3.176 \(\int \frac{1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=560 \[ -\frac{2 d^{5/2} (3 c-d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{\sqrt{a} c^2 f (c-d)^3 \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{d^3 \tan (e+f x)}{a c f (c-d)^2 (c+d) \sqrt{a \sec (e+f x)+a} (c+d \sec (e+f x))}-\frac{d^{5/2} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{\sqrt{a} c f (c-d)^2 (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x)}{2 a f (c-d)^2 (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} \sqrt{a} f (c-d)^2 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} (c-3 d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{a} f (c-d)^3 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

[Out]

-Tan[e + f*x]/(2*a*(c - d)^2*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]) + (2*ArcTanh[Sqrt[a - a*Sec[e + f*
x]]/Sqrt[a]]*Tan[e + f*x])/(Sqrt[a]*c^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (Sqrt[2]*(c - 3
*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(Sqrt[a]*(c - d)^3*f*Sqrt[a - a*Sec[e +
f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(2*Sqrt[2
]*Sqrt[a]*(c - d)^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a -
a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(Sqrt[a]*c*(c - d)^2*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e +
f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*(3*c - d)*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqr
t[c + d])]*Tan[e + f*x])/(Sqrt[a]*c^2*(c - d)^3*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]
]) - (d^3*Tan[e + f*x])/(a*c*(c - d)^2*(c + d)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

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Rubi [A]  time = 0.510682, antiderivative size = 560, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3940, 180, 63, 206, 51, 208} \[ -\frac{2 d^{5/2} (3 c-d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{\sqrt{a} c^2 f (c-d)^3 \sqrt{c+d} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{\sqrt{a} c^2 f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{d^3 \tan (e+f x)}{a c f (c-d)^2 (c+d) \sqrt{a \sec (e+f x)+a} (c+d \sec (e+f x))}-\frac{d^{5/2} \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a-a \sec (e+f x)}}{\sqrt{a} \sqrt{c+d}}\right )}{\sqrt{a} c f (c-d)^2 (c+d)^{3/2} \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x)}{2 a f (c-d)^2 (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{2 \sqrt{2} \sqrt{a} f (c-d)^2 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}-\frac{\sqrt{2} (c-3 d) \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{2} \sqrt{a}}\right )}{\sqrt{a} f (c-d)^3 \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sec[e + f*x])^(3/2)*(c + d*Sec[e + f*x])^2),x]

[Out]

-Tan[e + f*x]/(2*a*(c - d)^2*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]) + (2*ArcTanh[Sqrt[a - a*Sec[e + f*
x]]/Sqrt[a]]*Tan[e + f*x])/(Sqrt[a]*c^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (Sqrt[2]*(c - 3
*d)*ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(Sqrt[a]*(c - d)^3*f*Sqrt[a - a*Sec[e +
f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (ArcTanh[Sqrt[a - a*Sec[e + f*x]]/(Sqrt[2]*Sqrt[a])]*Tan[e + f*x])/(2*Sqrt[2
]*Sqrt[a]*(c - d)^2*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a -
a*Sec[e + f*x]])/(Sqrt[a]*Sqrt[c + d])]*Tan[e + f*x])/(Sqrt[a]*c*(c - d)^2*(c + d)^(3/2)*f*Sqrt[a - a*Sec[e +
f*x]]*Sqrt[a + a*Sec[e + f*x]]) - (2*(3*c - d)*d^(5/2)*ArcTanh[(Sqrt[d]*Sqrt[a - a*Sec[e + f*x]])/(Sqrt[a]*Sqr
t[c + d])]*Tan[e + f*x])/(Sqrt[a]*c^2*(c - d)^3*Sqrt[c + d]*f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]
]) - (d^3*Tan[e + f*x])/(a*c*(c - d)^2*(c + d)*f*Sqrt[a + a*Sec[e + f*x]]*(c + d*Sec[e + f*x]))

Rule 3940

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

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rule 51

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

Rule 208

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

Rubi steps

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

Mathematica [C]  time = 37.741, size = 581056, normalized size = 1037.6 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + a*Sec[e + f*x])^(3/2)*(c + d*Sec[e + f*x])^2),x]

[Out]

Result too large to show

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Maple [B]  time = 3.779, size = 164796, normalized size = 294.3 \begin{align*} \text{output 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))^(3/2)/(c+d*sec(f*x+e))^2,x)

[Out]

result too large to display

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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))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}} \left (c + d \sec{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))**(3/2)/(c+d*sec(f*x+e))**2,x)

[Out]

Integral(1/((a*(sec(e + f*x) + 1))**(3/2)*(c + d*sec(e + f*x))**2), x)

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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))^(3/2)/(c+d*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out