3.1337 \(\int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=277 \[ -\frac{2 \left (A b^2-a^2 (A+3 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 (3 a B+A b) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{3 d}+\frac{2 b C \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]
[Out]
(-2*(A*b^2 - a^2*(A + 3*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a*d*S
qrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b*C*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x
)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b + 3*a*B)*Sqrt[Cos[c + d*x]]*Ell
ipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*A
*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(3*d)
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Rubi [A] time = 1.02884, antiderivative size = 277, normalized size of antiderivative = 1.,
number of steps used = 13, number of rules used = 13, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.289, Rules
used = {4265, 4094, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ -\frac{2 \left (A b^2-a^2 (A+3 C)\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 (3 a B+A b) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 A \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{3 d}+\frac{2 b C \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(A*b^2 - a^2*(A + 3*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a*d*S
qrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*b*C*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x
)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b + 3*a*B)*Sqrt[Cos[c + d*x]]*Ell
ipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*A
*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*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 4094
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/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Rule 4108
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Rule 3859
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2807
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Rule 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 4035
Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
Rule 3856
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2655
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2653
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 3858
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 2663
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 2661
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rubi steps
\begin{align*} \int \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \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{\sqrt{a+b \sec (c+d x)} \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)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} (A b+3 a B)+\frac{1}{2} (3 b B+a (A+3 C)) \sec (c+d x)+\frac{3}{2} b C \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{1}{3} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} (A b+3 a B)+\frac{1}{2} (3 b B+a (A+3 C)) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx+\left (b C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}+\frac{\left ((A b+3 a B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a}-\frac{1}{3} \left (\left (\frac{A b^2}{a}-a (A+3 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{\left (b C \sqrt{b+a \cos (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{b+a \cos (c+d x)}} \, dx}{\sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ &=\frac{2 A \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac{\left (\left (\frac{A b^2}{a}-a (A+3 C)\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (b C \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{\sec (c+d x)}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{\sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left ((A b+3 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a \sqrt{b+a \cos (c+d x)}}\\ &=\frac{2 b C \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 A \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}-\frac{\left (\left (\frac{A b^2}{a}-a (A+3 C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left ((A b+3 a B) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 a \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (\frac{A b^2}{a}-a (A+3 C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 b C \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 (A b+3 a B) \sqrt{\cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{2 A \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [C] time = 33.2616, size = 43023, normalized size = 155.32 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
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Maple [C] time = 0.588, size = 1256, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/3/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^3*(A*sin(d*x+c)*(1/(cos(d*x+c)+1))
^(3/2)*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2+A*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)*((a-b)/(a+b))^(1/2)*cos(d*x+
c)*a^2+2*A*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a*b+3*B*sin(d*x+c)*(1/(cos(d*x+c
)+1))^(3/2)*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2+A*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)*((a-b)/(a+b))^(1/2)*a*b+A
*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)*((a-b)/(a+b))^(1/2)*b^2+3*B*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)*((a-b)/(a
+b))^(1/2)*a*b-A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)
/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+
c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2-A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)
*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2+A*(1/(a+b)*(b+a*cos(d*x+c)
)/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-3*B
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a
+b)/(a-b))^(1/2))*a^2+3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+
b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((
-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2-3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-3*C*(1/(a+b)*
(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))
^(1/2))*a^2+3*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/
sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b-6*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*
x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b)*cos(d*x+c)^(1/2)/a/((a-b)/(a+b))^
(1/2)/(b+a*cos(d*x+c))/(1/(cos(d*x+c)+1))^(3/2)/sin(d*x+c)^6
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Maxima [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 )} \sqrt{b \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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^(3/2), x)
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Fricas [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+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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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+B*sec(d*x+c)+C*sec(d*x+c)**2)*(a+b*sec(d*x+c))**(1/2),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 )} \sqrt{b \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(cos(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*cos(d*x + c)^(3/2), x)