3.850 \(\int \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=239 \[ \frac{16 b \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2+23 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 d}+\frac{22 a b \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 d} \]
[Out]
(16*b*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(15*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2 + 23*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)
]*Sqrt[a + b*Sec[c + d*x]])/(15*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (22*a*b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*
Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a^2*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d)
________________________________________________________________________________________
Rubi [A] time = 0.755703, antiderivative size = 239, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.4, Rules used = {4264, 3841, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac{16 b \left (a^2-b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2+23 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 d \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a^2 \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}}{5 d}+\frac{22 a b \sin (c+d x) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}{15 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(16*b*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(15*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2 + 23*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)
]*Sqrt[a + b*Sec[c + d*x]])/(15*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (22*a*b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*
Sec[c + d*x]]*Sin[c + d*x])/(15*d) + (2*a^2*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d)
Rule 4264
Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]
Rule 3841
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))
Rule 4104
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 +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
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{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{(a+b \sec (c+d x))^{5/2}}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{1}{5} \left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{11 a^2 b}{2}+\frac{3}{2} a \left (a^2+5 b^2\right ) \sec (c+d x)+\frac{1}{2} b \left (2 a^2+5 b^2\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{22 a b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}-\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} a^2 \left (9 a^2+23 b^2\right )-\frac{1}{4} a b \left (17 a^2+15 b^2\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{15 a}\\ &=\frac{22 a b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{1}{15} \left (8 b \left (a^2-b^2\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{1}{15} \left (\left (9 a^2+23 b^2\right ) \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\\ &=\frac{22 a b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{\left (8 b \left (a^2-b^2\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{15 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (9 a^2+23 b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{15 \sqrt{b+a \cos (c+d x)}}\\ &=\frac{22 a b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac{\left (8 b \left (a^2-b^2\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}{15 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (9 a^2+23 b^2\right ) \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}{15 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{16 b \left (a^2-b^2\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 )}{15 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2+23 b^2\right ) \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)}}{15 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}+\frac{22 a b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 d}+\frac{2 a^2 \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d}\\ \end{align*}
Mathematica [C] time = 11.6245, size = 391, normalized size = 1.64 \[ \frac{\cos ^{\frac{5}{2}}(c+d x) \left (\frac{1}{5} a^2 \sin (2 (c+d x))+\frac{22}{15} a b \sin (c+d x)\right ) (a+b \sec (c+d x))^{5/2}}{d (a \cos (c+d x)+b)^2}-\frac{2 \cos ^{\frac{3}{2}}(c+d x) \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} \left (i \left (17 a^2 b+9 a^3+23 a b^2+15 b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )-\left (9 a^2+23 b^2\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i \left (9 a^2 b+9 a^3+23 a b^2+23 b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{15 d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)*((22*a*b*Sin[c + d*x])/15 + (a^2*Sin[2*(c + d*x)])/5))/(d*(b +
a*Cos[c + d*x])^2) - (2*Cos[c + d*x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(a + b*Sec[c + d*x])^(5/2)*
((-I)*(9*a^3 + 9*a^2*b + 23*a*b^2 + 23*b^3)*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c +
d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + I*(9*a^3 + 17*a^2*b + 23*a*b^2 + 15*b^3)*E
llipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c +
d*x)/2]^2)/(a + b)] - (9*a^2 + 23*b^2)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(15
*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.312, size = 1921, normalized size = 8. \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)^(5/2)*(a+b*sec(d*x+c))^(5/2),x)
[Out]
-2/15/d/((a-b)/(a+b))^(1/2)*(-9*a^2*b*((a-b)/(a+b))^(1/2)-11*a*b^2*((a-b)/(a+b))^(1/2)-9*cos(d*x+c)*sin(d*x+c)
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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^3+9*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a
+b)/(a-b))^(1/2))*a^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-23*E
llipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3*(1/(a+b)*(b+a*cos(d*x+c))/(c
os(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-9*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*
x+c),(-(a+b)/(a-b))^(1/2))*a^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*
x+c)+15*cos(d*x+c)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(
1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*b^3+9*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/s
in(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3+3*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3+6*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*
a^3+23*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^3-9*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3-5*cos(d*x+c)*((a-b)/(a+b))^(1/2
)*a^2*b-23*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2+14*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b+34*cos(d*x+c)^2*((a-
b)/(a+b))^(1/2)*a*b^2+15*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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))*b^3*sin(d*x+c)+17*cos(d*x+c)*sin(d*x+c)*(1/
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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*b-23*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1)
)^(1/2)*(1/(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^2-9*cos(d*x+c)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^2*b+23*cos(d*x+c)*sin(d*x+c)*Ellip
ticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c
)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b^2-23*b^3*((a-b)/(a+b))^(1/2)-23*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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^3+17*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2
))*a^2*b*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-23*EllipticF((-1+
cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-9*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b
)/(a-b))^(1/2))*a^2*b*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+23*E
llipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*(1/(a+b)*(b+a*cos(d*x+c))/
(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2
)/(b+a*cos(d*x+c))/sin(d*x+c)
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((b^2*cos(d*x + c)^2*sec(d*x + c)^2 + 2*a*b*cos(d*x + c)^2*sec(d*x + c) + a^2*cos(d*x + c)^2)*sqrt(b*s
ec(d*x + c) + a)*sqrt(cos(d*x + c)), x)
________________________________________________________________________________________
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)**(5/2)*(a+b*sec(d*x+c))**(5/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2), x)