\(\int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [195]

Optimal result

Integrand size = 33, antiderivative size = 211 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {4 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]

[Out]

4/15*a^3*(20*A+21*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/5*a*B*(a+a*sec(d*x+c))^2*sin(d*x+c)*sec(d*x+c)^(1/2)/d+2/
15*(5*A+9*B)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)*sec(d*x+c)^(1/2)/d-4/5*a^3*(5*A+9*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)
/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4/3*a^3*(5*A+3*B
)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d
*x+c)^(1/2)/d

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4103, 4082, 3872, 3856, 2719, 2720} \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {4 a^3 (20 A+21 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{15 d}+\frac {2 (5 A+9 B) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {4 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}-\frac {4 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}{5 d} \]

[In]

Int[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]

[Out]

(-4*a^3*(5*A + 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*(5*A + 3*B
)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (4*a^3*(20*A + 21*B)*Sqrt[Sec[c + d
*x]]*Sin[c + d*x])/(15*d) + (2*a*B*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + (2*(5*A + 9
*B)*Sqrt[Sec[c + d*x]]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d)

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4103

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m +
n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d
*n) + (A*b*d*(m + n) + a*B*d*(2*m + n - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] &&
NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2}{5} \int \frac {(a+a \sec (c+d x))^2 \left (\frac {1}{2} a (5 A-B)+\frac {1}{2} a (5 A+9 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {4}{15} \int \frac {(a+a \sec (c+d x)) \left (\frac {1}{2} a^2 (5 A-6 B)+\frac {1}{2} a^2 (20 A+21 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {8}{15} \int \frac {-\frac {3}{4} a^3 (5 A+9 B)+\frac {5}{4} a^3 (5 A+3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{3} \left (2 a^3 (5 A+3 B)\right ) \int \sqrt {\sec (c+d x)} \, dx-\frac {1}{5} \left (2 a^3 (5 A+9 B)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx \\ & = \frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{3} \left (2 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx-\frac {1}{5} \left (2 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^3 (5 A+9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (5 A+3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {4 a^3 (20 A+21 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {2 (5 A+9 B) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 3.35 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.16 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {5}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (-90 i A \cos (c+d x)-162 i B \cos (c+d x)-30 i A \cos (3 (c+d x))-54 i B \cos (3 (c+d x))+40 (5 A+3 B) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 i (5 A+9 B) e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+45 A \sin (c+d x)+66 B \sin (c+d x)+10 A \sin (2 (c+d x))+30 B \sin (2 (c+d x))+45 A \sin (3 (c+d x))+54 B \sin (3 (c+d x))\right )}{30 d} \]

[In]

Integrate[((a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]

[Out]

(a^3*Sec[c + d*x]^(5/2)*(Cos[d*x] + I*Sin[d*x])*((-90*I)*A*Cos[c + d*x] - (162*I)*B*Cos[c + d*x] - (30*I)*A*Co
s[3*(c + d*x)] - (54*I)*B*Cos[3*(c + d*x)] + 40*(5*A + 3*B)*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + ((2
*I)*(5*A + 9*B)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^(I*(
c + d*x)) + 45*A*Sin[c + d*x] + 66*B*Sin[c + d*x] + 10*A*Sin[2*(c + d*x)] + 30*B*Sin[2*(c + d*x)] + 45*A*Sin[3
*(c + d*x)] + 54*B*Sin[3*(c + d*x)]))/(30*d*E^(I*d*x))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(915\) vs. \(2(239)=478\).

Time = 39.51 (sec) , antiderivative size = 916, normalized size of antiderivative = 4.34

method	result	size
default	\(\text {Expression too large to display}\)	\(916\)
parts	\(\text {Expression too large to display}\)	\(1061\)

[In]

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-4/15*a^3*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2
*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^3*(180*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-100*A*(sin
(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/
2*c)^4-60*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*sin(1/2*d*x+1/2*c)^4+216*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-60*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Ellipti
cF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-108*B*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-190*A*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+100*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))
*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+60*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2-246*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^4+60*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2
)*sin(1/2*d*x+1/2*c)^2+108*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x
+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^2+50*A*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-25*A*(2*sin(1/2*d*x+1/2*c
)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-15*A*(2*sin(1/2*d*x+1/2*c)^2-1
)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+72*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^2-15*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1
/2))-27*B*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))
*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.15 \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (5 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (5 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (9 \, {\left (5 \, A + 6 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 5 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 3 \, B a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )^{2}} \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(5*I*sqrt(2)*(5*A + 3*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) -
5*I*sqrt(2)*(5*A + 3*B)*a^3*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*I*sqr
t(2)*(5*A + 9*B)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x
+ c))) - 3*I*sqrt(2)*(5*A + 9*B)*a^3*cos(d*x + c)^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x
+ c) - I*sin(d*x + c))) - (9*(5*A + 6*B)*a^3*cos(d*x + c)^2 + 5*(A + 3*B)*a^3*cos(d*x + c) + 3*B*a^3)*sin(d*x
+ c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)

Giac [F]

\[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sqrt {\sec \left (d x + c\right )}} \,d x } \]

[In]

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^3/sqrt(sec(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx=\int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]

[In]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2),x)

[Out]

int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^3)/(1/cos(c + d*x))^(1/2), x)