3.696 \(\int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx\)

Optimal. Leaf size=261 \[ \frac{4 a b \left (a^2-b^2\right ) (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{\left (-6 a^2 b^2+a^4+b^4\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}-\frac{b^2 \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+3)}+\frac{b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}+\frac{2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)} \]

[Out]

-((b^2*(b^2*(3 + n) - a^2*(17 + 5*n))*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(3 + n))) + ((a^4 - 6*a^2*b^2 + b
^4)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (2*a
*b^3*(4 + n)*Tan[e + f*x]*(d*Tan[e + f*x])^(1 + n))/(d*f*(2 + n)*(3 + n)) + (4*a*b*(a^2 - b^2)*Hypergeometric2
F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(d^2*f*(2 + n)) + (b^2*(d*Tan[e + f*x])
^(1 + n)*(a + b*Tan[e + f*x])^2)/(d*f*(3 + n))

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Rubi [A]  time = 0.640625, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3566, 3637, 3630, 3538, 3476, 364} \[ \frac{4 a b \left (a^2-b^2\right ) (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{d^2 f (n+2)}+\frac{\left (-6 a^2 b^2+a^4+b^4\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{d f (n+1)}-\frac{b^2 \left (b^2 (n+3)-a^2 (5 n+17)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) (n+3)}+\frac{b^2 (a+b \tan (e+f x))^2 (d \tan (e+f x))^{n+1}}{d f (n+3)}+\frac{2 a b^3 (n+4) \tan (e+f x) (d \tan (e+f x))^{n+1}}{d f (n+2) (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(b^2*(3 + n) - a^2*(17 + 5*n))*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)*(3 + n))) + ((a^4 - 6*a^2*b^2 + b
^4)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n))/(d*f*(1 + n)) + (2*a
*b^3*(4 + n)*Tan[e + f*x]*(d*Tan[e + f*x])^(1 + n))/(d*f*(2 + n)*(3 + n)) + (4*a*b*(a^2 - b^2)*Hypergeometric2
F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(d^2*f*(2 + n)) + (b^2*(d*Tan[e + f*x])
^(1 + n)*(a + b*Tan[e + f*x])^2)/(d*f*(3 + n))

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3637

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(b*C*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] &&  !LtQ[n, -1]

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rule 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int (d \tan (e+f x))^n (a+b \tan (e+f x))^4 \, dx &=\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}+\frac{\int (d \tan (e+f x))^n (a+b \tan (e+f x)) \left (-a d \left (b^2 (1+n)-a^2 (3+n)\right )+b \left (3 a^2-b^2\right ) d (3+n) \tan (e+f x)+2 a b^2 d (4+n) \tan ^2(e+f x)\right ) \, dx}{d (3+n)}\\ &=\frac{2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\frac{\int (d \tan (e+f x))^n \left (a^2 d^2 (2+n) \left (b^2 (1+n)-a^2 (3+n)\right )-4 a b \left (a^2-b^2\right ) d^2 (2+n) (3+n) \tan (e+f x)+b^2 d^2 (2+n) \left (b^2 (3+n)-a^2 (17+5 n)\right ) \tan ^2(e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac{b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac{2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\frac{\int (d \tan (e+f x))^n \left (-\left (a^4-6 a^2 b^2+b^4\right ) d^2 (2+n) (3+n)-4 a b \left (a^2-b^2\right ) d^2 (2+n) (3+n) \tan (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac{b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac{2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}-\left (-a^4+6 a^2 b^2-b^4\right ) \int (d \tan (e+f x))^n \, dx+\frac{\left (4 a b \left (a^2-b^2\right )\right ) \int (d \tan (e+f x))^{1+n} \, dx}{d}\\ &=-\frac{b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac{2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}+\frac{\left (4 a b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}+\frac{\left (\left (a^4-6 a^2 b^2+b^4\right ) d\right ) \operatorname{Subst}\left (\int \frac{x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{f}\\ &=-\frac{b^2 \left (b^2 (3+n)-a^2 (17+5 n)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n) (3+n)}+\frac{\left (a^4-6 a^2 b^2+b^4\right ) \, _2F_1\left (1,\frac{1+n}{2};\frac{3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{d f (1+n)}+\frac{2 a b^3 (4+n) \tan (e+f x) (d \tan (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{4 a b \left (a^2-b^2\right ) \, _2F_1\left (1,\frac{2+n}{2};\frac{4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{d^2 f (2+n)}+\frac{b^2 (d \tan (e+f x))^{1+n} (a+b \tan (e+f x))^2}{d f (3+n)}\\ \end{align*}

Mathematica [A]  time = 1.51486, size = 191, normalized size = 0.73 \[ \frac{\tan (e+f x) (d \tan (e+f x))^n \left (\frac{(n+3) \left (-6 a^2 b^2+a^4+b^4\right ) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};-\tan ^2(e+f x)\right )}{n+1}+\frac{a^2 b^2 (5 n+17)-b^4 (n+3)}{n+1}+\frac{2 a b^3 (n+4) \tan (e+f x)}{n+2}+b^2 (a+b \tan (e+f x))^2+\frac{4 a b (n+3) (a-b) (a+b) \tan (e+f x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};-\tan ^2(e+f x)\right )}{n+2}\right )}{f (n+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Tan[e + f*x]*(d*Tan[e + f*x])^n*((-(b^4*(3 + n)) + a^2*b^2*(17 + 5*n))/(1 + n) + ((a^4 - 6*a^2*b^2 + b^4)*(3
+ n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2])/(1 + n) + (2*a*b^3*(4 + n)*Tan[e + f*x])/(2
+ n) + (4*a*(a - b)*b*(a + b)*(3 + n)*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*Tan[e + f*x]
)/(2 + n) + b^2*(a + b*Tan[e + f*x])^2))/(f*(3 + n))

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Maple [F]  time = 0.313, size = 0, normalized size = 0. \begin{align*} \int \left ( d\tan \left ( fx+e \right ) \right ) ^{n} \left ( a+b\tan \left ( fx+e \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x)

[Out]

int((d*tan(f*x+e))^n*(a+b*tan(f*x+e))^4,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^4*(d*tan(f*x + e))^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{4} \tan \left (f x + e\right )^{4} + 4 \, a b^{3} \tan \left (f x + e\right )^{3} + 6 \, a^{2} b^{2} \tan \left (f x + e\right )^{2} + 4 \, a^{3} b \tan \left (f x + e\right ) + a^{4}\right )} \left (d \tan \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^4*tan(f*x + e)^4 + 4*a*b^3*tan(f*x + e)^3 + 6*a^2*b^2*tan(f*x + e)^2 + 4*a^3*b*tan(f*x + e) + a^4)
*(d*tan(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))**n*(a+b*tan(f*x+e))**4,x)

[Out]

Integral((d*tan(e + f*x))**n*(a + b*tan(e + f*x))**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*tan(f*x + e) + a)^4*(d*tan(f*x + e))^n, x)