3.155 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=92 \[ -\frac {a 2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac {1}{2}-m} (a \sec (e+f x)+a)^{m-1} \, _2F_1\left (-\frac {3}{2},\frac {1}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{3 f (c-c \sec (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3961, 70, 69} \[ -\frac {a 2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac {1}{2}-m} (a \sec (e+f x)+a)^{m-1} \, _2F_1\left (-\frac {3}{2},\frac {1}{2}-m;-\frac {1}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{3 f (c-c \sec (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(1/2 + m)*a*Hypergeometric2F1[-3/2, 1/2 - m, -1/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(1/2 - m)*(a +
a*Sec[e + f*x])^(-1 + m)*Tan[e + f*x])/(3*f*(c - c*Sec[e + f*x])^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 70

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

Rule 3961

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

Rubi steps

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

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Mathematica [F]  time = 2.59, size = 0, normalized size = 0.00 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{c^{2} \sec \left (f x + e\right )^{2} - 2 \, c^{2} \sec \left (f x + e\right ) + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(f*x + e) + a)^m*sec(f*x + e)/(c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(c*sec(f*x + e) - c)^2, x)

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maple [F]  time = 1.97, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}}{\left (c -c \sec \left (f x +e \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^2,x)

[Out]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^2,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(c*sec(f*x + e) - c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^m/(cos(e + f*x)*(c - c/cos(e + f*x))^2),x)

[Out]

int((a + a/cos(e + f*x))^m/(cos(e + f*x)*(c - c/cos(e + f*x))^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**m/(c-c*sec(f*x+e))**2,x)

[Out]

Integral((a*sec(e + f*x) + a)**m*sec(e + f*x)/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x)/c**2

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