3.181 \(\int \frac {\sec ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=55 \[ \frac {\tanh ^{-1}(\sin (e+f x))}{b f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{b f \sqrt {a+b}} \]

[Out]

arctanh(sin(f*x+e))/b/f-arctanh(sin(f*x+e)*a^(1/2)/(a+b)^(1/2))*a^(1/2)/b/f/(a+b)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4147, 391, 206, 208} \[ \frac {\tanh ^{-1}(\sin (e+f x))}{b f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{b f \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[e + f*x]]/(b*f) - (Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(b*Sqrt[a + b]*f)

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 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]

Rule 391

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

Rule 4147

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rubi steps

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

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Mathematica [C]  time = 1.35, size = 1022, normalized size = 18.58 \[ \frac {(\cos (2 (e+f x)) a+a+2 b) \sec ^2(e+f x) \left (-4 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\sqrt {a} \cos (e) \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right )+\sqrt {a} \cos (e) \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right )-2 i \sqrt {a} \tan ^{-1}\left (\frac {2 \sin (e) \left (\sin (2 e) a+i a-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}+\sqrt {a+b} \cos (f x) \sqrt {(\cos (e)-i \sin (e))^2} \sqrt {a}-\sqrt {a+b} \cos (2 e+f x) \sqrt {(\cos (e)-i \sin (e))^2} \sqrt {a}+i b+i (a+b) \cos (2 e)+b \sin (2 e)\right )}{i (a+3 b) \cos (e)+i (a+b) \cos (3 e)+i a \cos (e+2 f x)+i a \cos (3 e+2 f x)+3 a \sin (e)+b \sin (e)+a \sin (3 e)+b \sin (3 e)+a \sin (e+2 f x)-a \sin (3 e+2 f x)}\right ) (\cos (e)-i \sin (e))+i \sqrt {a} \log \left (-\cos (2 (e+f x)) a-2 i \sin (2 e) a+a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}+2 (a+b) \cos (2 e)-2 i b \sin (2 e)\right ) \sin (e)-i \sqrt {a} \log \left (\cos (2 (e+f x)) a+2 i \sin (2 e) a-a+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (f x) \sqrt {a}+2 \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \sin (2 e+f x) \sqrt {a}-2 (a+b) \cos (2 e)+2 i b \sin (2 e)\right ) \sin (e)+2 \sqrt {a} \tan ^{-1}\left (\frac {(a+b) \sin (e)}{(a+b) \cos (e)-\sqrt {a} \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} (\cos (2 e)+i \sin (2 e)) \sin (e+f x)}\right ) (i \cos (e)+\sin (e))+4 \sqrt {a+b} \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {(\cos (e)-i \sin (e))^2}\right )}{8 b \sqrt {a+b} f \left (b \sec ^2(e+f x)+a\right ) \sqrt {(\cos (e)-i \sin (e))^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(-(Sqrt[a]*Cos[e]*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x
)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt
[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]) + Sqrt[a]*Cos[e]*Log[-a - 2*(a + b)*Cos[2*e] + a*
Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e] + (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin
[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]] - (2*I)*Sqrt[a]*ArcTan[(2*Sin[e]*(I*
a + I*b + I*(a + b)*Cos[2*e] + Sqrt[a]*Sqrt[a + b]*Cos[f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] - Sqrt[a]*Sqrt[a + b]*
Cos[2*e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + a*Sin[2*e] + b*Sin[2*e] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*
Sin[e])^2]*Sin[f*x] - I*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]))/(I*(a + 3*b)*Cos[e] +
I*(a + b)*Cos[3*e] + I*a*Cos[e + 2*f*x] + I*a*Cos[3*e + 2*f*x] + 3*a*Sin[e] + b*Sin[e] + a*Sin[3*e] + b*Sin[3
*e] + a*Sin[e + 2*f*x] - a*Sin[3*e + 2*f*x])]*(Cos[e] - I*Sin[e]) - 4*Sqrt[a + b]*Log[Cos[(e + f*x)/2] - Sin[(
e + f*x)/2]]*Sqrt[(Cos[e] - I*Sin[e])^2] + 4*Sqrt[a + b]*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sqrt[(Cos[e]
- I*Sin[e])^2] + I*Sqrt[a]*Log[a + 2*(a + b)*Cos[2*e] - a*Cos[2*(e + f*x)] - (2*I)*a*Sin[2*e] - (2*I)*b*Sin[2
*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[
e])^2]*Sin[2*e + f*x]]*Sin[e] - I*Sqrt[a]*Log[-a - 2*(a + b)*Cos[2*e] + a*Cos[2*(e + f*x)] + (2*I)*a*Sin[2*e]
+ (2*I)*b*Sin[2*e] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Sin[f*x] + 2*Sqrt[a]*Sqrt[a + b]*Sqrt[(
Cos[e] - I*Sin[e])^2]*Sin[2*e + f*x]]*Sin[e] + 2*Sqrt[a]*ArcTan[((a + b)*Sin[e])/((a + b)*Cos[e] - Sqrt[a]*Sqr
t[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*(Cos[2*e] + I*Sin[2*e])*Sin[e + f*x])]*(I*Cos[e] + Sin[e])))/(8*b*Sqrt[a
+ b]*f*(a + b*Sec[e + f*x]^2)*Sqrt[(Cos[e] - I*Sin[e])^2])

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fricas [A]  time = 1.96, size = 157, normalized size = 2.85 \[ \left [\frac {\sqrt {\frac {a}{a + b}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, {\left (a + b\right )} \sqrt {\frac {a}{a + b}} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + \log \left (\sin \left (f x + e\right ) + 1\right ) - \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, b f}, \frac {2 \, \sqrt {-\frac {a}{a + b}} \arctan \left (\sqrt {-\frac {a}{a + b}} \sin \left (f x + e\right )\right ) + \log \left (\sin \left (f x + e\right ) + 1\right ) - \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \, b f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a/(a + b))*log(-(a*cos(f*x + e)^2 + 2*(a + b)*sqrt(a/(a + b))*sin(f*x + e) - 2*a - b)/(a*cos(f*x +
e)^2 + b)) + log(sin(f*x + e) + 1) - log(-sin(f*x + e) + 1))/(b*f), 1/2*(2*sqrt(-a/(a + b))*arctan(sqrt(-a/(a
+ b))*sin(f*x + e)) + log(sin(f*x + e) + 1) - log(-sin(f*x + e) + 1))/(b*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*(-1/4/b*ln(abs(sin(f*x+exp(1))-1))+1/4/b*ln(abs(sin(f*x+ex
p(1))+1))+a/b*1/2/sqrt(-a^2-a*b)*atan(a*sin(f*x+exp(1))/sqrt(-a^2-a*b)))

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maple [A]  time = 0.62, size = 68, normalized size = 1.24 \[ -\frac {a \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{f b \sqrt {\left (a +b \right ) a}}-\frac {\ln \left (-1+\sin \left (f x +e \right )\right )}{2 f b}+\frac {\ln \left (1+\sin \left (f x +e \right )\right )}{2 f b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3/(a+b*sec(f*x+e)^2),x)

[Out]

-1/f/b*a/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2))-1/2/f/b*ln(-1+sin(f*x+e))+1/2/f/b*ln(1+sin(f*x+
e))

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maxima [A]  time = 0.44, size = 83, normalized size = 1.51 \[ \frac {\frac {a \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b} + \frac {\log \left (\sin \left (f x + e\right ) + 1\right )}{b} - \frac {\log \left (\sin \left (f x + e\right ) - 1\right )}{b}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

1/2*(a*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/(sqrt((a + b)*a)*b) + log(si
n(f*x + e) + 1)/b - log(sin(f*x + e) - 1)/b)/f

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mupad [B]  time = 4.58, size = 456, normalized size = 8.29 \[ \frac {\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )}{b\,f}+\frac {\mathrm {atan}\left (\frac {\frac {\left (2\,a^3\,\sin \left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2-\frac {\sin \left (e+f\,x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}\,1{}\mathrm {i}}{b^2+a\,b}+\frac {\left (2\,a^3\,\sin \left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2+\frac {\sin \left (e+f\,x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}\,1{}\mathrm {i}}{b^2+a\,b}}{\frac {\left (2\,a^3\,\sin \left (e+f\,x\right )+\frac {\left (2\,a^2\,b^2-\frac {\sin \left (e+f\,x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{b^2+a\,b}-\frac {\left (2\,a^3\,\sin \left (e+f\,x\right )-\frac {\left (2\,a^2\,b^2+\frac {\sin \left (e+f\,x\right )\,\left (16\,a^3\,b^2+8\,a^2\,b^3\right )\,\sqrt {a\,\left (a+b\right )}}{4\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{2\,\left (b^2+a\,b\right )}\right )\,\sqrt {a\,\left (a+b\right )}}{b^2+a\,b}}\right )\,\sqrt {a\,\left (a+b\right )}\,1{}\mathrm {i}}{f\,\left (b^2+a\,b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(sin(e + f*x))/(b*f) + (atan((((2*a^3*sin(e + f*x) + ((2*a^2*b^2 - (sin(e + f*x)*(8*a^2*b^3 + 16*a^3*b^2)
*(a*(a + b))^(1/2))/(4*(a*b + b^2)))*(a*(a + b))^(1/2))/(2*(a*b + b^2)))*(a*(a + b))^(1/2)*1i)/(a*b + b^2) + (
(2*a^3*sin(e + f*x) - ((2*a^2*b^2 + (sin(e + f*x)*(8*a^2*b^3 + 16*a^3*b^2)*(a*(a + b))^(1/2))/(4*(a*b + b^2)))
*(a*(a + b))^(1/2))/(2*(a*b + b^2)))*(a*(a + b))^(1/2)*1i)/(a*b + b^2))/(((2*a^3*sin(e + f*x) + ((2*a^2*b^2 -
(sin(e + f*x)*(8*a^2*b^3 + 16*a^3*b^2)*(a*(a + b))^(1/2))/(4*(a*b + b^2)))*(a*(a + b))^(1/2))/(2*(a*b + b^2)))
*(a*(a + b))^(1/2))/(a*b + b^2) - ((2*a^3*sin(e + f*x) - ((2*a^2*b^2 + (sin(e + f*x)*(8*a^2*b^3 + 16*a^3*b^2)*
(a*(a + b))^(1/2))/(4*(a*b + b^2)))*(a*(a + b))^(1/2))/(2*(a*b + b^2)))*(a*(a + b))^(1/2))/(a*b + b^2)))*(a*(a
+ b))^(1/2)*1i)/(f*(a*b + b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(sec(e + f*x)**3/(a + b*sec(e + f*x)**2), x)

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