3.2.76 \(\int \frac {1}{(a+a \sec (e+f x))^{3/2} (c+d \sec (e+f x))^2} \, dx\) [176]

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

[Out]

-1/2*tan(f*x+e)/a/(c-d)^2/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)-d^3*tan(f*x+e)/a/c/(c-d)^2/(c+d)/f/(c+d*sec(
f*x+e))/(a+a*sec(f*x+e))^(1/2)+2*arctanh((a-a*sec(f*x+e))^(1/2)/a^(1/2))*tan(f*x+e)/c^2/f/a^(1/2)/(a-a*sec(f*x
+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-d^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))*tan(f*x+
e)/c/(c-d)^2/(c+d)^(3/2)/f/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-1/4*arctanh(1/2*(a-a*sec(f*x+
e))^(1/2)*2^(1/2)/a^(1/2))*tan(f*x+e)/(c-d)^2/f*2^(1/2)/a^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-
(c-3*d)*arctanh(1/2*(a-a*sec(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*tan(f*x+e)/(c-d)^3/f/a^(1/2)/(a-a*sec(f*x+
e))^(1/2)/(a+a*sec(f*x+e))^(1/2)-2*(3*c-d)*d^(5/2)*arctanh(d^(1/2)*(a-a*sec(f*x+e))^(1/2)/a^(1/2)/(c+d)^(1/2))
*tan(f*x+e)/c^2/(c-d)^3/f/a^(1/2)/(c+d)^(1/2)/(a-a*sec(f*x+e))^(1/2)/(a+a*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.36, antiderivative size = 560, normalized size of antiderivative = 1.00, 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 = {4025, 186, 65, 212, 44, 214} \begin {gather*} -\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^{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 {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 {\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Tan[e + f*x]/(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]*S
qrt[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 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 186

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 212

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

Rule 214

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

Rule 4025

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]

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 ) \text {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 ) \text {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) \text {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)) \text {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) \text {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 ) \text {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 ) \text {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)) \text {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)) \text {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) \text {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 ) \text {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 ) \text {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) \text {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 ) \text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 37.91, size = 582620, normalized size = 1040.39 \begin {gather*} \text {Result too large to show} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(164977\) vs. \(2(480)=960\).
time = 8.36, size = 164978, normalized size = 294.60

method	result	size
default	\(\text {Expression too large to display}\)	\(164978\)

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,method=_RETURNVERBOSE)

[Out]

result too large to display

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

[Out]

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

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Fricas [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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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(-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(1/(a+a*sec(f*x+e))^(3/2)/(c+d*sec(f*x+e))^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 = -0.00 \begin {gather*} \int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a/cos(e + f*x))^(3/2)*(c + d/cos(e + f*x))^2),x)

[Out]

int(1/((a + a/cos(e + f*x))^(3/2)*(c + d/cos(e + f*x))^2), x)

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