\(\int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx\) [1490]

Optimal result

Integrand size = 27, antiderivative size = 151 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))}{16 d}-\frac {\left (16 a b-3 \left (a^2+5 b^2\right )\right ) \log (1+\sin (c+d x))}{16 d}-\frac {b^2 \sin (c+d x)}{d}-\frac {\sec ^2(c+d x) \left (12 a b+\left (5 a^2+7 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \] Output:

-1/16*(3*a^2+16*a*b+15*b^2)*ln(1-sin(d*x+c))/d-1/16*(-3*a^2+16*a*b-15*b^2) 
*ln(1+sin(d*x+c))/d-b^2*sin(d*x+c)/d-1/8*sec(d*x+c)^2*(12*a*b+(5*a^2+7*b^2 
)*sin(d*x+c))/d+1/4*sec(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {-\left (\left (3 a^2+16 a b+15 b^2\right ) \log (1-\sin (c+d x))\right )+\left (3 a^2-16 a b+15 b^2\right ) \log (1+\sin (c+d x))+\frac {(a+b)^2}{(-1+\sin (c+d x))^2}+\frac {(a+b) (5 a+9 b)}{-1+\sin (c+d x)}-16 b^2 \sin (c+d x)-\frac {(a-b)^2}{(1+\sin (c+d x))^2}+\frac {(5 a-9 b) (a-b)}{1+\sin (c+d x)}}{16 d} \] Input:

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

Output:

(-((3*a^2 + 16*a*b + 15*b^2)*Log[1 - Sin[c + d*x]]) + (3*a^2 - 16*a*b + 15 
*b^2)*Log[1 + Sin[c + d*x]] + (a + b)^2/(-1 + Sin[c + d*x])^2 + ((a + b)*( 
5*a + 9*b))/(-1 + Sin[c + d*x]) - 16*b^2*Sin[c + d*x] - (a - b)^2/(1 + Sin 
[c + d*x])^2 + ((5*a - 9*b)*(a - b))/(1 + Sin[c + d*x]))/(16*d)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.15, 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, 531, 25, 2176, 25, 2341, 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.

\(\displaystyle \int \tan ^4(c+d x) \sec (c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^4 (a+b \sin (c+d x))^2}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3316

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 531

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2176

\(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {8 \sin ^2(c+d x) b^6+16 a \sin (c+d x) b^5+\left (3 a^2+7 b^2\right ) b^4}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^2 (a+b \sin (c+d x)) \left (5 a b \sin (c+d x)+7 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x)) \left (5 a b \sin (c+d x)+7 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {8 \sin ^2(c+d x) b^6+16 a \sin (c+d x) b^5+\left (3 a^2+7 b^2\right ) b^4}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x)) \left (5 a b \sin (c+d x)+7 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (\frac {16 a \sin (c+d x) b^5+3 \left (a^2+5 b^2\right ) b^4}{b^2-b^2 \sin ^2(c+d x)}-8 b^4\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {b^3 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^2 (a+b \sin (c+d x)) \left (5 a b \sin (c+d x)+7 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {3 b^3 \left (a^2+5 b^2\right ) \text {arctanh}(\sin (c+d x))-8 a b^4 \log \left (b^2-b^2 \sin ^2(c+d x)\right )-8 b^5 \sin (c+d x)}{2 b^2}}{4 b^2}\right )}{d}\)

Input:

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

Output:

(b*((b^3*Sin[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2 
)^2) - (-1/2*(3*b^3*(a^2 + 5*b^2)*ArcTanh[Sin[c + d*x]] - 8*a*b^4*Log[b^2 
- b^2*Sin[c + d*x]^2] - 8*b^5*Sin[c + d*x])/b^2 + (b^2*(a + b*Sin[c + d*x] 
)*(7*b^2 + 5*a*b*Sin[c + d*x]))/(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 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

rule 2009

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

rule 2176

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

rule 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

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 = 5.02 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(200\)
default	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(200\)
risch	\(2 i x a b +\frac {i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{2 d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{2 d}+\frac {4 i a b c}{d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (5 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+32 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+32 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}-5 a^{2}-9 b^{2}+32 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\)	\(361\)

Input:

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

Output:

1/d*(a^2*(1/4*sin(d*x+c)^5/cos(d*x+c)^4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8* 
sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+2*a*b*(1/4*tan( 
d*x+c)^4-1/2*tan(d*x+c)^2-ln(cos(d*x+c)))+b^2*(1/4*sin(d*x+c)^7/cos(d*x+c) 
^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8*sin(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*si 
n(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 32 \, a b \cos \left (d x + c\right )^{2} + 8 \, a b - 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{4} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

1/16*((3*a^2 - 16*a*b + 15*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3* 
a^2 + 16*a*b + 15*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 32*a*b*cos( 
d*x + c)^2 + 8*a*b - 2*(8*b^2*cos(d*x + c)^4 + (5*a^2 + 9*b^2)*cos(d*x + c 
)^2 - 2*a^2 - 2*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.98 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right ) - {\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (16 \, a b \sin \left (d x + c\right )^{2} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 12 \, a b - {\left (3 \, a^{2} + 7 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \] Input:

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

Output:

-1/16*(16*b^2*sin(d*x + c) - (3*a^2 - 16*a*b + 15*b^2)*log(sin(d*x + c) + 
1) + (3*a^2 + 16*a*b + 15*b^2)*log(sin(d*x + c) - 1) - 2*(16*a*b*sin(d*x + 
 c)^2 + (5*a^2 + 9*b^2)*sin(d*x + c)^3 - 12*a*b - (3*a^2 + 7*b^2)*sin(d*x 
+ c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=-\frac {b^{2} \sin \left (d x + c\right )}{d} + \frac {{\left (3 \, a^{2} - 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (3 \, a^{2} + 16 \, a b + 15 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} + \frac {16 \, a b \sin \left (d x + c\right )^{2} + {\left (5 \, a^{2} + 9 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 12 \, a b - {\left (3 \, a^{2} + 7 \, b^{2}\right )} \sin \left (d x + c\right )}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:

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

Output:

-b^2*sin(d*x + c)/d + 1/16*(3*a^2 - 16*a*b + 15*b^2)*log(abs(sin(d*x + c) 
+ 1))/d - 1/16*(3*a^2 + 16*a*b + 15*b^2)*log(abs(sin(d*x + c) - 1))/d + 1/ 
8*(16*a*b*sin(d*x + c)^2 + (5*a^2 + 9*b^2)*sin(d*x + c)^3 - 12*a*b - (3*a^ 
2 + 7*b^2)*sin(d*x + c))/(d*(sin(d*x + c) + 1)^2*(sin(d*x + c) - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 20.29 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.20 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {3\,a^2}{8}-2\,a\,b+\frac {15\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {3\,a^2}{8}+2\,a\,b+\frac {15\,b^2}{8}\right )}{d}+\frac {\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {11\,a^2}{2}-\frac {9\,b^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (2\,a^2+10\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {3\,a^2}{4}-\frac {15\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:

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

Output:

(log(tan(c/2 + (d*x)/2) + 1)*((3*a^2)/8 - 2*a*b + (15*b^2)/8))/d - (log(ta 
n(c/2 + (d*x)/2) - 1)*(2*a*b + (3*a^2)/8 + (15*b^2)/8))/d + (tan(c/2 + (d* 
x)/2)^3*(2*a^2 + 10*b^2) + tan(c/2 + (d*x)/2)^7*(2*a^2 + 10*b^2) + tan(c/2 
 + (d*x)/2)^5*((11*a^2)/2 - (9*b^2)/2) - tan(c/2 + (d*x)/2)^9*((3*a^2)/4 + 
 (15*b^2)/4) - tan(c/2 + (d*x)/2)*((3*a^2)/4 + (15*b^2)/4) - 4*a*b*tan(c/2 
 + (d*x)/2)^2 + 12*a*b*tan(c/2 + (d*x)/2)^4 + 12*a*b*tan(c/2 + (d*x)/2)^6 
- 4*a*b*tan(c/2 + (d*x)/2)^8)/(d*(2*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d* 
x)/2)^2 + 2*tan(c/2 + (d*x)/2)^6 - 3*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x 
)/2)^10 + 1)) + (2*a*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 579, normalized size of antiderivative = 3.83 \[ \int \sec (c+d x) (a+b \sin (c+d x))^2 \tan ^4(c+d x) \, dx =\text {Too large to display} \] Input:

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

Output:

(16*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a*b - 32*log(tan((c + d*x 
)/2)**2 + 1)*sin(c + d*x)**2*a*b + 16*log(tan((c + d*x)/2)**2 + 1)*a*b - 3 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 16*log(tan((c + d*x)/2) 
- 1)*sin(c + d*x)**4*a*b - 15*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b* 
*2 + 6*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 32*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**2*a*b + 30*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**2*b**2 - 3*log(tan((c + d*x)/2) - 1)*a**2 - 16*log(tan((c + d*x)/2) - 1) 
*a*b - 15*log(tan((c + d*x)/2) - 1)*b**2 + 3*log(tan((c + d*x)/2) + 1)*sin 
(c + d*x)**4*a**2 - 16*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b + 15* 
log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 - 6*log(tan((c + d*x)/2) + 
1)*sin(c + d*x)**2*a**2 + 32*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b 
 - 30*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 + 3*log(tan((c + d*x) 
/2) + 1)*a**2 - 16*log(tan((c + d*x)/2) + 1)*a*b + 15*log(tan((c + d*x)/2) 
 + 1)*b**2 - 8*sin(c + d*x)**5*b**2 + 12*sin(c + d*x)**4*a*b + 5*sin(c + d 
*x)**3*a**2 + 25*sin(c + d*x)**3*b**2 - 8*sin(c + d*x)**2*a*b - 3*sin(c + 
d*x)*a**2 - 15*sin(c + d*x)*b**2)/(8*d*(sin(c + d*x)**4 - 2*sin(c + d*x)** 
2 + 1))