3.1257 \(\int \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=197 \[ \frac{a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (2 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}-\frac{a (2 A-3 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}{3 d} \]
[Out]
(a^(3/2)*(2*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/d + (a^2*(8*A + 6*B - 3*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - (a*(2*A -
3*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*(a + a*Sec[c +
d*x])^(3/2)*Sin[c + d*x])/(3*d)
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Rubi [A] time = 0.648714, antiderivative size = 197, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used =
{4265, 4086, 4018, 4015, 3801, 215} \[ \frac{a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{a^{3/2} (2 B+3 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{d}-\frac{a (2 A-3 C) \sin (c+d x) \sqrt{a \sec (c+d x)+a}}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} (a \sec (c+d x)+a)^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a^(3/2)*(2*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c +
d*x]])/d + (a^2*(8*A + 6*B - 3*C)*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - (a*(2*A -
3*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]) + (2*A*Sqrt[Cos[c + d*x]]*(a + a*Sec[c +
d*x])^(3/2)*Sin[c + d*x])/(3*d)
Rule 4265
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]
Rule 4086
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Rule 4018
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] && Ne
Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1]
Rule 4015
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[(A*b^2*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(a*f*n*Sqrt[a + b*Csc[e + f*x]]), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, 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 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 \cos ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+a \sec (c+d x))^{3/2} \left (\frac{3}{2} a (A+B)-\frac{1}{2} a (2 A-3 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=-\frac{a (2 A-3 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+a \sec (c+d x)} \left (\frac{1}{4} a^2 (8 A+6 B-3 C)+\frac{3}{4} a^2 (2 B+3 C) \sec (c+d x)\right )}{\sqrt{\sec (c+d x)}} \, dx}{3 a}\\ &=\frac{a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}+\frac{1}{2} \left (a (2 B+3 C) \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^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}-\frac{\left (a (2 B+3 C) \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 )}{d}\\ &=\frac{a^{3/2} (2 B+3 C) \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)}}{d}+\frac{a^2 (8 A+6 B-3 C) \sin (c+d x)}{3 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{a (2 A-3 C) \sqrt{a+a \sec (c+d x)} \sin (c+d x)}{3 d \sqrt{\cos (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 1.34133, size = 122, normalized size = 0.62 \[ \frac{a \sqrt{\cos (c+d x)} \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} \left (2 \sin \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) (2 (5 A+3 B) \cos (c+d x)+A \cos (2 (c+d x))+A+3 C)+3 \sqrt{2} (2 B+3 C) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{6 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(a*Sqrt[Cos[c + d*x]]*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(2*B + 3*C)*ArcTanh[Sqrt[2]*Sin[(
c + d*x)/2]] + 2*(A + 3*C + 2*(5*A + 3*B)*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]*Sin[(c + d*x)/2]))/(
6*d)
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Maple [B] time = 0.383, size = 366, normalized size = 1.9 \begin{align*} -{\frac{a \left ( -1+\cos \left ( dx+c \right ) \right ) }{6\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 4\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+20\,A\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+12\,B\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+6\,B\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 ) \cos \left ( dx+c \right ) -6\,B\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 ) \cos \left ( dx+c \right ) +9\,C\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 ) \cos \left ( dx+c \right ) -9\,C\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 ) \cos \left ( dx+c \right ) +6\,C\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sin \left ( dx+c \right ) \right ){\frac{1}{\sqrt{\cos \left ( dx+c \right ) }}}{\frac{1}{\sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
-1/6/d*a*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(4*A*cos(d*x+c)^2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^
(1/2)+20*A*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+12*B*cos(d*x+c)*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2
)+6*B*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)))*cos(d*x+c)-6*B*2^(1/2)*a
rctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)+9*C*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)))*cos(d*x+c)-9*C*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)))*cos(d*x+c)+6*C*(-2/(cos(d*x+c)+1))^(1/2)*sin(d*x+c))/sin(d*x+c)^2/cos
(d*x+c)^(1/2)/(-2/(cos(d*x+c)+1))^(1/2)
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Maxima [B] time = 2.4507, size = 2558, normalized size = 12.98 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
1/60*(20*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*A*sqrt(a) + 6*(2*sqrt(2)*a*sin(5/
2*d*x + 5/2*c) + 40*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 2*sqrt(2)*a*sin(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2
*d*x + 3/2*c))) - 20*sqrt(2)*a*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*a*log(2*cos(1/
3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sqrt(2)*cos(1/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*
x + 3/2*c))) + 2) + 5*a*log(2*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arcta
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*
x + 3/2*c))) + 2*sqrt(2)*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2) - 5*a*log(2*cos(1/3
*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c)))^2 - 2*sqrt(2)*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 2*sqrt(2)*sin(1/3*arct
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2))*B*sqrt(a) - 15*(2*sqrt(2)*a*cos(7/2*d*x + 7/2*c)*sin(2*
d*x + 2*c) + 6*sqrt(2)*a*cos(5/2*d*x + 5/2*c)*sin(2*d*x + 2*c) + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)
*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x
+ 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 +
2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*s
in(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1
/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c)
+ 2))*cos(2*d*x + 2*c)^2 + (2*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 6*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 3*a*log(2*c
os(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/
2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*s
qrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos
(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1
/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(2*d*x + 2*c)^2 - 4*sqrt(2)
*a*sin(3/2*d*x + 3/2*c) + 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) - 2*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 5*sqrt(2)*a*s
in(1/2*d*x + 1/2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1
/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d
*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2
))*cos(2*d*x + 2*c) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*s
qrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1
/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3*a*log(2*cos(1/2*d
*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2
) - 2*(sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*a)*sin(7/2*d*x + 7/2*c) - 6*(sqrt(2)*a*cos(2*d*x + 2*c) + sqrt(2)*
a)*sin(5/2*d*x + 5/2*c) + 2*(3*sqrt(2)*a*cos(3/2*d*x + 3/2*c) + sqrt(2)*a*cos(1/2*d*x + 1/2*c))*sin(2*d*x + 2*
c))*C*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d
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Fricas [A] time = 0.702067, size = 1119, normalized size = 5.68 \begin{align*} \left [\frac{4 \,{\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \,{\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\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 ) + 3 \,{\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\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 )}{12 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac{2 \,{\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \,{\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\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 ) + 3 \,{\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} +{\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )\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 )}{6 \,{\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
[1/12*(4*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*sqrt(cos(d*x + c))*sin(d*x + c) + 3*((2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*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)^2 + d*cos(d*x + c)),
1/6*(2*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*
sqrt(cos(d*x + c))*sin(d*x + c) + 3*((2*B + 3*C)*a*cos(d*x + c)^2 + (2*B + 3*C)*a*cos(d*x + c))*sqrt(-a)*arcta
n(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)^2 + d*cos(d*x + c))]
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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(cos(d*x+c)**(3/2)*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \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(cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(3/2), x)