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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3597, 1663, 845, 70} \[ \int \sin ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}-\frac {\cos ^2(c+d x) \left (a \left (5 a^2+b^2 (2 n+3)\right ) \tan (c+d x)+b \left (a^2 (7-n)+b^2 (n+5)\right )\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2}-\frac {\left (a b^2 n \left (5 a^2+b^2 (2 n+3)\right )+\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2+6 n+6\right )+b^4 \left (n^2+4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{16 b d (n+1) \left (a^2+b^2\right )^2 \left (a-\sqrt {-b^2}\right )}-\frac {\left (a b^2 n \left (5 a^2+b^2 (2 n+3)\right )-\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2+6 n+6\right )+b^4 \left (n^2+4 n+3\right )\right )\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{16 b d (n+1) \left (a^2+b^2\right )^2 \left (a+\sqrt {-b^2}\right )} \]

[In]

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

[Out]

-1/16*((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*(a^
2 + b^2)^2*(a - Sqrt[-b^2])*d*(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))/(16*b*(a^2 + b^2)^2*(a + Sqrt[-b^2])*d*(1 + n)) + (Cos[c + d*x]^4*(b + a*Tan[c +
d*x])*(a + b*Tan[c + d*x])^(1 + n))/(4*(a^2 + b^2)*d) - (Cos[c + d*x]^2*(a + b*Tan[c + d*x])^(1 + n)*(b*(a^2*(
7 - n) + b^2*(5 + n)) + a*(5*a^2 + b^2*(3 + 2*n))*Tan[c + d*x]))/(8*(a^2 + b^2)^2*d)

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 845

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m, (f + g*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !Ration
alQ[m]

Rule 1663

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(-(d + e*x)^(m + 1))*(a + c*x^2)^(p + 1)*((a*(e*f - d*g) + (c*d*f + a*e*g)*x)/(2*a*(p +
1)*(c*d^2 + a*e^2))), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandTo
Sum[2*a*(p + 1)*(c*d^2 + a*e^2)*Q + c*d^2*f*(2*p + 3) - a*e*(d*g*m - e*f*(m + 2*p + 3)) + e*(c*d*f + a*e*g)*(m
+ 2*p + 4)*x, x], x], x]] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1]
&&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 3597

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {b \text {Subst}\left (\int \frac {x^4 (a+x)^n}{\left (b^2+x^2\right )^3} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \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 {\text {Subst}\left (\int \frac {(a+x)^n \left (b^4 \left (a^2+b^2 (1+n)\right )-a b^4 (2-n) x-4 b^2 \left (a^2+b^2\right ) x^2\right )}{\left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{4 b \left (a^2+b^2\right ) d} \\ & = \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}+\frac {\text {Subst}\left (\int \frac {(a+x)^n \left (b^4 \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 )+a b^4 n \left (5 a^2+b^2 (3+2 n)\right ) x\right )}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{8 b^3 \left (a^2+b^2\right )^2 d} \\ & = \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}+\frac {\text {Subst}\left (\int \left (\frac {\left (-a b^6 n \left (5 a^2+b^2 (3+2 n)\right )+b^4 \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 ) (a+x)^n}{2 b^2 \left (\sqrt {-b^2}-x\right )}+\frac {\left (a b^6 n \left (5 a^2+b^2 (3+2 n)\right )+b^4 \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 ) (a+x)^n}{2 b^2 \left (\sqrt {-b^2}+x\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{8 b^3 \left (a^2+b^2\right )^2 d} \\ & = \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}-\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 ) \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}-x} \, dx,x,b \tan (c+d x)\right )}{16 b \left (a^2+b^2\right )^2 d}+\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 ) \text {Subst}\left (\int \frac {(a+x)^n}{\sqrt {-b^2}+x} \, dx,x,b \tan (c+d x)\right )}{16 b \left (a^2+b^2\right )^2 d} \\ & = -\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} \\ \end{align*}

Mathematica [B] (verified)

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

Time = 6.81 (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} \]

[In]

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

[Out]

(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 + 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)))/(2*b
^2*(a^2 + b^2))))/(4*(a^2 + b^2))))/d

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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