3.317 \(\int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^4} \, dx\)

Optimal. Leaf size=326 \[ \frac {i (2-n)^2 (4-n) n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{48 a^4 d^2 f (n+2)}+\frac {(1-n) (3-n) \left (n^2-4 n+1\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{48 a^4 d f (n+1)}+\frac {\left (n^2-7 n+13\right ) (d \tan (e+f x))^{n+1}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{n+1}}{48 a^4 d f (1+i \tan (e+f x))}+\frac {(5-n) (d \tan (e+f x))^{n+1}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(d \tan (e+f x))^{n+1}}{8 d f (a+i a \tan (e+f x))^4} \]

[Out]

1/48*(1-n)*(3-n)*(n^2-4*n+1)*hypergeom([1, 1/2+1/2*n],[3/2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e))^(1+n)/a^4/d/f/
(1+n)+1/48*(n^2-7*n+13)*(d*tan(f*x+e))^(1+n)/a^4/d/f/(1+I*tan(f*x+e))^2+1/48*(2-n)^2*(4-n)*(d*tan(f*x+e))^(1+n
)/a^4/d/f/(1+I*tan(f*x+e))+1/48*I*(2-n)^2*(4-n)*n*hypergeom([1, 1+1/2*n],[2+1/2*n],-tan(f*x+e)^2)*(d*tan(f*x+e
))^(2+n)/a^4/d^2/f/(2+n)+1/8*(d*tan(f*x+e))^(1+n)/d/f/(a+I*a*tan(f*x+e))^4+1/24*(5-n)*(d*tan(f*x+e))^(1+n)/a/d
/f/(a+I*a*tan(f*x+e))^3

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Rubi [A]  time = 0.97, antiderivative size = 326, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3559, 3596, 3538, 3476, 364} \[ \frac {i (2-n)^2 (4-n) n (d \tan (e+f x))^{n+2} \, _2F_1\left (1,\frac {n+2}{2};\frac {n+4}{2};-\tan ^2(e+f x)\right )}{48 a^4 d^2 f (n+2)}+\frac {(1-n) (3-n) \left (n^2-4 n+1\right ) (d \tan (e+f x))^{n+1} \, _2F_1\left (1,\frac {n+1}{2};\frac {n+3}{2};-\tan ^2(e+f x)\right )}{48 a^4 d f (n+1)}+\frac {\left (n^2-7 n+13\right ) (d \tan (e+f x))^{n+1}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{n+1}}{48 a^4 d f (1+i \tan (e+f x))}+\frac {(5-n) (d \tan (e+f x))^{n+1}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(d \tan (e+f x))^{n+1}}{8 d f (a+i a \tan (e+f x))^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - n)*(3 - n)*(1 - 4*n + n^2)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^
(1 + n))/(48*a^4*d*f*(1 + n)) + ((13 - 7*n + n^2)*(d*Tan[e + f*x])^(1 + n))/(48*a^4*d*f*(1 + I*Tan[e + f*x])^2
) + ((2 - n)^2*(4 - n)*(d*Tan[e + f*x])^(1 + n))/(48*a^4*d*f*(1 + I*Tan[e + f*x])) + ((I/48)*(2 - n)^2*(4 - n)
*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(2 + n))/(a^4*d^2*f*(2 + n)) +
(d*Tan[e + f*x])^(1 + n)/(8*d*f*(a + I*a*Tan[e + f*x])^4) + ((5 - n)*(d*Tan[e + f*x])^(1 + n))/(24*a*d*f*(a +
I*a*Tan[e + f*x])^3)

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])

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 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 3559

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

Rule 3596

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

Rubi steps

\begin {align*} \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^4} \, dx &=\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {\int \frac {(d \tan (e+f x))^n (a d (7-n)-i a d (3-n) \tan (e+f x))}{(a+i a \tan (e+f x))^3} \, dx}{8 a^2 d}\\ &=\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {\int \frac {(d \tan (e+f x))^n \left (2 a^2 d^2 \left (16-7 n+n^2\right )-2 i a^2 d^2 (2-n) (5-n) \tan (e+f x)\right )}{(a+i a \tan (e+f x))^2} \, dx}{48 a^4 d^2}\\ &=\frac {\left (13-7 n+n^2\right ) (d \tan (e+f x))^{1+n}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {\int \frac {(d \tan (e+f x))^n \left (4 a^3 d^3 \left (19-20 n+8 n^2-n^3\right )-4 i a^3 d^3 (1-n) \left (13-7 n+n^2\right ) \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{192 a^6 d^3}\\ &=\frac {\left (13-7 n+n^2\right ) (d \tan (e+f x))^{1+n}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{1+n}}{48 d f \left (a^4+i a^4 \tan (e+f x)\right )}+\frac {\int (d \tan (e+f x))^n \left (8 a^4 d^4 (1-n) (3-n) \left (1-4 n+n^2\right )+8 i a^4 d^4 (2-n)^2 (4-n) n \tan (e+f x)\right ) \, dx}{384 a^8 d^4}\\ &=\frac {\left (13-7 n+n^2\right ) (d \tan (e+f x))^{1+n}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{1+n}}{48 d f \left (a^4+i a^4 \tan (e+f x)\right )}+\frac {\left (i (2-n)^2 (4-n) n\right ) \int (d \tan (e+f x))^{1+n} \, dx}{48 a^4 d}+\frac {\left ((1-n) (3-n) \left (1-4 n+n^2\right )\right ) \int (d \tan (e+f x))^n \, dx}{48 a^4}\\ &=\frac {\left (13-7 n+n^2\right ) (d \tan (e+f x))^{1+n}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{1+n}}{48 d f \left (a^4+i a^4 \tan (e+f x)\right )}+\frac {\left (i (2-n)^2 (4-n) n\right ) \operatorname {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{48 a^4 f}+\frac {\left (d (1-n) (3-n) \left (1-4 n+n^2\right )\right ) \operatorname {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{48 a^4 f}\\ &=\frac {(1-n) (3-n) \left (1-4 n+n^2\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}}{48 a^4 d f (1+n)}+\frac {\left (13-7 n+n^2\right ) (d \tan (e+f x))^{1+n}}{48 a^4 d f (1+i \tan (e+f x))^2}+\frac {i (2-n)^2 (4-n) n \, _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}}{48 a^4 d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{8 d f (a+i a \tan (e+f x))^4}+\frac {(5-n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^3}+\frac {(2-n)^2 (4-n) (d \tan (e+f x))^{1+n}}{48 d f \left (a^4+i a^4 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [F]  time = 28.97, size = 0, normalized size = 0.00 \[ \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^4} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} {\left (e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-8 i \, f x - 8 i \, e\right )}}{16 \, a^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/16*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*(e^(8*I*f*x + 8*I*e) + 4*e^(6*I*f
*x + 6*I*e) + 6*e^(4*I*f*x + 4*I*e) + 4*e^(2*I*f*x + 2*I*e) + 1)*e^(-8*I*f*x - 8*I*e)/a^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(e + f*x))^n/(a + a*tan(e + f*x)*1i)^4,x)

[Out]

int((d*tan(e + f*x))^n/(a + a*tan(e + f*x)*1i)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{n}}{\tan ^{4}{\left (e + f x \right )} - 4 i \tan ^{3}{\left (e + f x \right )} - 6 \tan ^{2}{\left (e + f x \right )} + 4 i \tan {\left (e + f x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(e + f*x))**n/(tan(e + f*x)**4 - 4*I*tan(e + f*x)**3 - 6*tan(e + f*x)**2 + 4*I*tan(e + f*x) + 1
), x)/a**4

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