3.14.22 \(\int \frac {(c (d \tan (e+f x))^p)^n}{(a+i a \tan (e+f x))^2} \, dx\) [1322]

Optimal. Leaf size=227 \[ \frac {\left (1-4 n p+2 n^2 p^2\right ) \, _2F_1(1,1+n p;2+n p;-i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac {\, _2F_1(1,1+n p;2+n p;i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))} \]

[Out]

1/8*(2*n^2*p^2-4*n*p+1)*hypergeom([1, n*p+1],[n*p+2],-I*tan(f*x+e))*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/a^2/f/(n
*p+1)+1/8*hypergeom([1, n*p+1],[n*p+2],I*tan(f*x+e))*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/a^2/f/(n*p+1)+1/4*tan(f
*x+e)*(c*(d*tan(f*x+e))^p)^n/a^2/f/(1+I*tan(f*x+e))^2+1/4*(-n*p+2)*tan(f*x+e)*(c*(d*tan(f*x+e))^p)^n/a^2/f/(1+
I*tan(f*x+e))

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Rubi [A]
time = 0.23, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1970, 862, 105, 156, 162, 66} \begin {gather*} \frac {\left (2 n^2 p^2-4 n p+1\right ) \tan (e+f x) \, _2F_1(1,n p+1;n p+2;-i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac {\tan (e+f x) \, _2F_1(1,n p+1;n p+2;i \tan (e+f x)) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (n p+1)}+\frac {(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}+\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(d*Tan[e + f*x])^p)^n/(a + I*a*Tan[e + f*x])^2,x]

[Out]

((1 - 4*n*p + 2*n^2*p^2)*Hypergeometric2F1[1, 1 + n*p, 2 + n*p, (-I)*Tan[e + f*x]]*Tan[e + f*x]*(c*(d*Tan[e +
f*x])^p)^n)/(8*a^2*f*(1 + n*p)) + (Hypergeometric2F1[1, 1 + n*p, 2 + n*p, I*Tan[e + f*x]]*Tan[e + f*x]*(c*(d*T
an[e + f*x])^p)^n)/(8*a^2*f*(1 + n*p)) + (Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n)/(4*a^2*f*(1 + I*Tan[e + f*x])
^2) + ((2 - n*p)*Tan[e + f*x]*(c*(d*Tan[e + f*x])^p)^n)/(4*a^2*f*(1 + I*Tan[e + f*x]))

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 105

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

Rule 156

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f
))), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 1970

Int[(u_.)*((c_.)*((d_)*((a_.) + (b_.)*(x_)))^(q_))^(p_), x_Symbol] :> Dist[(c*(d*(a + b*x))^q)^p/(a + b*x)^(p*
q), Int[u*(a + b*x)^(p*q), x], x] /; FreeQ[{a, b, c, d, q, p}, x] &&  !IntegerQ[q] &&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\left (c (d \tan (e+f x))^p\right )^n}{(a+i a \tan (e+f x))^2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (c (d x)^p\right )^n}{(a+i a x)^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p}}{(a+i a x)^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p}}{\left (\frac {1}{a}-\frac {i x}{a}\right ) (a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}-\frac {\left (i (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p} (i d (3-n p)+d (1-n p) x)}{\left (\frac {1}{a}-\frac {i x}{a}\right ) (a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{4 a d f}\\ &=\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}-\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p} \left (-2 d^2 (1-n p)^2-2 i d^2 n p (2-n p) x\right )}{\left (\frac {1}{a}-\frac {i x}{a}\right ) (a+i a x)} \, dx,x,\tan (e+f x)\right )}{8 a^2 d^2 f}\\ &=\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}+\frac {\left ((d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p}}{\frac {1}{a}-\frac {i x}{a}} \, dx,x,\tan (e+f x)\right )}{8 a^3 f}+\frac {\left (\left (1-4 n p+2 n^2 p^2\right ) (d \tan (e+f x))^{-n p} \left (c (d \tan (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {(d x)^{n p}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{8 a f}\\ &=\frac {\left (1-4 n p+2 n^2 p^2\right ) \, _2F_1(1,1+n p;2+n p;-i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac {\, _2F_1(1,1+n p;2+n p;i \tan (e+f x)) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{8 a^2 f (1+n p)}+\frac {\tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(2-n p) \tan (e+f x) \left (c (d \tan (e+f x))^p\right )^n}{4 a^2 f (1+i \tan (e+f x))}\\ \end {align*}

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Mathematica [F]
time = 6.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c (d \tan (e+f x))^p\right )^n}{(a+i a \tan (e+f x))^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(c*(d*Tan[e + f*x])^p)^n/(a + I*a*Tan[e + f*x])^2,x]

[Out]

Integrate[(c*(d*Tan[e + f*x])^p)^n/(a + I*a*Tan[e + f*x])^2, x]

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \frac {\left (c \left (d \tan \left (f x +e \right )\right )^{p}\right )^{n}}{\left (a +i a \tan \left (f x +e \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*tan(f*x+e))^p)^n/(a+I*a*tan(f*x+e))^2,x)

[Out]

int((c*(d*tan(f*x+e))^p)^n/(a+I*a*tan(f*x+e))^2,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((c*(d*tan(f*x+e))^p)^n/(a+I*a*tan(f*x+e))^2,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((c*(d*tan(f*x+e))^p)^n/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

integral(1/4*(e^(4*I*f*x + 4*I*e) + 2*e^(2*I*f*x + 2*I*e) + 1)*e^(n*p*log((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^
(2*I*f*x + 2*I*e) + 1)) - 4*I*f*x + n*log(c) - 4*I*e)/a^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {\left (c \left (d \tan {\left (e + f x \right )}\right )^{p}\right )^{n}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*tan(f*x+e))**p)**n/(a+I*a*tan(f*x+e))**2,x)

[Out]

-Integral((c*(d*tan(e + f*x))**p)**n/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x)/a**2

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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((c*(d*tan(f*x+e))^p)^n/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate(((d*tan(f*x + e))^p*c)^n/(I*a*tan(f*x + e) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^p\right )}^n}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*tan(e + f*x))^p)^n/(a + a*tan(e + f*x)*1i)^2,x)

[Out]

int((c*(d*tan(e + f*x))^p)^n/(a + a*tan(e + f*x)*1i)^2, x)

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