3.4.16 \(\int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx\) [316]

Optimal. Leaf size=274 \[ \frac {(1-2 n) (1-n) (3-n) \, _2F_1\left (1,\frac {1+n}{2};\frac {3+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{1+n}}{24 a^3 d f (1+n)}+\frac {i (5-2 n) (2-n) n \, _2F_1\left (1,\frac {2+n}{2};\frac {4+n}{2};-\tan ^2(e+f x)\right ) (d \tan (e+f x))^{2+n}}{24 a^3 d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )} \]

[Out]

1/24*(1-2*n)*(1-n)*(3-n)*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^3/d/f/(1+n
)+1/24*I*(5-2*n)*(2-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^3/d^2/f/(2+n)+
1/6*(d*tan(f*x+e))^(1+n)/d/f/(a+I*a*tan(f*x+e))^3+1/24*(7-2*n)*(d*tan(f*x+e))^(1+n)/a/d/f/(a+I*a*tan(f*x+e))^2
+1/24*(5-2*n)*(2-n)*(d*tan(f*x+e))^(1+n)/d/f/(a^3+I*a^3*tan(f*x+e))

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Rubi [A]
time = 0.46, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3640, 3677, 3619, 3557, 371} \begin {gather*} \frac {i (5-2 n) (2-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 )}{24 a^3 d^2 f (n+2)}+\frac {(1-2 n) (1-n) (3-n) (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 )}{24 a^3 d f (n+1)}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{n+1}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {(7-2 n) (d \tan (e+f x))^{n+1}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(d \tan (e+f x))^{n+1}}{6 d f (a+i a \tan (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d*Tan[e + f*x])^n/(a + I*a*Tan[e + f*x])^3,x]

[Out]

((1 - 2*n)*(1 - n)*(3 - n)*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, -Tan[e + f*x]^2]*(d*Tan[e + f*x])^(1 + n
))/(24*a^3*d*f*(1 + n)) + ((I/24)*(5 - 2*n)*(2 - n)*n*Hypergeometric2F1[1, (2 + n)/2, (4 + n)/2, -Tan[e + f*x]
^2]*(d*Tan[e + f*x])^(2 + n))/(a^3*d^2*f*(2 + n)) + (d*Tan[e + f*x])^(1 + n)/(6*d*f*(a + I*a*Tan[e + f*x])^3)
+ ((7 - 2*n)*(d*Tan[e + f*x])^(1 + n))/(24*a*d*f*(a + I*a*Tan[e + f*x])^2) + ((5 - 2*n)*(2 - n)*(d*Tan[e + f*x
])^(1 + n))/(24*d*f*(a^3 + I*a^3*Tan[e + f*x]))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 3557

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 3619

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 3640

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 3677

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))^3} \, dx &=\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {\int \frac {(d \tan (e+f x))^n (a d (5-n)-i a d (2-n) \tan (e+f x))}{(a+i a \tan (e+f x))^2} \, dx}{6 a^2 d}\\ &=\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {\int \frac {(d \tan (e+f x))^n \left (a^2 d^2 \left (13-9 n+2 n^2\right )-i a^2 d^2 (7-2 n) (1-n) \tan (e+f x)\right )}{a+i a \tan (e+f x)} \, dx}{24 a^4 d^2}\\ &=\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {\int (d \tan (e+f x))^n \left (2 a^3 d^3 (1-2 n) (1-n) (3-n)+2 i a^3 d^3 (5-2 n) (2-n) n \tan (e+f x)\right ) \, dx}{48 a^6 d^3}\\ &=\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {((1-2 n) (1-n) (3-n)) \int (d \tan (e+f x))^n \, dx}{24 a^3}+\frac {(i (5-2 n) (2-n) n) \int (d \tan (e+f x))^{1+n} \, dx}{24 a^3 d}\\ &=\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )}+\frac {(d (1-2 n) (1-n) (3-n)) \text {Subst}\left (\int \frac {x^n}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{24 a^3 f}+\frac {(i (5-2 n) (2-n) n) \text {Subst}\left (\int \frac {x^{1+n}}{d^2+x^2} \, dx,x,d \tan (e+f x)\right )}{24 a^3 f}\\ &=\frac {(1-2 n) (1-n) (3-n) \, _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}}{24 a^3 d f (1+n)}+\frac {i (5-2 n) (2-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}}{24 a^3 d^2 f (2+n)}+\frac {(d \tan (e+f x))^{1+n}}{6 d f (a+i a \tan (e+f x))^3}+\frac {(7-2 n) (d \tan (e+f x))^{1+n}}{24 a d f (a+i a \tan (e+f x))^2}+\frac {(5-2 n) (2-n) (d \tan (e+f x))^{1+n}}{24 d f \left (a^3+i a^3 \tan (e+f x)\right )}\\ \end {align*}

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Mathematica [F]
time = 16.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d \tan (e+f x))^n}{(a+i a \tan (e+f x))^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(d*Tan[e + f*x])^n/(a + I*a*Tan[e + f*x])^3,x]

[Out]

Integrate[(d*Tan[e + f*x])^n/(a + I*a*Tan[e + f*x])^3, x]

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Maple [F]
time = 1.22, size = 0, normalized size = 0.00 \[\int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{3}}\, dx\]

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))^3,x)

[Out]

int((d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^3,x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^3,x, algorithm="fricas")

[Out]

integral(1/8*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*(e^(6*I*f*x + 6*I*e) + 3*e^(4*I*f*
x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1)*e^(-6*I*f*x - 6*I*e)/a^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{n}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx}{a^{3}} \end {gather*}

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))**3,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^3,x, algorithm="giac")

[Out]

integrate((d*tan(f*x + e))^n/(I*a*tan(f*x + e) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 )}^3} \,d x \end {gather*}

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)^3,x)

[Out]

int((d*tan(e + f*x))^n/(a + a*tan(e + f*x)*1i)^3, x)

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