3.51 \(\int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=142 \[ \frac{2 a^4 c^2 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^3 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{3/2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^2 c^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^(3/2)*c^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^2*c^2*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^3*c^2*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) + (2*a^4*c^2*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2))

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Rubi [A]  time = 0.170043, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3904, 3887, 459, 302, 203} \[ \frac{2 a^4 c^2 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^3 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{3/2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^2 c^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(3/2)*c^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^2*c^2*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^3*c^2*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) + (2*a^4*c^2*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2))

Rule 3904

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

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 459

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

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.84158, size = 112, normalized size = 0.79 \[ -\frac{a c^2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2(e+f x) \sqrt{a (\sec (e+f x)+1)} \left ((2 \cos (e+f x)+17 \cos (2 (e+f x))+11) \sqrt{\sec (e+f x)-1}-30 \cos ^2(e+f x) \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )\right )}{15 f \sqrt{\sec (e+f x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*c^2*(-30*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]^2 + (11 + 2*Cos[e + f*x] + 17*Cos[2*(e + f*x)])*Sqrt
[-1 + Sec[e + f*x]])*Sec[e + f*x]^2*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(15*f*Sqrt[-1 + Sec[e + f*x]]
)

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Maple [A]  time = 0.247, size = 232, normalized size = 1.6 \begin{align*}{\frac{{c}^{2}a}{30\,f \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) }\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+15\,{\it Artanh} \left ( 1/2\,{\frac{\sqrt{2}\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }\sqrt{-2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }}}} \right ) \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{3/2}\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) +68\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}-64\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-16\,\cos \left ( fx+e \right ) +12 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*c^2/f*a*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(15*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2
)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2+15*arctanh(1/2*2
^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(3/2)*2^(1/2
)*sin(f*x+e)*cos(f*x+e)+68*cos(f*x+e)^3-64*cos(f*x+e)^2-16*cos(f*x+e)+12)/cos(f*x+e)^2/sin(f*x+e)

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Maxima [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((a+a*sec(f*x+e))^(3/2)*(c-c*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.19576, size = 900, normalized size = 6.34 \begin{align*} \left [\frac{15 \,{\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt{-a} \log \left (\frac{2 \, a \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) - 2 \,{\left (17 \, a c^{2} \cos \left (f x + e\right )^{2} + a c^{2} \cos \left (f x + e\right ) - 3 \, a c^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}, -\frac{2 \,{\left (15 \,{\left (a c^{2} \cos \left (f x + e\right )^{3} + a c^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt{a} \sin \left (f x + e\right )}\right ) +{\left (17 \, a c^{2} \cos \left (f x + e\right )^{2} + a c^{2} \cos \left (f x + e\right ) - 3 \, a c^{2}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{15 \,{\left (f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/15*(15*(a*c^2*cos(f*x + e)^3 + a*c^2*cos(f*x + e)^2)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sqrt((a*
cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 2*(17*a*
c^2*cos(f*x + e)^2 + a*c^2*cos(f*x + e) - 3*a*c^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*co
s(f*x + e)^3 + f*cos(f*x + e)^2), -2/15*(15*(a*c^2*cos(f*x + e)^3 + a*c^2*cos(f*x + e)^2)*sqrt(a)*arctan(sqrt(
(a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (17*a*c^2*cos(f*x + e)^2 + a*c^2*cos
(f*x + e) - 3*a*c^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^3 + f*cos(f*x + e)^
2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*(Integral(a*sqrt(a*sec(e + f*x) + a), x) + Integral(-a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x), x) + Integr
al(-a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2, x) + Integral(a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**3, x))

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Giac [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((a+a*sec(f*x+e))^(3/2)*(c-c*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out