3.124 \(\int \frac{(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=220 \[ \frac{4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1)^2 \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{2 c^4 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c^4 \tan (e+f x) \log (\cos (e+f x))}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

(c^4*Log[Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (2*c^4*Log[1
+ Sec[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (4*c^4*Tan[e + f*x])
/(a^2*f*(1 + Sec[e + f*x])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (4*c^4*Tan[e + f*x])/(a^2*f*
(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.143807, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3912, 88} \[ \frac{4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^4 \tan (e+f x)}{a^2 f (\sec (e+f x)+1)^2 \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{2 c^4 \tan (e+f x) \log (\sec (e+f x)+1)}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c^4 \tan (e+f x) \log (\cos (e+f x))}{a^2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(c^4*Log[Cos[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (2*c^4*Log[1
+ Sec[e + f*x]]*Tan[e + f*x])/(a^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (4*c^4*Tan[e + f*x])
/(a^2*f*(1 + Sec[e + f*x])^2*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (4*c^4*Tan[e + f*x])/(a^2*f*
(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^(n - 1/2))/x, x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c-c x)^3}{x (a+a x)^3} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \left (\frac{c^3}{a^3 x}-\frac{8 c^3}{a^3 (1+x)^3}+\frac{4 c^3}{a^3 (1+x)^2}-\frac{2 c^3}{a^3 (1+x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=\frac{c^4 \log (\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{2 c^4 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x))^2 \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{4 c^4 \tan (e+f x)}{a^2 f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 2.39511, size = 157, normalized size = 0.71 \[ \frac{c^3 \cot \left (\frac{1}{2} (e+f x)\right ) \sqrt{c-c \sec (e+f x)} \left (4 \left (-4 \log \left (1+e^{i (e+f x)}\right )+\log \left (1+e^{2 i (e+f x)}\right )+i f x-2\right ) \cos (e+f x)+\left (-4 \log \left (1+e^{i (e+f x)}\right )+\log \left (1+e^{2 i (e+f x)}\right )+i f x\right ) (\cos (2 (e+f x))+3)\right )}{2 a^2 f (\cos (e+f x)+1)^2 \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sec[e + f*x])^(7/2)/(a + a*Sec[e + f*x])^(5/2),x]

[Out]

(c^3*Cot[(e + f*x)/2]*(4*Cos[e + f*x]*(-2 + I*f*x - 4*Log[1 + E^(I*(e + f*x))] + Log[1 + E^((2*I)*(e + f*x))])
+ (3 + Cos[2*(e + f*x)])*(I*f*x - 4*Log[1 + E^(I*(e + f*x))] + Log[1 + E^((2*I)*(e + f*x))]))*Sqrt[c - c*Sec[
e + f*x]])/(2*a^2*f*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [A]  time = 0.26, size = 335, normalized size = 1.5 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{4}}{f{a}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( -1+\cos \left ( fx+e \right ) \right ) } \left ( \ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( \cos \left ( fx+e \right ) \right ) ^{2}+2\,\cos \left ( fx+e \right ) \ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\cos \left ( fx+e \right ) \ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +2\,\cos \left ( fx+e \right ) \ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) -2\,\cos \left ( fx+e \right ) +\ln \left ({\frac{1-\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( -{\frac{-1+\cos \left ( fx+e \right ) +\sin \left ( fx+e \right ) }{\sin \left ( fx+e \right ) }} \right ) +\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) +1 \right ) \left ({\frac{c \left ( -1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{7}{2}}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x)

[Out]

1/f/a^3*(ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(
f*x+e)^2+ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+cos(f*x+e)^2+2*cos(f*x+e)*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+
2*cos(f*x+e)*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+2*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-2*cos(f*x+e)+ln((1-c
os(f*x+e)+sin(f*x+e))/sin(f*x+e))+ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))+ln(2/(1+cos(f*x+e)))+1)*(c*(-1+co
s(f*x+e))/cos(f*x+e))^(7/2)*cos(f*x+e)^4*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)/sin(f*x+e)^5/(-1+cos(f*x+e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (c^{3} \sec \left (f x + e\right )^{3} - 3 \, c^{3} \sec \left (f x + e\right )^{2} + 3 \, c^{3} \sec \left (f x + e\right ) - c^{3}\right )} \sqrt{a \sec \left (f x + e\right ) + a} \sqrt{-c \sec \left (f x + e\right ) + c}}{a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError