3.341 \(\int \sec ^4(e+f x) (a+a \sec (e+f x))^m \, dx\)

Optimal. Leaf size=211 \[ \frac {2^{m+\frac {1}{2}} m \left (m^2+3 m+5\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {m \tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f \left (m^2+5 m+6\right )}+\frac {\tan (e+f x) \sec ^2(e+f x) (a \sec (e+f x)+a)^m}{f (m+3)}+\frac {(m+4) \tan (e+f x) (a \sec (e+f x)+a)^m}{f (m+1) (m+2) (m+3)} \]

[Out]

(4+m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(m^3+6*m^2+11*m+6)+sec(f*x+e)^2*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(3+m)+2^
(1/2+m)*m*(m^2+3*m+5)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sec(f*x+e))*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e)
)^m*tan(f*x+e)/f/(m^3+6*m^2+11*m+6)+m*(a+a*sec(f*x+e))^(1+m)*tan(f*x+e)/a/f/(m^2+5*m+6)

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Rubi [A]  time = 0.35, antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3824, 4010, 4001, 3828, 3827, 69} \[ \frac {2^{m+\frac {1}{2}} m \left (m^2+3 m+5\right ) \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f (m+1) (m+2) (m+3)}+\frac {m \tan (e+f x) (a \sec (e+f x)+a)^{m+1}}{a f \left (m^2+5 m+6\right )}+\frac {\tan (e+f x) \sec ^2(e+f x) (a \sec (e+f x)+a)^m}{f (m+3)}+\frac {(m+4) \tan (e+f x) (a \sec (e+f x)+a)^m}{f (m+1) (m+2) (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(a + a*Sec[e + f*x])^m,x]

[Out]

((4 + m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + m)*(2 + m)*(3 + m)) + (Sec[e + f*x]^2*(a + a*Sec[e + f*x
])^m*Tan[e + f*x])/(f*(3 + m)) + (2^(1/2 + m)*m*(5 + 3*m + m^2)*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sec[
e + f*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + m)*(2 + m)*(3 + m)) +
(m*(a + a*Sec[e + f*x])^(1 + m)*Tan[e + f*x])/(a*f*(6 + 5*m + m^2))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 3824

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(d^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2))/(f*(m + n - 1)), x] + Dist[d^2/(b*(m + n - 1)),
Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*m*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d,
e, f, m}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 2] && NeQ[m + n - 1, 0] && IntegerQ[n]

Rule 3827

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^4(e+f x) (a+a \sec (e+f x))^m \, dx &=\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {\int \sec ^2(e+f x) (a+a \sec (e+f x))^m (2 a+a m \sec (e+f x)) \, dx}{a (3+m)}\\ &=\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac {\int \sec (e+f x) (a+a \sec (e+f x))^m \left (a^2 m (1+m)+a^2 (4+m) \sec (e+f x)\right ) \, dx}{a^2 (2+m) (3+m)}\\ &=\frac {(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac {\left (m \left (5+3 m+m^2\right )\right ) \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac {(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}+\frac {\left (m \left (5+3 m+m^2\right ) (1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int \sec (e+f x) (1+\sec (e+f x))^m \, dx}{(1+m) (2+m) (3+m)}\\ &=\frac {(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}-\frac {\left (m \left (5+3 m+m^2\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m}}{\sqrt {1-x}} \, dx,x,\sec (e+f x)\right )}{f (1+m) (2+m) (3+m) \sqrt {1-\sec (e+f x)}}\\ &=\frac {(4+m) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac {\sec ^2(e+f x) (a+a \sec (e+f x))^m \tan (e+f x)}{f (3+m)}+\frac {2^{\frac {1}{2}+m} m \left (5+3 m+m^2\right ) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+m) (2+m) (3+m)}+\frac {m (a+a \sec (e+f x))^{1+m} \tan (e+f x)}{a f \left (6+5 m+m^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.36, size = 154, normalized size = 0.73 \[ \frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a (\sec (e+f x)+1))^m \left (2^{m+\frac {3}{2}} m \left (m^2+3 m+5\right ) \, _2F_1\left (\frac {1}{2},-m-\frac {1}{2};\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )+\left (\left (2 m^2+5 m+2\right ) \sec ^2(e+f x)+(2 m+1) m \sec (e+f x)+m^2+m+4\right ) (\sec (e+f x)+1)^{m+\frac {1}{2}}\right )}{f (m+2) (m+3) (2 m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(a + a*Sec[e + f*x])^m,x]

[Out]

((1 + Sec[e + f*x])^(-1/2 - m)*(a*(1 + Sec[e + f*x]))^m*(2^(3/2 + m)*m*(5 + 3*m + m^2)*Hypergeometric2F1[1/2,
-1/2 - m, 3/2, (1 - Sec[e + f*x])/2] + (1 + Sec[e + f*x])^(1/2 + m)*(4 + m + m^2 + m*(1 + 2*m)*Sec[e + f*x] +
(2 + 5*m + 2*m^2)*Sec[e + f*x]^2))*Tan[e + f*x])/(f*(2 + m)*(3 + m)*(1 + 2*m))

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+a*sec(f*x+e))^m,x, algorithm="fricas")

[Out]

integral((a*sec(f*x + e) + a)^m*sec(f*x + e)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+a*sec(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)^4, x)

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maple [F]  time = 1.16, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{4}\left (f x +e \right )\right ) \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+a*sec(f*x+e))^m,x)

[Out]

int(sec(f*x+e)^4*(a+a*sec(f*x+e))^m,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+a*sec(f*x+e))^m,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^m/cos(e + f*x)^4,x)

[Out]

int((a + a/cos(e + f*x))^m/cos(e + f*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec ^{4}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**4*(a+a*sec(f*x+e))**m,x)

[Out]

Integral((a*(sec(e + f*x) + 1))**m*sec(e + f*x)**4, x)

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