3.57 \(\int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=212 \[ -\frac{2 a^7 c^3 \tan ^9(e+f x)}{9 f (a \sec (e+f x)+a)^{9/2}}-\frac{6 a^6 c^3 \tan ^7(e+f x)}{7 f (a \sec (e+f x)+a)^{7/2}}-\frac{2 a^5 c^3 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^4 c^3 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c^3 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^(5/2)*c^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^3*c^3*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^4*c^3*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) - (2*a^5*c^3*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2)) - (6*a^6*c^3*Tan[e + f*x]^7)/(7*f*(a + a*Sec[e + f*x])^(7/2)) - (2*a^7*c^3*T
an[e + f*x]^9)/(9*f*(a + a*Sec[e + f*x])^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.190908, antiderivative size = 212, 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} \[ -\frac{2 a^7 c^3 \tan ^9(e+f x)}{9 f (a \sec (e+f x)+a)^{9/2}}-\frac{6 a^6 c^3 \tan ^7(e+f x)}{7 f (a \sec (e+f x)+a)^{7/2}}-\frac{2 a^5 c^3 \tan ^5(e+f x)}{5 f (a \sec (e+f x)+a)^{5/2}}+\frac{2 a^4 c^3 \tan ^3(e+f x)}{3 f (a \sec (e+f x)+a)^{3/2}}+\frac{2 a^{5/2} c^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{f}-\frac{2 a^3 c^3 \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^3,x]

[Out]

(2*a^(5/2)*c^3*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/f - (2*a^3*c^3*Tan[e + f*x])/(f*Sqrt[a
+ a*Sec[e + f*x]]) + (2*a^4*c^3*Tan[e + f*x]^3)/(3*f*(a + a*Sec[e + f*x])^(3/2)) - (2*a^5*c^3*Tan[e + f*x]^5)
/(5*f*(a + a*Sec[e + f*x])^(5/2)) - (6*a^6*c^3*Tan[e + f*x]^7)/(7*f*(a + a*Sec[e + f*x])^(7/2)) - (2*a^7*c^3*T
an[e + f*x]^9)/(9*f*(a + a*Sec[e + f*x])^(9/2))

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))^3 \, dx &=-\left (\left (a^3 c^3\right ) \int \frac{\tan ^6(e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx\right )\\ &=\frac{\left (2 a^6 c^3\right ) \operatorname{Subst}\left (\int \frac{x^6 \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^6 c^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a^3}-\frac{x^2}{a^2}+\frac{x^4}{a}+3 x^6+a x^8-\frac{1}{a^3 \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^3 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^4 c^3 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac{2 a^5 c^3 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}-\frac{6 a^6 c^3 \tan ^7(e+f x)}{7 f (a+a \sec (e+f x))^{7/2}}-\frac{2 a^7 c^3 \tan ^9(e+f x)}{9 f (a+a \sec (e+f x))^{9/2}}-\frac{\left (2 a^3 c^3\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^3 \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{f}-\frac{2 a^3 c^3 \tan (e+f x)}{f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^4 c^3 \tan ^3(e+f x)}{3 f (a+a \sec (e+f x))^{3/2}}-\frac{2 a^5 c^3 \tan ^5(e+f x)}{5 f (a+a \sec (e+f x))^{5/2}}-\frac{6 a^6 c^3 \tan ^7(e+f x)}{7 f (a+a \sec (e+f x))^{7/2}}-\frac{2 a^7 c^3 \tan ^9(e+f x)}{9 f (a+a \sec (e+f x))^{9/2}}\\ \end{align*}

Mathematica [A]  time = 1.21355, size = 134, normalized size = 0.63 \[ -\frac{a^2 c^3 \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^4(e+f x) \sqrt{a (\sec (e+f x)+1)} \left ((164 \cos (e+f x)+1004 \cos (2 (e+f x))+68 \cos (3 (e+f x))+383 \cos (4 (e+f x))+901) \sqrt{\sec (e+f x)-1}-2520 \cos ^4(e+f x) \tan ^{-1}\left (\sqrt{\sec (e+f x)-1}\right )\right )}{1260 f \sqrt{\sec (e+f x)-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)*(c - c*Sec[e + f*x])^3,x]

[Out]

-(a^2*c^3*(-2520*ArcTan[Sqrt[-1 + Sec[e + f*x]]]*Cos[e + f*x]^4 + (901 + 164*Cos[e + f*x] + 1004*Cos[2*(e + f*
x)] + 68*Cos[3*(e + f*x)] + 383*Cos[4*(e + f*x)])*Sqrt[-1 + Sec[e + f*x]])*Sec[e + f*x]^4*Sqrt[a*(1 + Sec[e +
f*x])]*Tan[(e + f*x)/2])/(1260*f*Sqrt[-1 + Sec[e + f*x]])

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Maple [B]  time = 0.28, size = 483, normalized size = 2.3 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

-1/5040*c^3/f*a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(315*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*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*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+1260*sin(f
*x+e)*cos(f*x+e)^3*2^(1/2)*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
*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+1890*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+c
os(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(9/2)+1260*sin(f*x+e)*cos(f*x+e)*2^(1/
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*cos(f*x+e)/(1+cos(f*x+e
)))^(9/2)+315*2^(1/2)*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*cos(
f*x+e)/(1+cos(f*x+e)))^(9/2)*sin(f*x+e)-12256*cos(f*x+e)^5+11168*cos(f*x+e)^4+5312*cos(f*x+e)^3-4064*cos(f*x+e
)^2-1280*cos(f*x+e)+1120)/cos(f*x+e)^4/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))^(5/2)*(c-c*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.30711, size = 1098, normalized size = 5.18 \begin{align*} \left [\frac{315 \,{\left (a^{2} c^{3} \cos \left (f x + e\right )^{5} + a^{2} c^{3} \cos \left (f x + e\right )^{4}\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 (383 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 34 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 132 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 5 \, a^{2} c^{3} \cos \left (f x + e\right ) + 35 \, a^{2} c^{3}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{315 \,{\left (f \cos \left (f x + e\right )^{5} + f \cos \left (f x + e\right )^{4}\right )}}, -\frac{2 \,{\left (315 \,{\left (a^{2} c^{3} \cos \left (f x + e\right )^{5} + a^{2} c^{3} \cos \left (f x + e\right )^{4}\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 (383 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 34 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} - 132 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} - 5 \, a^{2} c^{3} \cos \left (f x + e\right ) + 35 \, a^{2} c^{3}\right )} \sqrt{\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{315 \,{\left (f \cos \left (f x + e\right )^{5} + f \cos \left (f x + e\right )^{4}\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))^3,x, algorithm="fricas")

[Out]

[1/315*(315*(a^2*c^3*cos(f*x + e)^5 + a^2*c^3*cos(f*x + e)^4)*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*
(383*a^2*c^3*cos(f*x + e)^4 + 34*a^2*c^3*cos(f*x + e)^3 - 132*a^2*c^3*cos(f*x + e)^2 - 5*a^2*c^3*cos(f*x + e)
+ 35*a^2*c^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^5 + f*cos(f*x + e)^4), -2/
315*(315*(a^2*c^3*cos(f*x + e)^5 + a^2*c^3*cos(f*x + e)^4)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x +
e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) + (383*a^2*c^3*cos(f*x + e)^4 + 34*a^2*c^3*cos(f*x + e)^3 - 132*a^2*c
^3*cos(f*x + e)^2 - 5*a^2*c^3*cos(f*x + e) + 35*a^2*c^3)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))
/(f*cos(f*x + e)^5 + f*cos(f*x + e)^4)]

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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))**3,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))^3,x, algorithm="giac")

[Out]

Timed out