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

Optimal. Leaf size=177 \[ \frac{2 a^6 c^2 \tan ^7(e+f x)}{7 f (a \sec (e+f x)+a)^{7/2}}+\frac{6 a^5 c^2 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^4 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^(5/2)*c^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^3*c^2*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^4*c^2*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) + (6*a^5*c^2*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2)) + (2*a^6*c^2*Tan[e + f*x]^7)/(7*f*(a + a*Sec[e + f*x])^(7/2))

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Rubi [A]  time = 0.174306, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3904, 3887, 461, 203} \[ \frac{2 a^6 c^2 \tan ^7(e+f x)}{7 f (a \sec (e+f x)+a)^{7/2}}+\frac{6 a^5 c^2 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^4 c^2 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c^2 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*c^2*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^3*c^2*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^4*c^2*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) + (6*a^5*c^2*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2)) + (2*a^6*c^2*Tan[e + f*x]^7)/(7*f*(a + a*Sec[e + f*x])^(7/2))

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 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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))^{5/2} (c-c \sec (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \sqrt{a+a \sec (e+f x)} \tan ^4(e+f x) \, dx\\ &=-\frac{\left (2 a^5 c^2\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (2+a x^2\right )^2}{1+a x^2} \, dx,x,-\frac{\tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}\\ &=-\frac{\left (2 a^5 c^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}+\frac{x^2}{a}+3 x^4+a x^6+\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^3 c^2 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^4 c^2 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}+\frac{6 a^5 c^2 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}+\frac{2 a^6 c^2 \tan ^7(e+f x)}{7 f (a+a \sec (e+f x))^{7/2}}-\frac{\left (2 a^3 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^{5/2} c^2 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}-\frac{2 a^3 c^2 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^4 c^2 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}+\frac{6 a^5 c^2 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}+\frac{2 a^6 c^2 \tan ^7(e+f x)}{7 f (a+a \sec (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.921237, size = 124, normalized size = 0.7 \[ -\frac{2 a^2 c^2 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \left ((51 \cos (e+f x)+23 \cos (2 (e+f x))+23 \cos (3 (e+f x))+8) \sqrt{\sec (e+f x)-1}-105 \cos ^3(e+f x) \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )\right )}{105 f \sqrt{\sec (e+f x)-1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*c^2*(-105*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]^3 + (8 + 51*Cos[e + f*x] + 23*Cos[2*(e + f*x)]
+ 23*Cos[3*(e + f*x)])*Sqrt[-1 + Sec[e + f*x]])*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(1
05*f*Sqrt[-1 + Sec[e + f*x]])

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Maple [B]  time = 0.26, size = 323, normalized size = 1.8 \begin{align*} -{\frac{{c}^{2}{a}^{2}}{420\,f\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( 1+\cos \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) }}} \left ( 105\,\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{3} \left ( -2\,{\frac{\cos \left ( fx+e \right ) }{1+\cos \left ( fx+e \right ) }} \right ) ^{5/2}{\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 ) +210\,\sqrt{2}\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}{\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 ) ^{5/2}+105\,\sqrt{2}\sin \left ( fx+e \right ) \cos \left ( fx+e \right ){\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 ) ^{5/2}-736\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+368\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}+512\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-24\,\cos \left ( fx+e \right ) -120 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*c^2/f*a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(105*2^(1/2)*sin(f*x+e)*cos(f*x+e)^3*(-2*cos(f*x+e)/(1+
cos(f*x+e)))^(5/2)*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))+210*2^(1/2)
*sin(f*x+e)*cos(f*x+e)^2*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*c
os(f*x+e)/(1+cos(f*x+e)))^(5/2)+105*2^(1/2)*sin(f*x+e)*cos(f*x+e)*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)))^(5/2)-736*cos(f*x+e)^4+368*cos(f*x+e)^3+512
*cos(f*x+e)^2-24*cos(f*x+e)-120)/sin(f*x+e)/cos(f*x+e)^3

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

[Out]

Timed out

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Fricas [A]  time = 1.23743, size = 1019, normalized size = 5.76 \begin{align*} \left [\frac{105 \,{\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\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 (92 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} + 46 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - 18 \, a^{2} c^{2} \cos \left (f x + e\right ) - 15 \, a^{2} 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 )}{105 \,{\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}, -\frac{2 \,{\left (105 \,{\left (a^{2} c^{2} \cos \left (f x + e\right )^{4} + a^{2} c^{2} \cos \left (f x + e\right )^{3}\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 (92 \, a^{2} c^{2} \cos \left (f x + e\right )^{3} + 46 \, a^{2} c^{2} \cos \left (f x + e\right )^{2} - 18 \, a^{2} c^{2} \cos \left (f x + e\right ) - 15 \, a^{2} 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 )}}{105 \,{\left (f \cos \left (f x + e\right )^{4} + f \cos \left (f x + e\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/105*(105*(a^2*c^2*cos(f*x + e)^4 + a^2*c^2*cos(f*x + e)^3)*sqrt(-a)*log((2*a*cos(f*x + e)^2 - 2*sqrt(-a)*sq
rt((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*
(92*a^2*c^2*cos(f*x + e)^3 + 46*a^2*c^2*cos(f*x + e)^2 - 18*a^2*c^2*cos(f*x + e) - 15*a^2*c^2)*sqrt((a*cos(f*x
+ e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3), -2/105*(105*(a^2*c^2*cos(f*x + e
)^4 + a^2*c^2*cos(f*x + e)^3)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin
(f*x + e))) + (92*a^2*c^2*cos(f*x + e)^3 + 46*a^2*c^2*cos(f*x + e)^2 - 18*a^2*c^2*cos(f*x + e) - 15*a^2*c^2)*s
qrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 + f*cos(f*x + e)^3)]

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

[Out]

Timed out

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

[Out]

Timed out