3.213 \(\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=169 \[ -\frac {e^3 (e \tan (c+d x))^{m-3} \, _2F_1\left (1,\frac {m-3}{2};\frac {m-1}{2};-\tan ^2(c+d x)\right )}{a^2 d (3-m)}+\frac {2 e^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-3} \, _2F_1\left (\frac {m-3}{2},\frac {m-2}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m)}-\frac {e^3 (e \tan (c+d x))^{m-3}}{a^2 d (3-m)} \]

[Out]

-e^3*(e*tan(d*x+c))^(-3+m)/a^2/d/(3-m)-e^3*hypergeom([1, -3/2+1/2*m],[-1/2+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c)
)^(-3+m)/a^2/d/(3-m)+2*e^3*(cos(d*x+c)^2)^(-1+1/2*m)*hypergeom([-1+1/2*m, -3/2+1/2*m],[-1/2+1/2*m],sin(d*x+c)^
2)*sec(d*x+c)*(e*tan(d*x+c))^(-3+m)/a^2/d/(3-m)

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Rubi [A]  time = 0.27, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3888, 3886, 3476, 364, 2617, 2607, 32} \[ -\frac {e^3 (e \tan (c+d x))^{m-3} \, _2F_1\left (1,\frac {m-3}{2};\frac {m-1}{2};-\tan ^2(c+d x)\right )}{a^2 d (3-m)}+\frac {2 e^3 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-3} \, _2F_1\left (\frac {m-3}{2},\frac {m-2}{2};\frac {m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m)}-\frac {e^3 (e \tan (c+d x))^{m-3}}{a^2 d (3-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]

[Out]

-((e^3*(e*Tan[c + d*x])^(-3 + m))/(a^2*d*(3 - m))) - (e^3*Hypergeometric2F1[1, (-3 + m)/2, (-1 + m)/2, -Tan[c
+ d*x]^2]*(e*Tan[c + d*x])^(-3 + m))/(a^2*d*(3 - m)) + (2*e^3*(Cos[c + d*x]^2)^((-2 + m)/2)*Hypergeometric2F1[
(-3 + m)/2, (-2 + m)/2, (-1 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[c + d*x])^(-3 + m))/(a^2*d*(3 - m))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2617

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*Sec[e +
f*x])^m*(b*Tan[e + f*x])^(n + 1)*(Cos[e + f*x]^2)^((m + n + 1)/2)*Hypergeometric2F1[(n + 1)/2, (m + n + 1)/2,
(n + 3)/2, Sin[e + f*x]^2])/(b*f*(n + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !IntegerQ[(n - 1)/2] &&  !In
tegerQ[m/2]

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 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx &=\frac {e^4 \int (-a+a \sec (c+d x))^2 (e \tan (c+d x))^{-4+m} \, dx}{a^4}\\ &=\frac {e^4 \int \left (a^2 (e \tan (c+d x))^{-4+m}-2 a^2 \sec (c+d x) (e \tan (c+d x))^{-4+m}+a^2 \sec ^2(c+d x) (e \tan (c+d x))^{-4+m}\right ) \, dx}{a^4}\\ &=\frac {e^4 \int (e \tan (c+d x))^{-4+m} \, dx}{a^2}+\frac {e^4 \int \sec ^2(c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2}-\frac {\left (2 e^4\right ) \int \sec (c+d x) (e \tan (c+d x))^{-4+m} \, dx}{a^2}\\ &=\frac {2 e^3 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-3+m),\frac {1}{2} (-2+m);\frac {1}{2} (-1+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}+\frac {e^4 \operatorname {Subst}\left (\int (e x)^{-4+m} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac {e^5 \operatorname {Subst}\left (\int \frac {x^{-4+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^2 d}\\ &=-\frac {e^3 (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac {e^3 \, _2F_1\left (1,\frac {1}{2} (-3+m);\frac {1}{2} (-1+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}+\frac {2 e^3 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-3+m),\frac {1}{2} (-2+m);\frac {1}{2} (-1+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-3+m}}{a^2 d (3-m)}\\ \end {align*}

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Mathematica [C]  time = 6.92, size = 329, normalized size = 1.95 \[ \frac {\tan \left (\frac {1}{2} (c+d x)\right ) (e \tan (c+d x))^m \left ((m+1) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \, _2F_1\left (m,\frac {m+3}{2};\frac {m+5}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-3 (m+3) \, _2F_1\left (m,\frac {m+1}{2};\frac {m+3}{2};\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^m}{2 a^2 d (m+1) (m+3)}+\frac {i 2^{1-m} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^2(c+d x) \left (2^m \, _2F_1\left (1,m;m+1;-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-\left (1+e^{2 i (c+d x)}\right )^m \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right ) \tan ^{-m}(c+d x) (e \tan (c+d x))^m}{d m (a \sec (c+d x)+a)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]

[Out]

((Cos[c + d*x]*Sec[(c + d*x)/2]^2)^m*Tan[(c + d*x)/2]*(-3*(3 + m)*Hypergeometric2F1[m, (1 + m)/2, (3 + m)/2, T
an[(c + d*x)/2]^2] + (1 + m)*Hypergeometric2F1[m, (3 + m)/2, (5 + m)/2, Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2
)*(e*Tan[c + d*x])^m)/(2*a^2*d*(1 + m)*(3 + m)) + (I*2^(1 - m)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I
)*(c + d*x))))^m*Cos[c/2 + (d*x)/2]^4*(2^m*Hypergeometric2F1[1, m, 1 + m, -((-1 + E^((2*I)*(c + d*x)))/(1 + E^
((2*I)*(c + d*x))))] - (1 + E^((2*I)*(c + d*x)))^m*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2]
)*Sec[c + d*x]^2*(e*Tan[c + d*x])^m)/(d*m*(a + a*Sec[c + d*x])^2*Tan[c + d*x]^m)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

integral((e*tan(d*x + c))^m/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^2, x)

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maple [F]  time = 1.72, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x)

[Out]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^2\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e*tan(c + d*x))^m)/(a^2*(cos(c + d*x) + 1)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c))**2,x)

[Out]

Integral((e*tan(c + d*x))**m/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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