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

Optimal. Leaf size=434 \[ \frac{b \left (\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}-\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}+\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}+\frac{b \cos ^2(c+d x) \left (a b \left (\frac{3 a^2}{b^2}-2 n+5\right ) \tan (c+d x)+a^2 (n+1)+b^2 (3-n)\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2} \]

[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)

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Rubi [A]  time = 0.686767, antiderivative size = 434, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3506, 741, 823, 831, 68} \[ \frac{b \left (\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}-\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a-\sqrt{-b^2}\right )}+\frac{b \left (\frac{\sqrt{-b^2} \left (a^2 b^2 \left (-n^2-2 n+6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )}{b^6}+\frac{a n \left (\frac{3 a^2}{b^2}-2 n+5\right )}{b^2}\right ) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{16 d (n+1) \left (\frac{a^2}{b^2}+1\right )^2 \left (a+\sqrt{-b^2}\right )}+\frac{\cos ^4(c+d x) (a \tan (c+d x)+b) (a+b \tan (c+d x))^{n+1}}{4 d \left (a^2+b^2\right )}+\frac{b \cos ^2(c+d x) \left (a b \left (\frac{3 a^2}{b^2}-2 n+5\right ) \tan (c+d x)+a^2 (n+1)+b^2 (3-n)\right ) (a+b \tan (c+d x))^{n+1}}{8 d \left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[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 3506

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]

Rule 741

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 823

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] + Di
st[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 831

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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac{\operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \, _2F_1\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 ) \, _2F_1\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]  time = 4.34034, size = 360, normalized size = 0.83 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (\frac{\frac{\left (\sqrt{-b^2} \left (a^2 b^2 \left (n^2+2 n-6\right )-3 a^4-b^4 \left (n^2-4 n+3\right )\right )+a b^2 n \left (3 a^2+b^2 (5-2 n)\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-\sqrt{-b^2}}\right )}{a-\sqrt{-b^2}}+\frac{\left (\sqrt{-b^2} \left (-a^2 b^2 \left (n^2+2 n-6\right )+3 a^4+b^4 \left (n^2-4 n+3\right )\right )+a b^2 n \left (3 a^2+b^2 (5-2 n)\right )\right ) \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+\sqrt{-b^2}}\right )}{a+\sqrt{-b^2}}}{(n+1) \left (a^2+b^2\right )}-\frac{2 b \cos ^2(c+d x) \left (-a \left (3 a^2+b^2 (5-2 n)\right ) \tan (c+d x)-a^2 b (n+1)+b^3 (n-3)\right )}{a^2+b^2}+4 b \cos ^4(c+d x) (a \tan (c+d x)+b)\right )}{16 b d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{4} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)