin(c + d*x)**2*a*b**4 + 3*log(tan((c + d*x)/2) + 1)*a**5 
+ 8*log(tan((c + d*x)/2) + 1)*a**4*b + 6*log(tan((c + d*x)/2) + 1)*a**3*b* 
*2 - log(tan((c + d*x)/2) + 1)*a*b**4 - 8*log(tan((c + d*x)/2)**2*a + 2*ta 
n((c + d*x)/2)*b + a)*sin(c + d*x)**4*a**4*b + 16*log(tan((c + d*x)/2)**2* 
a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**4*b - 8*log(tan((c + d*x) 
/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**4*b - 2*sin(c + d*x)**4*a**4*b + 4 
*sin(c + d*x)**4*a**2*b**3 - 2*sin(c + d*x)**4*b**5 + 5*sin(c + d*x)**3...