3.85 \(\int \frac{1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=269 \[ \frac{43 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{96 a^4 c^2 f}+\frac{21 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{64 a^3 c^2 f}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} c^2 f}-\frac{107 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{64 \sqrt{2} a^{5/2} c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a \sec (e+f x)+a)^{3/2}}{16 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a \sec (e+f x)+a)^{3/2}}{32 a^4 c^2 f} \]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*c^2*f) - (107*ArcTan[(Sqrt[a]*Tan[e + f*x
])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(64*Sqrt[2]*a^(5/2)*c^2*f) + (21*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]]
)/(64*a^3*c^2*f) + (43*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(96*a^4*c^2*f) - (15*Cos[e + f*x]*Cot[e + f*
x]^3*Sec[(e + f*x)/2]^2*(a + a*Sec[e + f*x])^(3/2))/(32*a^4*c^2*f) - (Cos[e + f*x]^2*Cot[e + f*x]^3*Sec[(e + f
*x)/2]^4*(a + a*Sec[e + f*x])^(3/2))/(16*a^4*c^2*f)

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Rubi [A]  time = 0.335235, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3904, 3887, 472, 579, 583, 522, 203} \[ \frac{43 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{96 a^4 c^2 f}+\frac{21 \cot (e+f x) \sqrt{a \sec (e+f x)+a}}{64 a^3 c^2 f}+\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} c^2 f}-\frac{107 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{64 \sqrt{2} a^{5/2} c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a \sec (e+f x)+a)^{3/2}}{16 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a \sec (e+f x)+a)^{3/2}}{32 a^4 c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^2),x]

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*c^2*f) - (107*ArcTan[(Sqrt[a]*Tan[e + f*x
])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(64*Sqrt[2]*a^(5/2)*c^2*f) + (21*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]]
)/(64*a^3*c^2*f) + (43*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(96*a^4*c^2*f) - (15*Cos[e + f*x]*Cot[e + f*
x]^3*Sec[(e + f*x)/2]^2*(a + a*Sec[e + f*x])^(3/2))/(32*a^4*c^2*f) - (Cos[e + f*x]^2*Cot[e + f*x]^3*Sec[(e + f
*x)/2]^4*(a + a*Sec[e + f*x])^(3/2))/(16*a^4*c^2*f)

Rule 3904

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

Rule 3887

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

Rule 472

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 579

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*g*n*(b*c - a*d)*(p +
1)), x] + Dist[1/(a*n*(b*c - a*d)*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f)*(
m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^2} \, dx &=\frac{\int \frac{\cot ^4(e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{a^2 c^2}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^4 c^2 f}\\ &=-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}-\frac{\operatorname{Subst}\left (\int \frac{a-7 a^2 x^2}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{4 a^5 c^2 f}\\ &=-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{32 a^4 c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}-\frac{\operatorname{Subst}\left (\int \frac{-43 a^2-75 a^3 x^2}{x^4 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{16 a^6 c^2 f}\\ &=\frac{43 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{96 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{32 a^4 c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}+\frac{\operatorname{Subst}\left (\int \frac{63 a^3-129 a^4 x^2}{x^2 \left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{96 a^6 c^2 f}\\ &=\frac{21 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{64 a^3 c^2 f}+\frac{43 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{96 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{32 a^4 c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}-\frac{\operatorname{Subst}\left (\int \frac{447 a^4+63 a^5 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{192 a^6 c^2 f}\\ &=\frac{21 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{64 a^3 c^2 f}+\frac{43 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{96 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{32 a^4 c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}+\frac{107 \operatorname{Subst}\left (\int \frac{1}{2+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{64 a^2 c^2 f}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^2 c^2 f}\\ &=\frac{2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^{5/2} c^2 f}-\frac{107 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{64 \sqrt{2} a^{5/2} c^2 f}+\frac{21 \cot (e+f x) \sqrt{a+a \sec (e+f x)}}{64 a^3 c^2 f}+\frac{43 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{96 a^4 c^2 f}-\frac{15 \cos (e+f x) \cot ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{32 a^4 c^2 f}-\frac{\cos ^2(e+f x) \cot ^3(e+f x) \sec ^4\left (\frac{1}{2} (e+f x)\right ) (a+a \sec (e+f x))^{3/2}}{16 a^4 c^2 f}\\ \end{align*}

Mathematica [C]  time = 24.0523, size = 5660, normalized size = 21.04 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^2),x]

[Out]

Result too large to show

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Maple [B]  time = 0.326, size = 725, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x)

[Out]

1/384/c^2/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(1+cos(f*x+e))*(-1+cos(f*x+e))^2*(384*2^(1/2)*cos(f*x+e)
^3*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*si
n(f*x+e)/cos(f*x+e))+321*sin(f*x+e)*cos(f*x+e)^3*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+co
s(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))+384*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+co
s(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))+321*sin(f*x+e
)*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*
x+e)+1)/sin(f*x+e))-384*2^(1/2)*cos(f*x+e)*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)
*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))-321*sin(f*x+e)*cos(f*x+e)*(-2*cos(f*x+e)/(1+cos(f
*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))-384*2^(1/2)*sin(f*
x+e)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/(1+cos(f*x
+e)))^(1/2)-410*cos(f*x+e)^4-321*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(((-2*cos(f*x+e)/(1+cos(f*x
+e)))^(1/2)*sin(f*x+e)-cos(f*x+e)+1)/sin(f*x+e))-142*cos(f*x+e)^3+298*cos(f*x+e)^2+126*cos(f*x+e))/sin(f*x+e)^
7

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**2,x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out