3.193 \(\int \frac{\sec ^5(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\)

Optimal. Leaf size=102 \[ -\frac{\sqrt{a} (2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 b^2 f (a+b)^{3/2}}-\frac{a \sin (e+f x)}{2 b f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\tanh ^{-1}(\sin (e+f x))}{b^2 f} \]

[Out]

ArcTanh[Sin[e + f*x]]/(b^2*f) - (Sqrt[a]*(2*a + 3*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*b^2*(a +
b)^(3/2)*f) - (a*Sin[e + f*x])/(2*b*(a + b)*f*(a + b - a*Sin[e + f*x]^2))

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Rubi [A]  time = 0.140251, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4147, 414, 522, 206, 208} \[ -\frac{\sqrt{a} (2 a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (e+f x)}{\sqrt{a+b}}\right )}{2 b^2 f (a+b)^{3/2}}-\frac{a \sin (e+f x)}{2 b f (a+b) \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\tanh ^{-1}(\sin (e+f x))}{b^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[e + f*x]]/(b^2*f) - (Sqrt[a]*(2*a + 3*b)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(2*b^2*(a +
b)^(3/2)*f) - (a*Sin[e + f*x])/(2*b*(a + b)*f*(a + b - a*Sin[e + f*x]^2))

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]

Rule 414

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

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 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]

Rubi steps

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

Mathematica [C]  time = 4.42183, size = 980, normalized size = 9.61 \[ \frac{(\cos (2 (e+f x)) a+a+2 b) \sec ^3(e+f x) \left (-8 b \sqrt{a+b} \sqrt{(\cos (e)-i \sin (e))^2} \tan (e+f x) a^{3/2}-2 i (2 a+3 b) \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 (2 (e+f x)) a+a+2 b) \sec (e+f x) (\cos (e)-i \sin (e)) a-(2 a+3 b) (\cos (2 (e+f x)) a+a+2 b) \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 ) \sec (e+f x) (\cos (e)-i \sin (e)) a+(2 a+3 b) (\cos (2 (e+f x)) a+a+2 b) \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 ) \sec (e+f x) (\cos (e)-i \sin (e)) a+2 (2 a+3 b) \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 ) (\cos (2 (e+f x)) a+a+2 b) \sec (e+f x) (i \cos (e)+\sin (e)) a-8 (a+b)^{3/2} (\cos (2 (e+f x)) a+a+2 b) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sec (e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}+8 (a+b)^{3/2} (\cos (2 (e+f x)) a+a+2 b) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )+\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sec (e+f x) \sqrt{(\cos (e)-i \sin (e))^2} \sqrt{a}\right )}{32 \sqrt{a} b^2 (a+b)^{3/2} f \left (b \sec ^2(e+f x)+a\right )^2 \sqrt{(\cos (e)-i \sin (e))^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^3*((-2*I)*a*(2*a + 3*b)*ArcTan[(2*Sin[e]*(I*a + I*b + I*(a + b)*C
os[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])]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*(Cos[e] - I*Sin[e]) - a*(2*a + 3*b)*(a +
2*b + a*Cos[2*(e + f*x)])*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]]*Sec[e + f*x]*(Cos[e] - I*Sin[e]) + a*(2*a + 3*b)*(a + 2*b + a*Cos[2*(e + f*x)])*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[(Co
s[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]]*Sec[e + f*x]*
(Cos[e] - I*Sin[e]) - 8*Sqrt[a]*(a + b)^(3/2)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[Cos[(e + f*x)/2] - Sin[(e + f
*x)/2]]*Sec[e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + 8*Sqrt[a]*(a + b)^(3/2)*(a + 2*b + a*Cos[2*(e + f*x)])*Log[
Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*Sec[e + f*x]*Sqrt[(Cos[e] - I*Sin[e])^2] + 2*a*(2*a + 3*b)*ArcTan[((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])]*(a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]*(I*Cos[e] + Sin[e]) - 8*a^(3/2)*b*Sqrt[a + b]*Sqrt[(Cos[e] -
I*Sin[e])^2]*Tan[e + f*x]))/(32*Sqrt[a]*b^2*(a + b)^(3/2)*f*(a + b*Sec[e + f*x]^2)^2*Sqrt[(Cos[e] - I*Sin[e])^
2])

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Maple [A]  time = 0.087, size = 151, normalized size = 1.5 \begin{align*}{\frac{\ln \left ( \sin \left ( fx+e \right ) +1 \right ) }{2\,f{b}^{2}}}+{\frac{\sin \left ( fx+e \right ) a}{2\,fb \left ( a+b \right ) \left ( -a-b+a \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{{a}^{2}}{f{b}^{2} \left ( a+b \right ) }{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{3\,a}{2\,fb \left ( a+b \right ) }{\it Artanh} \left ({\sin \left ( fx+e \right ) a{\frac{1}{\sqrt{ \left ( a+b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) a}}}}-{\frac{\ln \left ( \sin \left ( fx+e \right ) -1 \right ) }{2\,f{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^5/(a+b*sec(f*x+e)^2)^2,x)

[Out]

1/2/f/b^2*ln(sin(f*x+e)+1)+1/2/f*a/b/(a+b)*sin(f*x+e)/(-a-b+a*sin(f*x+e)^2)-1/f*a^2/b^2/(a+b)/((a+b)*a)^(1/2)*
arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2))-3/2/f*a/b/(a+b)/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2))-1/
2/f/b^2*ln(sin(f*x+e)-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.707565, size = 950, normalized size = 9.31 \begin{align*} \left [-\frac{2 \, a b \sin \left (f x + e\right ) -{\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + 3 \, b^{2}\right )} \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 ) - 2 \,{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + 2 \,{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{4 \,{\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a b^{3} + b^{4}\right )} f\right )}}, -\frac{a b \sin \left (f x + e\right ) -{\left ({\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{2} + 2 \, a b + 3 \, b^{2}\right )} \sqrt{-\frac{a}{a + b}} \arctan \left (\sqrt{-\frac{a}{a + b}} \sin \left (f x + e\right )\right ) -{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) +{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right )}{2 \,{\left ({\left (a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} +{\left (a b^{3} + b^{4}\right )} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*b*sin(f*x + e) - ((2*a^2 + 3*a*b)*cos(f*x + e)^2 + 2*a*b + 3*b^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)) - 2*((a^2 + a*b)*cos(f*x + e
)^2 + a*b + b^2)*log(sin(f*x + e) + 1) + 2*((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)*log(-sin(f*x + e) + 1))/((
a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a*b^3 + b^4)*f), -1/2*(a*b*sin(f*x + e) - ((2*a^2 + 3*a*b)*cos(f*x + e)^2
+ 2*a*b + 3*b^2)*sqrt(-a/(a + b))*arctan(sqrt(-a/(a + b))*sin(f*x + e)) - ((a^2 + a*b)*cos(f*x + e)^2 + a*b +
b^2)*log(sin(f*x + e) + 1) + ((a^2 + a*b)*cos(f*x + e)^2 + a*b + b^2)*log(-sin(f*x + e) + 1))/((a^2*b^2 + a*b
^3)*f*cos(f*x + e)^2 + (a*b^3 + b^4)*f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**5/(a+b*sec(f*x+e)**2)**2,x)

[Out]

Integral(sec(e + f*x)**5/(a + b*sec(e + f*x)**2)**2, x)

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Giac [A]  time = 1.27204, size = 177, normalized size = 1.74 \begin{align*} \frac{\frac{{\left (2 \, a^{2} + 3 \, a b\right )} \arctan \left (\frac{a \sin \left (f x + e\right )}{\sqrt{-a^{2} - a b}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt{-a^{2} - a b}} + \frac{a \sin \left (f x + e\right )}{{\left (a \sin \left (f x + e\right )^{2} - a - b\right )}{\left (a b + b^{2}\right )}} + \frac{\log \left (\sin \left (f x + e\right ) + 1\right )}{b^{2}} - \frac{\log \left (-\sin \left (f x + e\right ) + 1\right )}{b^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*a^2 + 3*a*b)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b))/((a*b^2 + b^3)*sqrt(-a^2 - a*b)) + a*sin(f*x + e)
/((a*sin(f*x + e)^2 - a - b)*(a*b + b^2)) + log(sin(f*x + e) + 1)/b^2 - log(-sin(f*x + e) + 1)/b^2)/f