\(\int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx\) [650]

Optimal result

Integrand size = 21, antiderivative size = 434 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {b \left (\frac {a \left (5+\frac {3 a^2}{b^2}-2 n\right ) n}{b^2}-\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 \left (1+\frac {a^2}{b^2}\right )^2 \left (a-\sqrt {-b^2}\right ) d (1+n)}+\frac {b \left (\frac {a \left (5+\frac {3 a^2}{b^2}-2 n\right ) n}{b^2}+\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (6-2 n-n^2\right )+b^4 \left (3-4 n+n^2\right )\right )}{b^6}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right ) (a+b \tan (c+d x))^{1+n}}{16 \left (1+\frac {a^2}{b^2}\right )^2 \left (a+\sqrt {-b^2}\right ) d (1+n)}+\frac {\cos ^4(c+d x) (b+a \tan (c+d x)) (a+b \tan (c+d x))^{1+n}}{4 \left (a^2+b^2\right ) d}+\frac {b \cos ^2(c+d x) (a+b \tan (c+d x))^{1+n} \left (b^2 (3-n)+a^2 (1+n)+a b \left (5+\frac {3 a^2}{b^2}-2 n\right ) \tan (c+d x)\right )}{8 \left (a^2+b^2\right )^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 434, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3587, 755, 837, 845, 70} \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}+\frac {b \cos ^2(c+d x) \left (a b \left (\frac {3 a^2}{b^2}-2 n+5\right ) \tan (c+d x)+a^2 (n+1)+b^2 (3-n)\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2}+\frac {b \left (\frac {a n \left (\frac {3 a^2}{b^2}-2 n+5\right )}{b^2}-\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2-2 n+6\right )+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{16 d (n+1) \left (\frac {a^2}{b^2}+1\right )^2 \left (a-\sqrt {-b^2}\right )}+\frac {b \left (\frac {a n \left (\frac {3 a^2}{b^2}-2 n+5\right )}{b^2}+\frac {\sqrt {-b^2} \left (3 a^4+a^2 b^2 \left (-n^2-2 n+6\right )+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{16 d (n+1) \left (\frac {a^2}{b^2}+1\right )^2 \left (a+\sqrt {-b^2}\right )} \]

[In]

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

[Out]

(b*((a*(5 + (3*a^2)/b^2 - 2*n)*n)/b^2 - (Sqrt[-b^2]*(3*a^4 + a^2*b^2*(6 - 2*n - n^2) + b^4*(3 - 4*n + n^2)))/b
^6)*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])]*(a + b*Tan[c + d*x])^(1 + n))/(1
6*(1 + a^2/b^2)^2*(a - Sqrt[-b^2])*d*(1 + n)) + (b*((a*(5 + (3*a^2)/b^2 - 2*n)*n)/b^2 + (Sqrt[-b^2]*(3*a^4 + a
^2*b^2*(6 - 2*n - n^2) + b^4*(3 - 4*n + n^2)))/b^6)*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*(1 + 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) + (b*Cos[c + d*x]^2*(a + b*Tan[c + d*x]
)^(1 + n)*(b^2*(3 - n) + a^2*(1 + n) + a*b*(5 + (3*a^2)/b^2 - 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 755

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(a*e + c*d*x)*
((a + c*x^2)^(p + 1)/(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*Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[
{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 837

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(d + e*x)^(
m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] +
Dist[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^
2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g},
x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 3587

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(b*f), Subst
[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && NeQ[a^2 + b
^2, 0] && IntegerQ[m/2]

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

Mathematica [A] (verified)

Time = 4.71 (sec) , antiderivative size = 360, normalized size of antiderivative = 0.83 \[ \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx=\frac {(a+b \tan (c+d x))^{1+n} \left (\frac {\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n+\sqrt {-b^2} \left (-3 a^4-b^4 \left (3-4 n+n^2\right )+a^2 b^2 \left (-6+2 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a-\sqrt {-b^2}}\right )}{a-\sqrt {-b^2}}+\frac {\left (a b^2 \left (3 a^2+b^2 (5-2 n)\right ) n+\sqrt {-b^2} \left (3 a^4+b^4 \left (3-4 n+n^2\right )-a^2 b^2 \left (-6+2 n+n^2\right )\right )\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {a+b \tan (c+d x)}{a+\sqrt {-b^2}}\right )}{a+\sqrt {-b^2}}}{\left (a^2+b^2\right ) (1+n)}+4 b \cos ^4(c+d x) (b+a \tan (c+d x))-\frac {2 b \cos ^2(c+d x) \left (b^3 (-3+n)-a^2 b (1+n)-a \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)\right )}{a^2+b^2}\right )}{16 b \left (a^2+b^2\right ) d} \]

[In]

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

[Out]

((a + b*Tan[c + d*x])^(1 + n)*((((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sqrt[-b^2]*(-3*a^4 - b^4*(3 - 4*n + n^2) +
a^2*b^2*(-6 + 2*n + n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a - Sqrt[-b^2])])/(a - Sq
rt[-b^2]) + ((a*b^2*(3*a^2 + b^2*(5 - 2*n))*n + Sqrt[-b^2]*(3*a^4 + b^4*(3 - 4*n + n^2) - a^2*b^2*(-6 + 2*n +
n^2)))*Hypergeometric2F1[1, 1 + n, 2 + n, (a + b*Tan[c + d*x])/(a + Sqrt[-b^2])])/(a + Sqrt[-b^2]))/((a^2 + b^
2)*(1 + n)) + 4*b*Cos[c + d*x]^4*(b + a*Tan[c + d*x]) - (2*b*Cos[c + d*x]^2*(b^3*(-3 + n) - a^2*b*(1 + n) - a*
(3*a^2 + b^2*(5 - 2*n))*Tan[c + d*x]))/(a^2 + b^2)))/(16*b*(a^2 + b^2)*d)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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