\(\int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) [84]

Optimal result

Integrand size = 21, antiderivative size = 435 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx=-\frac {\left (a b^2 n \left (5 a^2+b^2 (3+2 n)\right )+\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6+6 n-n^2\right )+b^4 \left (3+4 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 b \left (a^2+b^2\right )^2 \left (a-\sqrt {-b^2}\right ) d (1+n)}-\frac {\left (a b^2 n \left (5 a^2+b^2 (3+2 n)\right )-\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6+6 n-n^2\right )+b^4 \left (3+4 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 b \left (a^2+b^2\right )^2 \left (a+\sqrt {-b^2}\right ) d (1+n)}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}-\frac {\cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b \left (a^2 (7-n)+b^2 (5+n)\right )+a \left (5 a^2+b^2 (3+2 n)\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \] Output:

-1/16*(a*b^2*n*(5*a^2+b^2*(3+2*n))+(-b^2)^(1/2)*(3*a^4+a^2*b^2*(-n^2+6*n+6 
)+b^4*(n^2+4*n+3)))*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a-(-b^2)^(1 
/2)))*(a+b*tan(d*x+c))^(1+n)/b/(a^2+b^2)^2/(a-(-b^2)^(1/2))/d/(1+n)-1/16*( 
a*b^2*n*(5*a^2+b^2*(3+2*n))-(-b^2)^(1/2)*(3*a^4+a^2*b^2*(-n^2+6*n+6)+b^4*( 
n^2+4*n+3)))*hypergeom([1, 1+n],[2+n],(a+b*tan(d*x+c))/(a+(-b^2)^(1/2)))*( 
a+b*tan(d*x+c))^(1+n)/b/(a^2+b^2)^2/(a+(-b^2)^(1/2))/d/(1+n)+1/4*cos(d*x+c 
)^4*(b+a*tan(d*x+c))*(a+b*tan(d*x+c))^(1+n)/(a^2+b^2)/d-1/8*cos(d*x+c)^2*( 
a+b*tan(d*x+c))^(1+n)*(b*(a^2*(7-n)+b^2*(5+n))+a*(5*a^2+b^2*(3+2*n))*tan(d 
*x+c))/(a^2+b^2)^2/d
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(910\) vs. \(2(435)=870\).

Time = 6.50 (sec) , antiderivative size = 910, normalized size of antiderivative = 2.09 \[ \int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b \left (\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a-\sqrt {-b^2}\right ) (1+n)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 \sqrt {-b^2} \left (a+\sqrt {-b^2}\right ) (1+n)}-\frac {\cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2+a b \tan (c+d x)\right )}{b^2 \left (a^2+b^2\right )}+\frac {\cos ^4(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2+a b \tan (c+d x)\right )}{4 b^2 \left (a^2+b^2\right )}+\frac {\frac {\left (\sqrt {-b^2} \left (a^2+b^2 (1-n)\right )-a b^2 n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{b^2 \left (a-\sqrt {-b^2}\right ) (1+n)}-\frac {\left (a^2 \sqrt {-b^2}-\left (-b^2\right )^{3/2} (1-n)+a b^2 n\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{b^2 \left (a+\sqrt {-b^2}\right ) (1+n)}}{2 \left (a^2+b^2\right )}-\frac {b^2 \left (\frac {\cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 \left (-3 a^2-b^2 (3-n)\right )+a^2 b^2 (2-n)+b \left (a \left (-3 a^2-b^2 (3-n)\right )-a b^2 (2-n)\right ) \tan (c+d x)\right )}{2 b^4 \left (a^2+b^2\right )}-\frac {\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n-\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 b^2 \left (a-\sqrt {-b^2}\right ) (1+n)}+\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n+\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{2 b^2 \left (a+\sqrt {-b^2}\right ) (1+n)}}{2 b^2 \left (a^2+b^2\right )}\right )}{4 \left (a^2+b^2\right )}\right )}{d} \] Input:

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

Output:

(b*((Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^ 
2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a - Sqrt[-b^2])*(1 + n)) 
 - (Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2 
])]*(a + b*Tan[c + d*x])^(1 + n))/(2*Sqrt[-b^2]*(a + Sqrt[-b^2])*(1 + n)) 
- (Cos[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n)*(b^2 + a*b*Tan[c + d*x]))/( 
b^2*(a^2 + b^2)) + (Cos[c + d*x]^4*(a + b*Tan[c + d*x])^(1 + n)*(b^2 + a*b 
*Tan[c + d*x]))/(4*b^2*(a^2 + b^2)) + (((Sqrt[-b^2]*(a^2 + b^2*(1 - n)) - 
a*b^2*n)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt 
[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(b^2*(a - Sqrt[-b^2])*(1 + n)) - (( 
a^2*Sqrt[-b^2] - (-b^2)^(3/2)*(1 - n) + a*b^2*n)*Hypergeometric2F1[1, 1 + 
n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + 
 n))/(b^2*(a + Sqrt[-b^2])*(1 + n)))/(2*(a^2 + b^2)) - (b^2*((Cos[c + d*x] 
^2*(a + b*Tan[c + d*x])^(1 + n)*(b^2*(-3*a^2 - b^2*(3 - n)) + a^2*b^2*(2 - 
 n) + b*(a*(-3*a^2 - b^2*(3 - n)) - a*b^2*(2 - n))*Tan[c + d*x]))/(2*b^4*( 
a^2 + b^2)) - (((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n - Sqrt[-b^2]*(3*a^4 + a^2 
*b^2*(6 - 2*n - n^2) + b^4*(3 - 4*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 
 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n)) 
/(2*b^2*(a - Sqrt[-b^2])*(1 + n)) + ((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sq 
rt[-b^2]*(3*a^4 + a^2*b^2*(6 - 2*n - n^2) + b^4*(3 - 4*n + n^2)))*Hypergeo 
metric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + ...
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 476, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3999, 602, 2180, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (c+d x)^4 (a+b \tan (c+d x))^ndx\)

\(\Big \downarrow \) 3999

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

\(\Big \downarrow \) 602

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

\(\Big \downarrow \) 2180

\(\displaystyle \frac {b \left (\frac {b^2 \left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {b^2 (a+b \tan (c+d x))^{n+1} \left (a b \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x)+b^2 \left (a^2 (7-n)+b^2 (n+5)\right )\right )}{2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {\int \frac {b^4 (a+b \tan (c+d x))^n \left (3 a^4+b^2 \left (-n^2+6 n+6\right ) a^2+b n \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x) a+b^4 \left (n^2+4 n+3\right )\right )}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b \left (\frac {b^2 \left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {b^2 (a+b \tan (c+d x))^{n+1} \left (a b \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x)+b^2 \left (a^2 (7-n)+b^2 (n+5)\right )\right )}{2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {b^2 \int \frac {(a+b \tan (c+d x))^n \left (3 a^4+b^2 \left (-n^2+6 n+6\right ) a^2+b n \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x) a+b^4 \left (n^2+4 n+3\right )\right )}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\)

\(\Big \downarrow \) 657

\(\displaystyle \frac {b \left (\frac {b^2 \left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {b^2 (a+b \tan (c+d x))^{n+1} \left (a b \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x)+b^2 \left (a^2 (7-n)+b^2 (n+5)\right )\right )}{2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {b^2 \int \left (\frac {\left (\sqrt {-b^2} \left (3 a^4+b^2 \left (-n^2+6 n+6\right ) a^2+b^4 \left (n^2+4 n+3\right )\right )-a b^2 n \left (5 a^2+b^2 (2 n+3)\right )\right ) (a+b \tan (c+d x))^n}{2 b^2 \left (\sqrt {-b^2}-b \tan (c+d x)\right )}+\frac {\left (a n \left (5 a^2+b^2 (2 n+3)\right ) b^2+\sqrt {-b^2} \left (3 a^4+b^2 \left (-n^2+6 n+6\right ) a^2+b^4 \left (n^2+4 n+3\right )\right )\right ) (a+b \tan (c+d x))^n}{2 b^2 \left (b \tan (c+d x)+\sqrt {-b^2}\right )}\right )d(b \tan (c+d x))}{2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b \left (\frac {b^2 \left (a b \tan (c+d x)+b^2\right ) (a+b \tan (c+d x))^{n+1}}{4 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )^2}-\frac {\frac {b^2 (a+b \tan (c+d x))^{n+1} \left (a b \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x)+b^2 \left (a^2 (7-n)+b^2 (n+5)\right )\right )}{2 \left (a^2+b^2\right ) \left (b^2 \tan ^2(c+d x)+b^2\right )}-\frac {b^2 \left (-\frac {\left (a b^2 n \left (5 a^2+b^2 (2 n+3)\right )+\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2+6 n+6\right )+b^4 \left (n^2+4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{2 b^2 (n+1) \left (a-\sqrt {-b^2}\right )}-\frac {\left (a b^2 n \left (5 a^2+b^2 (2 n+3)\right )-\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2+6 n+6\right )+b^4 \left (n^2+4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{2 b^2 (n+1) \left (a+\sqrt {-b^2}\right )}\right )}{2 \left (a^2+b^2\right )}}{4 b^2 \left (a^2+b^2\right )}\right )}{d}\)

Input:

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

Output:

(b*((b^2*(a + b*Tan[c + d*x])^(1 + n)*(b^2 + a*b*Tan[c + d*x]))/(4*(a^2 + 
b^2)*(b^2 + b^2*Tan[c + d*x]^2)^2) - ((b^2*(a + b*Tan[c + d*x])^(1 + n)*(b 
^2*(a^2*(7 - n) + b^2*(5 + n)) + a*b*(5*a^2 + b^2*(3 + 2*n))*Tan[c + d*x]) 
)/(2*(a^2 + b^2)*(b^2 + b^2*Tan[c + d*x]^2)) - (b^2*(-1/2*((a*b^2*n*(5*a^2 
 + b^2*(3 + 2*n)) + Sqrt[-b^2]*(3*a^4 + a^2*b^2*(6 + 6*n - n^2) + b^4*(3 + 
 4*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - 
 Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(b^2*(a - Sqrt[-b^2])*(1 + n)) 
 - ((a*b^2*n*(5*a^2 + b^2*(3 + 2*n)) - Sqrt[-b^2]*(3*a^4 + a^2*b^2*(6 + 6* 
n - n^2) + b^4*(3 + 4*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b 
*Tan[c + d*x])/(a + Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(2*b^2*(a + 
 Sqrt[-b^2])*(1 + n))))/(2*(a^2 + b^2)))/(4*b^2*(a^2 + b^2))))/d
 

Defintions of rubi rules used

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 602

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomia 
lRemainder[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 + 1))*(a + b*x^2)^(p + 1)*((a*(d*e 
 - c*f) + (b*c*e + a*d*f)*x)/(2*a*(p + 1)*(b*c^2 + a*d^2))), x] + Simp[1/(2 
*a*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b*x^2)^(p + 1)*ExpandToS 
um[2*a*(p + 1)*(b*c^2 + a*d^2)*Qx + e*(b*c^2*(2*p + 3) + a*d^2*(n + 2*p + 3 
)) - a*c*d*f*n + d*(b*c*e + a*d*f)*(n + 2*p + 4)*x, x], x], x]] /; FreeQ[{a 
, b, c, d, n}, x] && IGtQ[m, 1] && LtQ[p, -1] && 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 2180

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 + 1))*(a + b*x^2)^(p + 1)*((a*(e*R - d*S) 
 + (b*d*R + a*e*S)*x)/(2*a*(p + 1)*(b*d^2 + a*e^2))), x] + Simp[1/(2*a*(p + 
 1)*(b*d^2 + a*e^2))   Int[(d + e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a* 
(p + 1)*(b*d^2 + a*e^2)*Qx + b*d^2*R*(2*p + 3) - a*e*(d*S*m - e*R*(m + 2*p 
+ 3)) + e*(b*d*R + a*e*S)*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, d, e 
, m}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] &&  !(IGtQ[ 
m, 0] && RationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

rule 3042

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

rule 3999

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[b/f   Subst[Int[x^m*((a + x)^n/(b^2 + x^2)^(m/2 + 1)), 
 x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/2]
 

Maple [F]

\[\int \sin \left (d x +c \right )^{4} \left (a +b \tan \left (d x +c \right )\right )^{n}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*(b*tan(d*x + c) + a)^n, x 
)
 

Sympy [F(-1)]

Timed out. \[ \int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

integrate((b*tan(d*x + c) + a)^n*sin(d*x + c)^4, x)
 

Giac [F]

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

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

Output:

integrate((b*tan(d*x + c) + a)^n*sin(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\int {\sin \left (c+d\,x\right )}^4\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int((tan(c + d*x)*b + a)**n*sin(c + d*x)**4,x)