3.214 \(\int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=252 \[ \frac {e^5 (e \tan (c+d x))^{m-5} \, _2F_1\left (1,\frac {m-5}{2};\frac {m-3}{2};-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-2}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-4}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac {3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \]

[Out]

3*e^5*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)+e^5*hypergeom([1, -5/2+1/2*m],[-3/2+1/2*m],-tan(d*x+c)^2)*(e*tan(d*x+c
))^(-5+m)/a^3/d/(5-m)-3*e^5*(cos(d*x+c)^2)^(-2+1/2*m)*hypergeom([-2+1/2*m, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)
^2)*sec(d*x+c)*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)-e^5*(cos(d*x+c)^2)^(-1+1/2*m)*hypergeom([-1+1/2*m, -5/2+1/2*m
],[-3/2+1/2*m],sin(d*x+c)^2)*sec(d*x+c)^3*(e*tan(d*x+c))^(-5+m)/a^3/d/(5-m)

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Rubi [A]  time = 0.34, antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {e^5 (e \tan (c+d x))^{m-5} \, _2F_1\left (1,\frac {m-5}{2};\frac {m-3}{2};-\tan ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {e^5 \sec ^3(c+d x) \cos ^2(c+d x)^{\frac {m-2}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-2}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}-\frac {3 e^5 \sec (c+d x) \cos ^2(c+d x)^{\frac {m-4}{2}} (e \tan (c+d x))^{m-5} \, _2F_1\left (\frac {m-5}{2},\frac {m-4}{2};\frac {m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m)}+\frac {3 e^5 (e \tan (c+d x))^{m-5}}{a^3 d (5-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]

[Out]

(3*e^5*(e*Tan[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) + (e^5*Hypergeometric2F1[1, (-5 + m)/2, (-3 + m)/2, -Tan[c +
d*x]^2]*(e*Tan[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) - (3*e^5*(Cos[c + d*x]^2)^((-4 + m)/2)*Hypergeometric2F1[(
-5 + m)/2, (-4 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*Sec[c + d*x]*(e*Tan[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) - (
e^5*(Cos[c + d*x]^2)^((-2 + m)/2)*Hypergeometric2F1[(-5 + m)/2, (-2 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*Sec[c
+ d*x]^3*(e*Tan[c + d*x])^(-5 + m))/(a^3*d*(5 - 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))^3} \, dx &=\frac {e^6 \int (-a+a \sec (c+d x))^3 (e \tan (c+d x))^{-6+m} \, dx}{a^6}\\ &=\frac {e^6 \int \left (-a^3 (e \tan (c+d x))^{-6+m}+3 a^3 \sec (c+d x) (e \tan (c+d x))^{-6+m}-3 a^3 \sec ^2(c+d x) (e \tan (c+d x))^{-6+m}+a^3 \sec ^3(c+d x) (e \tan (c+d x))^{-6+m}\right ) \, dx}{a^6}\\ &=-\frac {e^6 \int (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {e^6 \int \sec ^3(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}+\frac {\left (3 e^6\right ) \int \sec (c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}-\frac {\left (3 e^6\right ) \int \sec ^2(c+d x) (e \tan (c+d x))^{-6+m} \, dx}{a^3}\\ &=-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {\left (3 e^6\right ) \operatorname {Subst}\left (\int (e x)^{-6+m} \, dx,x,\tan (c+d x)\right )}{a^3 d}-\frac {e^7 \operatorname {Subst}\left (\int \frac {x^{-6+m}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{a^3 d}\\ &=\frac {3 e^5 (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \, _2F_1\left (1,\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);-\tan ^2(c+d x)\right ) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {3 e^5 \cos ^2(c+d x)^{\frac {1}{2} (-4+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-4+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec (c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}-\frac {e^5 \cos ^2(c+d x)^{\frac {1}{2} (-2+m)} \, _2F_1\left (\frac {1}{2} (-5+m),\frac {1}{2} (-2+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) \sec ^3(c+d x) (e \tan (c+d x))^{-5+m}}{a^3 d (5-m)}\\ \end {align*}

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Mathematica [F]  time = 11.60, size = 0, normalized size = 0.00 \[ \int \frac {(e \tan (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]

[Out]

Integrate[(e*Tan[c + d*x])^m/(a + a*Sec[c + d*x])^3, x]

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \tan \left (d x + c\right )\right )^{m}}{a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((e*tan(d*x + c))^m/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)

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maple [F]  time = 1.85, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x)

[Out]

int((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \tan \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^m/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

integrate((e*tan(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^m/(a + a/cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^3*(e*tan(c + d*x))^m)/(a^3*(cos(c + d*x) + 1)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (e \tan {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))**m/(a+a*sec(d*x+c))**3,x)

[Out]

Integral((e*tan(c + d*x))**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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