3.494 \(\int \frac{(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt{\cos (c+d x)}} \, dx\)
Optimal. Leaf size=159 \[ \frac{4 a^2 (2 A+B) \text{EllipticF}\left (\frac{1}{2} (c+d x),2\right )}{3 d}-\frac{4 a^2 (5 A+4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+4 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
[Out]
(-4*a^2*(5*A + 4*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(2*A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*
a^2*(5*A + 7*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (4*a^2*(5*A + 4*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d
*x]]) + (2*B*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
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Rubi [A] time = 0.380816, antiderivative size = 159, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used =
{2954, 2975, 2968, 3021, 2748, 2636, 2639, 2641} \[ \frac{4 a^2 (2 A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}-\frac{4 a^2 (5 A+4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+4 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d \cos ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sqrt[Cos[c + d*x]],x]
[Out]
(-4*a^2*(5*A + 4*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(2*A + B)*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*
a^2*(5*A + 7*B)*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)) + (4*a^2*(5*A + 4*B)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d
*x]]) + (2*B*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2))
Rule 2954
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(
d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && I
ntegerQ[m] && IntegerQ[n]
Rule 2975
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])
Rule 2968
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Rule 3021
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Rule 2748
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Rule 2636
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]
Rule 2639
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]
Rule 2641
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{(a+a \sec (c+d x))^2 (A+B \sec (c+d x))}{\sqrt{\cos (c+d x)}} \, dx &=\int \frac{(a+a \cos (c+d x))^2 (B+A \cos (c+d x))}{\cos ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(a+a \cos (c+d x)) \left (\frac{1}{2} a (5 A+7 B)+\frac{1}{2} a (5 A+B) \cos (c+d x)\right )}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{1}{2} a^2 (5 A+7 B)+\left (\frac{1}{2} a^2 (5 A+B)+\frac{1}{2} a^2 (5 A+7 B)\right ) \cos (c+d x)+\frac{1}{2} a^2 (5 A+B) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4}{15} \int \frac{\frac{3}{2} a^2 (5 A+4 B)+\frac{5}{2} a^2 (2 A+B) \cos (c+d x)}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{1}{3} \left (2 a^2 (2 A+B)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx+\frac{1}{5} \left (2 a^2 (5 A+4 B)\right ) \int \frac{1}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 (2 A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+4 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}-\frac{1}{5} \left (2 a^2 (5 A+4 B)\right ) \int \sqrt{\cos (c+d x)} \, dx\\ &=-\frac{4 a^2 (5 A+4 B) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d}+\frac{4 a^2 (2 A+B) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 d}+\frac{2 a^2 (5 A+7 B) \sin (c+d x)}{15 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{4 a^2 (5 A+4 B) \sin (c+d x)}{5 d \sqrt{\cos (c+d x)}}+\frac{2 B \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d \cos ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [C] time = 6.51973, size = 1025, normalized size = 6.45 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x]))/Sqrt[Cos[c + d*x]],x]
[Out]
(Cos[c + d*x]^(7/2)*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*(((5*A + 4*B)*Csc[c]*Sec[
c])/(5*d) + (B*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(10*d) + (Sec[c]*Sec[c + d*x]^2*(3*B*Sin[c] + 5*A*Sin[d*x] + 10
*B*Sin[d*x]))/(30*d) + (Sec[c]*Sec[c + d*x]*(5*A*Sin[c] + 10*B*Sin[c] + 30*A*Sin[d*x] + 24*B*Sin[d*x]))/(30*d)
))/(B + A*Cos[c + d*x]) - (2*A*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot
[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - S
in[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcT
an[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) - (B*Cos[c + d*x]^3*Csc[c]*HypergeometricPFQ[{1/4,
1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*Se
c[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[
Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]) + (A*Cos[c + d*x
]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4},
{3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqr
t[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2])
- ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan
[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(B + A*Cos[c +
d*x])) + (2*B*Cos[c + d*x]^3*Csc[c]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x])*((Hyperg
eometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[
d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[
c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + Ar
cTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]
^2]]))/(5*d*(B + A*Cos[c + d*x]))
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Maple [B] time = 6.256, size = 741, normalized size = 4.7 \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(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x)
[Out]
-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+(1/4*A+1/2*B)*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x
+1/2*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-1/20*B/(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)^2*(12*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)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1
/2*c)-12*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)^(1/2)*s
in(1/2*d*x+1/2*c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*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)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*
d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(1/2*A+1/4*B)*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-
1)^(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*(-2*sin(
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)^2/(2*
sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
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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(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B a^{2} \sec \left (d x + c\right )^{3} +{\left (A + 2 \, B\right )} a^{2} \sec \left (d x + c\right )^{2} +{\left (2 \, A + B\right )} a^{2} \sec \left (d x + c\right ) + A a^{2}}{\sqrt{\cos \left (d x + c\right )}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
integral((B*a^2*sec(d*x + c)^3 + (A + 2*B)*a^2*sec(d*x + c)^2 + (2*A + B)*a^2*sec(d*x + c) + A*a^2)/sqrt(cos(d
*x + c)), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \frac{A}{\sqrt{\cos{\left (c + d x \right )}}}\, dx + \int \frac{2 A \sec{\left (c + d x \right )}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx + \int \frac{A \sec ^{2}{\left (c + d x \right )}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx + \int \frac{B \sec{\left (c + d x \right )}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx + \int \frac{2 B \sec ^{2}{\left (c + d x \right )}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx + \int \frac{B \sec ^{3}{\left (c + d x \right )}}{\sqrt{\cos{\left (c + d x \right )}}}\, dx\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**2*(A+B*sec(d*x+c))/cos(d*x+c)**(1/2),x)
[Out]
a**2*(Integral(A/sqrt(cos(c + d*x)), x) + Integral(2*A*sec(c + d*x)/sqrt(cos(c + d*x)), x) + Integral(A*sec(c
+ d*x)**2/sqrt(cos(c + d*x)), x) + Integral(B*sec(c + d*x)/sqrt(cos(c + d*x)), x) + Integral(2*B*sec(c + d*x)*
*2/sqrt(cos(c + d*x)), x) + Integral(B*sec(c + d*x)**3/sqrt(cos(c + d*x)), x))
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^2*(A+B*sec(d*x+c))/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^2/sqrt(cos(d*x + c)), x)