3.521 \(\int \frac{\sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=151 \[ \frac{a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{a} (4 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]
[Out]
(Sqrt[a]*(4*A + 3*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(4*d) + (a*B*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(4*A + 3*B)*Sin[c + d
*x])/(4*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])
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Rubi [A] time = 0.331359, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{2955, 4016, 3803, 3801, 215} \[ \frac{a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{\sqrt{a} (4 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 d}+\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
(Sqrt[a]*(4*A + 3*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/(4*d) + (a*B*Sin[c + d*x])/(2*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]]) + (a*(4*A + 3*B)*Sin[c + d
*x])/(4*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])
Rule 2955
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*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
Rule 4016
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(-2*b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]
), x] + Dist[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n, x], x]
/; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n
, 0] && !LtQ[n, 0]
Rule 3803
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*b*d
*Cot[e + f*x]*(d*Csc[e + f*x])^(n - 1))/(f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[(2*a*d*(n - 1))/(b*(
2*n - 1)), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a
^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
Rule 3801
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]
Rule 215
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]
Rubi steps
\begin{align*} \int \frac{\sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x))}{\cos ^{\frac{3}{2}}(c+d x)} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} (A+B \sec (c+d x)) \, dx\\ &=\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{1}{4} \left ((4 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{1}{8} \left ((4 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx\\ &=\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left ((4 A+3 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac{\sqrt{a} (4 A+3 B) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{4 d}+\frac{a B \sin (c+d x)}{2 d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{a (4 A+3 B) \sin (c+d x)}{4 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 0.622837, size = 106, normalized size = 0.7 \[ \frac{\sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (\sqrt{2} (4 A+3 B) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (4 A+2 B \sec (c+d x)+3 B)\right )}{8 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[a + a*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Cos[c + d*x]^(3/2),x]
[Out]
(Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(Sqrt[2]*(4*A + 3*B)*ArcTanh[Sqrt[2]*Sin[(c +
d*x)/2]] + 2*Sec[c + d*x]*(4*A + 3*B + 2*B*Sec[c + d*x])*Sin[(c + d*x)/2]))/(8*d)
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Maple [B] time = 0.344, size = 342, normalized size = 2.3 \begin{align*} -{\frac{-1+\cos \left ( dx+c \right ) }{8\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}} \left ( -4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) -3\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1-\sin \left ( dx+c \right ) \right ) \right ) +3\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sqrt{2}\arctan \left ( 1/4\,\sqrt{2}\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \left ( \cos \left ( dx+c \right ) +1+\sin \left ( dx+c \right ) \right ) \right ) +8\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+6\,B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+4\,B\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}} \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x)
[Out]
-1/8/d*(-1+cos(d*x+c))*(-4*A*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-s
in(d*x+c)))+4*A*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))-3
*B*cos(d*x+c)^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))+3*B*cos(d*x+c)
^2*2^(1/2)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+8*A*cos(d*x+c)*sin(d*x+c)*(
-2/(cos(d*x+c)+1))^(1/2)+6*B*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+4*B*(-2/(cos(d*x+c)+1))^(1/2)*sin
(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^2/cos(d*x+c)^(3/2)
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Maxima [B] time = 2.27221, size = 2601, normalized size = 17.23 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="maxima")
[Out]
-1/16*(4*(4*sqrt(2)*cos(3/2*arctan2(sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - 4*sqrt(2)*cos(1/2*arctan2(
sin(d*x + c), cos(d*x + c)))*sin(2*d*x + 2*c) - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c)
+ 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 +
2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)
)) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c
), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c
), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - (cos(2*d*x + 2*c)^2 + sin(2*
d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan
2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*
arctan2(sin(d*x + c), cos(d*x + c))) + 2) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)
*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*s
qrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) +
2) - 4*(sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(2*d
*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*A*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1) + (12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(7/2*arctan
2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(5/2*arctan2(sin
(d*x + c), cos(d*x + c))) - 4*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(3/2*arctan2(sin(d*x
+ c), cos(d*x + c))) - 12*(sqrt(2)*sin(4*d*x + 4*c) + 2*sqrt(2)*sin(2*d*x + 2*c))*cos(1/2*arctan2(sin(d*x + c)
, cos(d*x + c))) - 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2
+ sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*lo
g(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt
(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2)
+ 3*(2*(2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*
c)^2 + 4*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arct
an2(sin(d*x + c), cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 + 2*sqrt(2)*cos(1/2*arct
an2(sin(d*x + c), cos(d*x + c))) - 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 3*(2*(2*cos(2
*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d
*x + 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c)
, cos(d*x + c)))^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c)
, cos(d*x + c))) + 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) + 3*(2*(2*cos(2*d*x + 2*c) + 1)
*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x + 4*c)*sin(2*
d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1)*log(2*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))
^2 + 2*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c)))^2 - 2*sqrt(2)*cos(1/2*arctan2(sin(d*x + c), cos(d*x + c)))
- 2*sqrt(2)*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))) + 2) - 12*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)*cos(
2*d*x + 2*c) + sqrt(2))*sin(7/2*arctan2(sin(d*x + c), cos(d*x + c))) - 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sqrt(2)
*cos(2*d*x + 2*c) + sqrt(2))*sin(5/2*arctan2(sin(d*x + c), cos(d*x + c))) + 4*(sqrt(2)*cos(4*d*x + 4*c) + 2*sq
rt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(3/2*arctan2(sin(d*x + c), cos(d*x + c))) + 12*(sqrt(2)*cos(4*d*x + 4*c)
+ 2*sqrt(2)*cos(2*d*x + 2*c) + sqrt(2))*sin(1/2*arctan2(sin(d*x + c), cos(d*x + c))))*B*sqrt(a)/(2*(2*cos(2*d*
x + 2*c) + 1)*cos(4*d*x + 4*c) + cos(4*d*x + 4*c)^2 + 4*cos(2*d*x + 2*c)^2 + sin(4*d*x + 4*c)^2 + 4*sin(4*d*x
+ 4*c)*sin(2*d*x + 2*c) + 4*sin(2*d*x + 2*c)^2 + 4*cos(2*d*x + 2*c) + 1))/d
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Fricas [A] time = 0.681025, size = 1035, normalized size = 6.85 \begin{align*} \left [\frac{4 \,{\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{a} \log \left (\frac{a \cos \left (d x + c\right )^{3} - 4 \, \sqrt{a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}}{\left (\cos \left (d x + c\right ) - 2\right )} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}, \frac{2 \,{\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left ({\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (4 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{-a} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \,{\left (d \cos \left (d x + c\right )^{3} + d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="fricas")
[Out]
[1/16*(4*((4*A + 3*B)*cos(d*x + c) + 2*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x +
c) + ((4*A + 3*B)*cos(d*x + c)^3 + (4*A + 3*B)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3 - 4*sqrt(a)*sqrt
((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 +
8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2), 1/8*(2*((4*A + 3*B)*cos(d*x +
c) + 2*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((4*A + 3*B)*cos(d*x + c)^
3 + (4*A + 3*B)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*
x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(d*cos(d*x + c)^3 + d*cos(d*x + c)^2)]
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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+B*sec(d*x+c))*(a+a*sec(d*x+c))**(1/2)/cos(d*x+c)**(3/2),x)
[Out]
Timed out
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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 )} \sqrt{a \sec \left (d x + c\right ) + a}}{\cos \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*(a+a*sec(d*x+c))^(1/2)/cos(d*x+c)^(3/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(a*sec(d*x + c) + a)/cos(d*x + c)^(3/2), x)