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

Optimal. Leaf size=258 \[ -\frac{2 \left (c^2+8 c d+8 d^2\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{7/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 a d (2 c+5 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}}-\frac{2 d^2 \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^3*(c + 2*d)*(3*c + 2*d)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*c^2*ArcTanh[Sqrt[a - a*Se
c[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (2*a*d*(2*c + 5*d)*
(a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(a - a*Sec[e + f*x])^3*Tan[e + f*
x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) - (2*(c^2 + 8*c*d + 8*d^2)*(a^3 - a^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt
[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.17731, antiderivative size = 258, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3940, 180, 63, 206} \[ -\frac{2 \left (c^2+8 c d+8 d^2\right ) \tan (e+f x) \left (a^3-a^3 \sec (e+f x)\right )}{3 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{7/2} c^2 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^3 (c+2 d) (3 c+2 d) \tan (e+f x)}{f \sqrt{a \sec (e+f x)+a}}+\frac{2 a d (2 c+5 d) \tan (e+f x) (a-a \sec (e+f x))^2}{5 f \sqrt{a \sec (e+f x)+a}}-\frac{2 d^2 \tan (e+f x) (a-a \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(c + 2*d)*(3*c + 2*d)*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^(7/2)*c^2*ArcTanh[Sqrt[a - a*Se
c[e + f*x]]/Sqrt[a]]*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]) + (2*a*d*(2*c + 5*d)*
(a - a*Sec[e + f*x])^2*Tan[e + f*x])/(5*f*Sqrt[a + a*Sec[e + f*x]]) - (2*d^2*(a - a*Sec[e + f*x])^3*Tan[e + f*
x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) - (2*(c^2 + 8*c*d + 8*d^2)*(a^3 - a^3*Sec[e + f*x])*Tan[e + f*x])/(3*f*Sqrt
[a + a*Sec[e + f*x]])

Rule 3940

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

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 2.68358, size = 191, normalized size = 0.74 \[ \frac{a^2 \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (4 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\left (420 c^2+987 c d+465 d^2\right ) \cos (e+f x)+\left (35 c^2+196 c d+115 d^2\right ) \cos (2 (e+f x))+140 c^2 \cos (3 (e+f x))+35 c^2+301 c d \cos (3 (e+f x))+196 c d+115 d^2 \cos (3 (e+f x))+145 d^2\right )+420 \sqrt{2} c^2 \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \cos ^{\frac{7}{2}}(e+f x)\right )}{420 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*(420*Sqrt[2]*c^2*ArcSin[Sqrt[2]*Sin[(e + f*x)/
2]]*Cos[e + f*x]^(7/2) + 4*(35*c^2 + 196*c*d + 145*d^2 + (420*c^2 + 987*c*d + 465*d^2)*Cos[e + f*x] + (35*c^2
+ 196*c*d + 115*d^2)*Cos[2*(e + f*x)] + 140*c^2*Cos[3*(e + f*x)] + 301*c*d*Cos[3*(e + f*x)] + 115*d^2*Cos[3*(e
+ f*x)])*Sin[(e + f*x)/2]))/(420*f)

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Maple [B]  time = 0.283, size = 504, normalized size = 2. \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+d*sec(f*x+e))^2,x)

[Out]

1/840/f*a^2*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(105*sin(f*x+e)*cos(f*x+e)^3*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f
*x+e)))^(7/2)*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))*c^2+315*sin(f*x+
e)*cos(f*x+e)^2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*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))*c^2+315*sin(f*x+e)*cos(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(7/2)*arct
anh(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))*c^2+105*(-2*cos(f*x+e)/(1+cos(f*x+
e)))^(7/2)*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^(1/2)*c^2*sin(f*x
+e)-4480*cos(f*x+e)^4*c^2-9632*cos(f*x+e)^4*c*d-3680*cos(f*x+e)^4*d^2+3920*cos(f*x+e)^3*c^2+6496*cos(f*x+e)^3*
c*d+1840*cos(f*x+e)^3*d^2+560*cos(f*x+e)^2*c^2+2464*cos(f*x+e)^2*c*d+880*cos(f*x+e)^2*d^2+672*cos(f*x+e)*c*d+7
20*cos(f*x+e)*d^2+240*d^2)/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+d*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.10297, size = 1233, normalized size = 4.78 \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 (15 \, a^{2} d^{2} + 2 \,{\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\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 (15 \, a^{2} d^{2} + 2 \,{\left (140 \, a^{2} c^{2} + 301 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{3} +{\left (35 \, a^{2} c^{2} + 196 \, a^{2} c d + 115 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (7 \, a^{2} c d + 10 \, a^{2} d^{2}\right )} \cos \left (f x + e\right )\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+d*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*
(15*a^2*d^2 + 2*(140*a^2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + (35*a^2*c^2 + 196*a^2*c*d + 115*a^2
*d^2)*cos(f*x + e)^2 + 6*(7*a^2*c*d + 10*a^2*d^2)*cos(f*x + e))*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)*s
qrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - (15*a^2*d^2 + 2*(
140*a^2*c^2 + 301*a^2*c*d + 115*a^2*d^2)*cos(f*x + e)^3 + (35*a^2*c^2 + 196*a^2*c*d + 115*a^2*d^2)*cos(f*x + e
)^2 + 6*(7*a^2*c*d + 10*a^2*d^2)*cos(f*x + e))*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)]

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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+d*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+d*sec(f*x+e))^2,x, algorithm="giac")

[Out]

Timed out