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

Optimal. Leaf size=86 \[ \frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{b^2 f \sqrt {a+b}}-\frac {(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {\tan (e+f x) \sec (e+f x)}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 86, normalized size of antiderivative = 1.00, 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} \[ \frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{b^2 f \sqrt {a+b}}-\frac {(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {\tan (e+f x) \sec (e+f x)}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a - b)*ArcTanh[Sin[e + f*x]])/(2*b^2*f) + (a^(3/2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]])/(b^2*Sqrt
[a + b]*f) + (Sec[e + f*x]*Tan[e + f*x])/(2*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 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 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 ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2 \left (a+b-a x^2\right )} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sec (e+f x) \tan (e+f x)}{2 b f}+\frac {\operatorname {Subst}\left (\int \frac {-a+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 f}\\ &=\frac {\sec (e+f x) \tan (e+f x)}{2 b f}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{b^2 f}-\frac {(2 a-b) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{2 b^2 f}\\ &=-\frac {(2 a-b) \tanh ^{-1}(\sin (e+f x))}{2 b^2 f}+\frac {a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{b^2 \sqrt {a+b} f}+\frac {\sec (e+f x) \tan (e+f x)}{2 b f}\\ \end {align*}

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Mathematica [C]  time = 6.12, size = 1195, normalized size = 13.90 \[ \frac {(\cos (2 (e+f x)) a+a+2 b) \sec ^2(e+f x) \left (\frac {2 i \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 ) \sqrt {(\cos (e)-i \sin (e))^2} (\cos (e)+i \sin (e)) a^{3/2}}{\sqrt {a+b}}-\frac {i \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) a^{3/2}}{\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {i \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) a^{3/2}}{\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}}+\frac {2 \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 ) \sqrt {(\cos (e)-i \sin (e))^2} (\sin (e)-i \cos (e)) a^{3/2}}{\sqrt {a+b}}+\frac {\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 ) a^{3/2}}{\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}}-\frac {\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 ) a^{3/2}}{\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}}+4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) a-4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) a-2 b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {b}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}-\frac {b}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}\right )}{8 b^2 f \left (b \sec ^2(e+f x)+a\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 1.99, size = 272, normalized size = 3.16 \[ \left [\frac {2 \, a \sqrt {\frac {a}{a + b}} \cos \left (f x + e\right )^{2} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, {\left (a + b\right )} \sqrt {\frac {a}{a + b}} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) - {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) + {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, b \sin \left (f x + e\right )}{4 \, b^{2} f \cos \left (f x + e\right )^{2}}, -\frac {4 \, a \sqrt {-\frac {a}{a + b}} \arctan \left (\sqrt {-\frac {a}{a + b}} \sin \left (f x + e\right )\right ) \cos \left (f x + e\right )^{2} + {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) - {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, b \sin \left (f x + e\right )}{4 \, b^{2} f \cos \left (f x + e\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^5/(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*(-a^2*1/2/b^2/sqrt(-a^2-a*b)*atan(a*sin(f*x+exp(1))/sqrt(-
a^2-a*b))-(-2*a+b)*1/8/b^2*ln(abs(sin(f*x+exp(1))-1))+(-2*a+b)*1/8/b^2*ln(abs(sin(f*x+exp(1))+1))-sin(f*x+exp(
1))*1/4/b/(sin(f*x+exp(1))^2-1))

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maple [A]  time = 0.64, size = 141, normalized size = 1.64 \[ \frac {a^{2} \arctanh \left (\frac {a \sin \left (f x +e \right )}{\sqrt {\left (a +b \right ) a}}\right )}{f \,b^{2} \sqrt {\left (a +b \right ) a}}-\frac {1}{4 f b \left (-1+\sin \left (f x +e \right )\right )}+\frac {\ln \left (-1+\sin \left (f x +e \right )\right ) a}{2 f \,b^{2}}-\frac {\ln \left (-1+\sin \left (f x +e \right )\right )}{4 f b}-\frac {1}{4 f b \left (1+\sin \left (f x +e \right )\right )}-\frac {\ln \left (1+\sin \left (f x +e \right )\right ) a}{2 f \,b^{2}}+\frac {\ln \left (1+\sin \left (f x +e \right )\right )}{4 f b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/b^2*a^2/((a+b)*a)^(1/2)*arctanh(a*sin(f*x+e)/((a+b)*a)^(1/2))-1/4/f/b/(-1+sin(f*x+e))+1/2/f/b^2*ln(-1+sin(
f*x+e))*a-1/4/f/b*ln(-1+sin(f*x+e))-1/4/f/b/(1+sin(f*x+e))-1/2/f/b^2*ln(1+sin(f*x+e))*a+1/4/f/b*ln(1+sin(f*x+e
))

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maxima [A]  time = 0.45, size = 124, normalized size = 1.44 \[ -\frac {\frac {2 \, a^{2} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{2}} + \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (f x + e\right ) + 1\right )}{b^{2}} - \frac {{\left (2 \, a - b\right )} \log \left (\sin \left (f x + e\right ) - 1\right )}{b^{2}} + \frac {2 \, \sin \left (f x + e\right )}{b \sin \left (f x + e\right )^{2} - b}}{4 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^2*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + sqrt((a + b)*a)))/(sqrt((a + b)*a)*b^2) +
(2*a - b)*log(sin(f*x + e) + 1)/b^2 - (2*a - b)*log(sin(f*x + e) - 1)/b^2 + 2*sin(f*x + e)/(b*sin(f*x + e)^2
- b))/f

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mupad [B]  time = 4.95, size = 591, normalized size = 6.87 \[ \frac {b\,\left (a\,\sin \left (e+f\,x\right )-a\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )+a\,{\sin \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\right )+b^2\,\left (\sin \left (e+f\,x\right )+\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )-{\sin \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )\right )-2\,a^2\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )+2\,a^2\,{\sin \left (e+f\,x\right )}^2\,\mathrm {atanh}\left (\sin \left (e+f\,x\right )\right )+\mathrm {atan}\left (\frac {-a\,\sin \left (e+f\,x\right )\,{\left (a^4+b\,a^3\right )}^{3/2}\,8{}\mathrm {i}-b\,\sin \left (e+f\,x\right )\,{\left (a^4+b\,a^3\right )}^{3/2}\,4{}\mathrm {i}+a^5\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,8{}\mathrm {i}-a^2\,b^3\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,2{}\mathrm {i}+a^3\,b^2\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,1{}\mathrm {i}+a\,b^4\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,1{}\mathrm {i}+a^4\,b\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,12{}\mathrm {i}}{3\,a^5\,b^2+5\,a^4\,b^3+a^3\,b^4-a^2\,b^5}\right )\,\sqrt {a^4+b\,a^3}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {-a\,\sin \left (e+f\,x\right )\,{\left (a^4+b\,a^3\right )}^{3/2}\,8{}\mathrm {i}-b\,\sin \left (e+f\,x\right )\,{\left (a^4+b\,a^3\right )}^{3/2}\,4{}\mathrm {i}+a^5\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,8{}\mathrm {i}-a^2\,b^3\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,2{}\mathrm {i}+a^3\,b^2\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,1{}\mathrm {i}+a\,b^4\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,1{}\mathrm {i}+a^4\,b\,\sin \left (e+f\,x\right )\,\sqrt {a^4+b\,a^3}\,12{}\mathrm {i}}{3\,a^5\,b^2+5\,a^4\,b^3+a^3\,b^4-a^2\,b^5}\right )\,{\sin \left (e+f\,x\right )}^2\,\sqrt {a^4+b\,a^3}\,2{}\mathrm {i}}{f\,\left (-2\,b^3\,{\sin \left (e+f\,x\right )}^2+2\,b^3-2\,a\,b^2\,{\sin \left (e+f\,x\right )}^2+2\,a\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^5*(a + b/cos(e + f*x)^2)),x)

[Out]

(b*(a*sin(e + f*x) - a*atanh(sin(e + f*x)) + a*sin(e + f*x)^2*atanh(sin(e + f*x))) + atan((a^5*sin(e + f*x)*(a
^3*b + a^4)^(1/2)*8i - b*sin(e + f*x)*(a^3*b + a^4)^(3/2)*4i - a*sin(e + f*x)*(a^3*b + a^4)^(3/2)*8i - a^2*b^3
*sin(e + f*x)*(a^3*b + a^4)^(1/2)*2i + a^3*b^2*sin(e + f*x)*(a^3*b + a^4)^(1/2)*1i + a*b^4*sin(e + f*x)*(a^3*b
+ a^4)^(1/2)*1i + a^4*b*sin(e + f*x)*(a^3*b + a^4)^(1/2)*12i)/(a^3*b^4 - a^2*b^5 + 5*a^4*b^3 + 3*a^5*b^2))*(a
^3*b + a^4)^(1/2)*2i + b^2*(sin(e + f*x) + atanh(sin(e + f*x)) - sin(e + f*x)^2*atanh(sin(e + f*x))) - 2*a^2*a
tanh(sin(e + f*x)) - atan((a^5*sin(e + f*x)*(a^3*b + a^4)^(1/2)*8i - b*sin(e + f*x)*(a^3*b + a^4)^(3/2)*4i - a
*sin(e + f*x)*(a^3*b + a^4)^(3/2)*8i - a^2*b^3*sin(e + f*x)*(a^3*b + a^4)^(1/2)*2i + a^3*b^2*sin(e + f*x)*(a^3
*b + a^4)^(1/2)*1i + a*b^4*sin(e + f*x)*(a^3*b + a^4)^(1/2)*1i + a^4*b*sin(e + f*x)*(a^3*b + a^4)^(1/2)*12i)/(
a^3*b^4 - a^2*b^5 + 5*a^4*b^3 + 3*a^5*b^2))*sin(e + f*x)^2*(a^3*b + a^4)^(1/2)*2i + 2*a^2*sin(e + f*x)^2*atanh
(sin(e + f*x)))/(f*(2*a*b^2 + 2*b^3 - 2*b^3*sin(e + f*x)^2 - 2*a*b^2*sin(e + f*x)^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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