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

Optimal. Leaf size=204 \[ \frac{c^4 \tan (e+f x) \sec ^2(e+f x)}{2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^4 \tan (e+f x) \sec (e+f x)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{8 c^4 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c^4 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

[Out]

(c^4*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (8*c^4*Log[1 + Se
c[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (4*c^4*Sec[e + f*x]*Tan[e +
f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (c^4*Sec[e + f*x]^2*Tan[e + f*x])/(2*f*Sqrt[a +
a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

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Rubi [A]  time = 0.123592, antiderivative size = 204, 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, 72} \[ \frac{c^4 \tan (e+f x) \sec ^2(e+f x)}{2 f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}-\frac{4 c^4 \tan (e+f x) \sec (e+f x)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{8 c^4 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}}+\frac{c^4 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt{a \sec (e+f x)+a} \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^4*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (8*c^4*Log[1 + Se
c[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (4*c^4*Sec[e + f*x]*Tan[e +
f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (c^4*Sec[e + f*x]^2*Tan[e + f*x])/(2*f*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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

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

Mathematica [C]  time = 15.3406, size = 153, normalized size = 0.75 \[ \frac{c^3 \cot \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{c-c \sec (e+f x)} \left (-16 \log \left (1+e^{i (e+f x)}\right )+7 \log \left (1+e^{2 i (e+f x)}\right )+8 \cos (e+f x)+\left (-16 \log \left (1+e^{i (e+f x)}\right )+7 \log \left (1+e^{2 i (e+f x)}\right )+i f x\right ) \cos (2 (e+f x))+i f x-1\right )}{2 f \sqrt{a (\sec (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Cot[(e + f*x)/2]*(-1 + I*f*x + 8*Cos[e + f*x] - 16*Log[1 + E^(I*(e + f*x))] + 7*Log[1 + E^((2*I)*(e + f*x
))] + Cos[2*(e + f*x)]*(I*f*x - 16*Log[1 + E^(I*(e + f*x))] + 7*Log[1 + E^((2*I)*(e + f*x))]))*Sec[e + f*x]^2*
Sqrt[c - c*Sec[e + f*x]])/(2*f*Sqrt[a*(1 + Sec[e + f*x])])

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Maple [A]  time = 0.309, size = 189, normalized size = 0.9 \begin{align*}{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{2\,af\sin \left ( fx+e \right ) \left ( -1+\cos \left ( fx+e \right ) \right ) ^{3}} \left ( 14\,\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}+14\,\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}+2\,\ln \left ( 2\, \left ( 1+\cos \left ( fx+e \right ) \right ) ^{-1} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+9\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}+8\,\cos \left ( fx+e \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))^(1/2),x)

[Out]

1/2/f/a*(14*ln((1-cos(f*x+e)+sin(f*x+e))/sin(f*x+e))*cos(f*x+e)^2+14*ln(-(-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e)
)*cos(f*x+e)^2+2*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+9*cos(f*x+e)^2+8*cos(f*x+e)-1)*(c*(-1+cos(f*x+e))/cos(f*x+e
))^(7/2)*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*cos(f*x+e)^2/sin(f*x+e)/(-1+cos(f*x+e))^3

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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))^(1/2),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{-c \sec \left (f x + e\right ) + c}}{\sqrt{a \sec \left (f x + e\right ) + a}}, 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))^(1/2),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(-c*sec(f*x + e) + c)/sqr
t(a*sec(f*x + e) + a), 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))**(1/2),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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError