3.210 \(\int (a+a \sec (c+d x))^2 (e \tan (c+d x))^m \, dx\)

Optimal. Leaf size=161 \[ \frac {a^2 (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+2}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {a^2 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3886, 3476, 364, 2617, 2607, 32} \[ \frac {a^2 (e \tan (c+d x))^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{d e (m+1)}+\frac {2 a^2 \sec (c+d x) \cos ^2(c+d x)^{\frac {m+2}{2}} (e \tan (c+d x))^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {m+2}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac {a^2 (e \tan (c+d x))^{m+1}}{d e (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 3.20, size = 358, normalized size = 2.22 \[ \frac {a^2 (\cos (c+d x)+1)^2 \csc (c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\tan ^2(c+d x)\right )^{\frac {1-m}{2}} (e \tan (c+d x))^m \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3}{2};\sec ^2(c+d x)\right )}{2 d}+\frac {a^2 2^{-m-3} (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tan ^{-m}(c+d x) (e \tan (c+d x))^m \left (i 2^m (m+1) \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos (c+d x) \, _2F_1\left (1,m;m+1;-\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-i (m+1) \left (1+e^{2 i (c+d x)}\right )^m \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \cos (c+d x) \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+2^{m+1} m \sin (c+d x) \tan ^m(c+d x)\right )}{d m (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}\right )} \left (e \tan \left (d x + c\right )\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.37, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{2} \left (e \tan \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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