3.2.61 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^{5/2}} \, dx\) [161]

Optimal. Leaf size=74 \[ -\frac {\, _2F_1\left (3,\frac {1}{2}+m;\frac {3}{2}+m;\frac {1}{2} (1+\sec (e+f x))\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{4 c^2 f (1+2 m) \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {4046, 70} \begin {gather*} -\frac {\tan (e+f x) (a \sec (e+f x)+a)^m \, _2F_1\left (3,m+\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sec (e+f x)+1)\right )}{4 c^2 f (2 m+1) \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^m)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

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

Rule 70

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

Rule 4046

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

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Mathematica [F]
time = 1.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^m}{(c-c \sec (e+f x))^{5/2}} \, dx \end {gather*}

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])^(5/2),x]

[Out]

Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^m)/(c - c*Sec[e + f*x])^(5/2), x]

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Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \frac {\sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}}{\left (c -c \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx\]

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))^(5/2),x)

[Out]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^(5/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^m*sec(f*x + e)/(-c*sec(f*x + e) + c)^(5/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-c*sec(f*x + e) + c)*(a*sec(f*x + e) + a)^m*sec(f*x + e)/(c^3*sec(f*x + e)^3 - 3*c^3*sec(f*x +
e)^2 + 3*c^3*sec(f*x + e) - c^3), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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))**(5/2),x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 )}^{5/2}} \,d x \end {gather*}

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))^(5/2)),x)

[Out]

int((a + a/cos(e + f*x))^m/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)), x)

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