3.1.76 \(\int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))} \, dx\) [76]

Optimal. Leaf size=177 \[ \frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{3/2} c f}-\frac {7 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{4 \sqrt {2} a^{3/2} c f}+\frac {\cot (e+f x) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f} \]

[Out]

2*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(3/2)/c/f-7/8*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/(a+a
*sec(f*x+e))^(1/2))/a^(3/2)/c/f*2^(1/2)+1/4*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/a^2/c/f+1/4*cos(f*x+e)*cot(f*x+e
)*sec(1/2*f*x+1/2*e)^2*(a+a*sec(f*x+e))^(1/2)/a^2/c/f

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Rubi [A]
time = 0.16, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3989, 3972, 483, 597, 536, 209} \begin {gather*} \frac {2 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{3/2} c f}-\frac {7 \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{4 \sqrt {2} a^{3/2} c f}+\frac {\cot (e+f x) \sqrt {a \sec (e+f x)+a}}{4 a^2 c f}+\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a \sec (e+f x)+a}}{4 a^2 c f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(3/2)*c*f) - (7*ArcTan[(Sqrt[a]*Tan[e + f*x])/(
Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(4*Sqrt[2]*a^(3/2)*c*f) + (Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]])/(4*a^2*c
*f) + (Cos[e + f*x]*Cot[e + f*x]*Sec[(e + f*x)/2]^2*Sqrt[a + a*Sec[e + f*x]])/(4*a^2*c*f)

Rule 209

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

Rule 483

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 536

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 597

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 3972

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 3989

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])

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))} \, dx &=-\frac {\int \frac {\cot ^2(e+f x)}{\sqrt {a+a \sec (e+f x)}} \, dx}{a c}\\ &=\frac {2 \text {Subst}\left (\int \frac {1}{x^2 \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 )}{a^2 c f}\\ &=\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {\text {Subst}\left (\int \frac {a-3 a^2 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 )}{2 a^3 c f}\\ &=\frac {\cot (e+f x) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}-\frac {\text {Subst}\left (\int \frac {9 a^2+a^3 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 )}{4 a^3 c f}\\ &=\frac {\cot (e+f x) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {7 \text {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{4 a c f}-\frac {2 \text {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 c f}\\ &=\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{3/2} c f}-\frac {7 \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{4 \sqrt {2} a^{3/2} c f}+\frac {\cot (e+f x) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}+\frac {\cos (e+f x) \cot (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {a+a \sec (e+f x)}}{4 a^2 c f}\\ \end {align*}

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Mathematica [A]
time = 1.25, size = 154, normalized size = 0.87 \begin {gather*} \frac {\left (1+3 \cos (e+f x)-7 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {-1+\sec (e+f x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {-1+\sec (e+f x)}+8 \text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right ) (1+\cos (e+f x)) \sqrt {-1+\sec (e+f x)}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tan \left (\frac {1}{2} (e+f x)\right )}{2 a c f (-1+\cos (e+f x))^2 \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + 3*Cos[e + f*x] - 7*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*Cos[(e + f*x)/2]^2*Sqrt[-1 + Sec[e +
f*x]] + 8*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*(1 + Cos[e + f*x])*Sqrt[-1 + Sec[e + f*x]])*Sin[(e + f*x)/2]^2*Tan[(
e + f*x)/2])/(2*a*c*f*(-1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(376\) vs. \(2(151)=302\).
time = 0.21, size = 377, normalized size = 2.13

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (8 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}+7 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-7 \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right )-6 \left (\cos ^{3}\left (f x +e \right )\right )+4 \left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )\right )}{8 c f \sin \left (f x +e \right )^{3} a^{2}}\)	\(377\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/8/c/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(8*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)
/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)+7*(-2*cos(f*x+e)/(cos(f*
x+e)+1))^(1/2)*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*cos(f*x+e)^2*sin(
f*x+e)-8*2^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e
)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)-7*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*ln((sin(f*x+e)*(-2*cos(f*
x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))-6*cos(f*x+e)^3+4*cos(f*x+e)^2+2*cos(f*x+e))/sin(f*x+e)^3/
a^2

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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-c*sec(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.48, size = 560, normalized size = 3.16 \begin {gather*} \left [-\frac {7 \, \sqrt {2} \sqrt {-a} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {-a} {\left (\cos \left (f x + e\right ) + 1\right )} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} + 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) - 4 \, {\left (3 \, \cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{16 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}, \frac {7 \, \sqrt {2} \sqrt {a} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {a} {\left (\cos \left (f x + e\right ) + 1\right )} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{8 \, {\left (a^{2} c f \cos \left (f x + e\right ) + a^{2} c f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e)),x, algorithm="fricas")

[Out]

[-1/16*(7*sqrt(2)*sqrt(-a)*(cos(f*x + e) + 1)*log(-(2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))
*cos(f*x + e)*sin(f*x + e) - 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e) + a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1))
*sin(f*x + e) + 8*sqrt(-a)*(cos(f*x + e) + 1)*log(-(8*a*cos(f*x + e)^3 + 4*(2*cos(f*x + e)^2 - cos(f*x + e))*s
qrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f
*x + e) - 4*(3*cos(f*x + e)^2 + cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((a^2*c*f*cos(f*x + e)
+ a^2*c*f)*sin(f*x + e)), 1/8*(7*sqrt(2)*sqrt(a)*(cos(f*x + e) + 1)*arctan(sqrt(2)*sqrt((a*cos(f*x + e) + a)/c
os(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))*sin(f*x + e) + 8*sqrt(a)*(cos(f*x + e) + 1)*arctan(2*sqrt(a)
*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*
sin(f*x + e) + 2*(3*cos(f*x + e)^2 + cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((a^2*c*f*cos(f*x
+ e) + a^2*c*f)*sin(f*x + e))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {1}{a \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} - a \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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-c*sec(f*x+e)),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 {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )} \,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 - c/cos(e + f*x))),x)

[Out]

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

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