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

Optimal. Leaf size=241 \[ \frac{2 a^2 \tan (e+f x) \left (d \left (24 c^2+111 c d+52 d^2\right ) \sec (e+f x)+2 \left (243 c^2 d+36 c^3+189 c d^2+52 d^3\right )\right )}{105 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{5/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c+d \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 (6 c+13 d) \tan (e+f x) (c+d \sec (e+f x))^2}{35 f \sqrt{a \sec (e+f x)+a}} \]

[Out]

(2*a^(5/2)*c^3*ArcTanh[Sqrt[a - a*Sec[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^2*(6*c + 13*d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(35*f*Sqrt[a + a*Sec[e + f*x]]) + (2
*a^2*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(2*(36*c^3 + 243*c^2*d + 189
*c*d^2 + 52*d^3) + d*(24*c^2 + 111*c*d + 52*d^2)*Sec[e + f*x])*Tan[e + f*x])/(105*f*Sqrt[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.199716, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3940, 153, 147, 63, 206} \[ \frac{2 a^2 \tan (e+f x) \left (d \left (24 c^2+111 c d+52 d^2\right ) \sec (e+f x)+2 \left (243 c^2 d+36 c^3+189 c d^2+52 d^3\right )\right )}{105 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^{5/2} c^3 \tan (e+f x) \tanh ^{-1}\left (\frac{\sqrt{a-a \sec (e+f x)}}{\sqrt{a}}\right )}{f \sqrt{a-a \sec (e+f x)} \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c+d \sec (e+f x))^3}{7 f \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 (6 c+13 d) \tan (e+f x) (c+d \sec (e+f x))^2}{35 f \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^(5/2)*c^3*ArcTanh[Sqrt[a - a*Sec[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^2*(6*c + 13*d)*(c + d*Sec[e + f*x])^2*Tan[e + f*x])/(35*f*Sqrt[a + a*Sec[e + f*x]]) + (2
*a^2*(c + d*Sec[e + f*x])^3*Tan[e + f*x])/(7*f*Sqrt[a + a*Sec[e + f*x]]) + (2*a^2*(2*(36*c^3 + 243*c^2*d + 189
*c*d^2 + 52*d^3) + d*(24*c^2 + 111*c*d + 52*d^2)*Sec[e + f*x])*Tan[e + f*x])/(105*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 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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))^{3/2} (c+d \sec (e+f x))^3 \, dx &=-\frac{\left (a^2 \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(a+a x) (c+d x)^3}{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^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{(2 a \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c+d x)^2 \left (-\frac{7 a^2 c}{2}-\frac{1}{2} a^2 (6 c+13 d) x\right )}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{7 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (6 c+13 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}-\frac{(4 \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c+d x) \left (\frac{35 a^3 c^2}{4}+\frac{1}{4} a^3 \left (24 c^2+111 c d+52 d^2\right ) x\right )}{x \sqrt{a-a x}} \, dx,x,\sec (e+f x)\right )}{35 f \sqrt{a-a \sec (e+f x)} \sqrt{a+a \sec (e+f x)}}\\ &=\frac{2 a^2 (6 c+13 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (36 c^3+243 c^2 d+189 c d^2+52 d^3\right )+d \left (24 c^2+111 c d+52 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{105 f \sqrt{a+a \sec (e+f x)}}-\frac{\left (a^3 c^3 \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^2 (6 c+13 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (36 c^3+243 c^2 d+189 c d^2+52 d^3\right )+d \left (24 c^2+111 c d+52 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{105 f \sqrt{a+a \sec (e+f x)}}+\frac{\left (2 a^2 c^3 \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^{5/2} c^3 \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^2 (6 c+13 d) (c+d \sec (e+f x))^2 \tan (e+f x)}{35 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 (c+d \sec (e+f x))^3 \tan (e+f x)}{7 f \sqrt{a+a \sec (e+f x)}}+\frac{2 a^2 \left (2 \left (36 c^3+243 c^2 d+189 c d^2+52 d^3\right )+d \left (24 c^2+111 c d+52 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{105 f \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 3.86925, size = 219, normalized size = 0.91 \[ \frac{a \sec \left (\frac{1}{2} (e+f x)\right ) \sec ^3(e+f x) \sqrt{a (\sec (e+f x)+1)} \left (2 \sin \left (\frac{1}{2} (e+f x)\right ) \left (9 \left (175 c^2 d+35 c^3+154 c d^2+52 d^3\right ) \cos (e+f x)+2 d \left (105 c^2+189 c d+52 d^2\right ) \cos (2 (e+f x))+525 c^2 d \cos (3 (e+f x))+210 c^2 d+105 c^3 \cos (3 (e+f x))+378 c d^2 \cos (3 (e+f x))+378 c d^2+104 d^3 \cos (3 (e+f x))+164 d^3\right )+420 \sqrt{2} c^3 \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (e+f x)\right )\right ) \cos ^{\frac{7}{2}}(e+f x)\right )}{420 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[(e + f*x)/2]*Sec[e + f*x]^3*Sqrt[a*(1 + Sec[e + f*x])]*(420*Sqrt[2]*c^3*ArcSin[Sqrt[2]*Sin[(e + f*x)/2]
]*Cos[e + f*x]^(7/2) + 2*(210*c^2*d + 378*c*d^2 + 164*d^3 + 9*(35*c^3 + 175*c^2*d + 154*c*d^2 + 52*d^3)*Cos[e
+ f*x] + 2*d*(105*c^2 + 189*c*d + 52*d^2)*Cos[2*(e + f*x)] + 105*c^3*Cos[3*(e + f*x)] + 525*c^2*d*Cos[3*(e + f
*x)] + 378*c*d^2*Cos[3*(e + f*x)] + 104*d^3*Cos[3*(e + f*x)])*Sin[(e + f*x)/2]))/(420*f)

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Maple [B]  time = 0.316, size = 539, normalized size = 2.2 \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))^(3/2)*(c+d*sec(f*x+e))^3,x)

[Out]

1/840/f*a*(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*arctanh(1/2*2^(1/2)*(-2*c
os(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)))^(7/2)*c^3+315*2^(1/2)*si
n(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*cos(
f*x+e)/(1+cos(f*x+e)))^(7/2)*c^3+315*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)))^(7/2)*c^3+105*(-2*cos(f*x+e)/(1+cos(f*x+e)
))^(7/2)*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))*c^3*sin(f*x+e
)-1680*cos(f*x+e)^4*c^3-8400*cos(f*x+e)^4*c^2*d-6048*cos(f*x+e)^4*c*d^2-1664*cos(f*x+e)^4*d^3+1680*cos(f*x+e)^
3*c^3+6720*cos(f*x+e)^3*c^2*d+3024*cos(f*x+e)^3*c*d^2+832*cos(f*x+e)^3*d^3+1680*cos(f*x+e)^2*c^2*d+2016*cos(f*
x+e)^2*c*d^2+208*cos(f*x+e)^2*d^3+1008*cos(f*x+e)*c*d^2+384*cos(f*x+e)*d^3+240*d^3)/cos(f*x+e)^3/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+d*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 1.34508, size = 1230, normalized size = 5.1 \begin{align*} \left [\frac{105 \,{\left (a c^{3} \cos \left (f x + e\right )^{4} + a c^{3} \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 d^{3} +{\left (105 \, a c^{3} + 525 \, a c^{2} d + 378 \, a c d^{2} + 104 \, a d^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (105 \, a c^{2} d + 189 \, a c d^{2} + 52 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (21 \, a c d^{2} + 13 \, a d^{3}\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 c^{3} \cos \left (f x + e\right )^{4} + a c^{3} \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 d^{3} +{\left (105 \, a c^{3} + 525 \, a c^{2} d + 378 \, a c d^{2} + 104 \, a d^{3}\right )} \cos \left (f x + e\right )^{3} +{\left (105 \, a c^{2} d + 189 \, a c d^{2} + 52 \, a d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \,{\left (21 \, a c d^{2} + 13 \, a d^{3}\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))^(3/2)*(c+d*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

[1/105*(105*(a*c^3*cos(f*x + e)^4 + a*c^3*cos(f*x + e)^3)*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*(15*
a*d^3 + (105*a*c^3 + 525*a*c^2*d + 378*a*c*d^2 + 104*a*d^3)*cos(f*x + e)^3 + (105*a*c^2*d + 189*a*c*d^2 + 52*a
*d^3)*cos(f*x + e)^2 + 3*(21*a*c*d^2 + 13*a*d^3)*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*c^3*cos(f*x + e)^4 + a*c^3*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))) - (15*a*d^3 + (105*a*c^3
+ 525*a*c^2*d + 378*a*c*d^2 + 104*a*d^3)*cos(f*x + e)^3 + (105*a*c^2*d + 189*a*c*d^2 + 52*a*d^3)*cos(f*x + e)
^2 + 3*(21*a*c*d^2 + 13*a*d^3)*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))**(3/2)*(c+d*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))^(3/2)*(c+d*sec(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out