3.832 \(\int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=442 \[ \frac{2 \sqrt{a+b} \left (a^2 b (45 B-23 C)+15 a^3 C-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b d}-\frac{2 (a-b) \sqrt{a+b} \left (23 a^2 C+35 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b d}-\frac{2 a^2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 b (8 a C+5 b B) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]
[Out]
(-2*(a - b)*Sqrt[a + b]*(35*a*b*B + 23*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(15*b*d) + (2*Sqrt[a + b]*(a^2*b*(45*B - 23*C) - a*b^2*(35*B - 17*C) + b^3*(5*B - 9*C) + 15*a^3*C)*Cot[c + d*x
]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)
]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*b*d) - (2*a^2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/a,
ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*
(1 + Sec[c + d*x]))/(a - b))])/d + (2*b*(5*b*B + 8*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*b*C
*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
________________________________________________________________________________________
Rubi [A] time = 0.715349, antiderivative size = 442, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{4072, 3918, 4056, 4058, 3921, 3784, 3832, 4004} \[ \frac{2 \sqrt{a+b} \left (a^2 b (45 B-23 C)+15 a^3 C-a b^2 (35 B-17 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b d}-\frac{2 (a-b) \sqrt{a+b} \left (23 a^2 C+35 a b B+9 b^2 C\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b d}-\frac{2 a^2 B \sqrt{a+b} \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{d}+\frac{2 b (8 a C+5 b B) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 b C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*(35*a*b*B + 23*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(15*b*d) + (2*Sqrt[a + b]*(a^2*b*(45*B - 23*C) - a*b^2*(35*B - 17*C) + b^3*(5*B - 9*C) + 15*a^3*C)*Cot[c + d*x
]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)
]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(15*b*d) - (2*a^2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/a,
ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*
(1 + Sec[c + d*x]))/(a - b))])/d + (2*b*(5*b*B + 8*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*b*C
*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
Rule 4072
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Rule 3918
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 2)*Simp[a^2*c
*m + (b^2*d*(m - 1) + 2*a*b*c*m + a^2*d*m)*Csc[e + f*x] + b*(b*c*m + a*d*(2*m - 1))*Csc[e + f*x]^2, x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]
Rule 4056
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_.), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int
[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m)*Csc[e + f*x] + (b*B*(m + 1) + a
*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
Rule 4058
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_
.) + (a_)], x_Symbol] :> Int[(A + (B - C)*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Dist[C, Int[(Csc[e + f*
x]*(1 + Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0
]
Rule 3921
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c, In
t[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[d, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Rule 3784
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(2*Rt[a + b, 2]*Sqrt[(b*(1 - Csc[c + d*x])
)/(a + b)]*Sqrt[-((b*(1 + Csc[c + d*x]))/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[c + d*x]]/Rt[a
+ b, 2]], (a + b)/(a - b)])/(a*d*Cot[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4004
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rubi steps
\begin{align*} \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int (a+b \sec (c+d x))^{5/2} (B+C \sec (c+d x)) \, dx\\ &=\frac{2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{2}{5} \int \sqrt{a+b \sec (c+d x)} \left (\frac{5 a^2 B}{2}+\frac{1}{2} \left (10 a b B+5 a^2 C+3 b^2 C\right ) \sec (c+d x)+\frac{1}{2} b (5 b B+8 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 b (5 b B+8 a C) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{15 a^3 B}{4}+\frac{1}{4} \left (45 a^2 b B+5 b^3 B+15 a^3 C+17 a b^2 C\right ) \sec (c+d x)+\frac{1}{4} b \left (35 a b B+23 a^2 C+9 b^2 C\right ) \sec ^2(c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 b (5 b B+8 a C) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{4}{15} \int \frac{\frac{15 a^3 B}{4}+\left (-\frac{1}{4} b \left (35 a b B+23 a^2 C+9 b^2 C\right )+\frac{1}{4} \left (45 a^2 b B+5 b^3 B+15 a^3 C+17 a b^2 C\right )\right ) \sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{15} \left (b \left (35 a b B+23 a^2 C+9 b^2 C\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} \left (35 a b B+23 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac{2 b (5 b B+8 a C) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\left (a^3 B\right ) \int \frac{1}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{15} \left (a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)+15 a^3 C\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} \left (35 a b B+23 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac{2 \sqrt{a+b} \left (a^2 b (45 B-23 C)-a b^2 (35 B-17 C)+b^3 (5 B-9 C)+15 a^3 C\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac{2 a^2 \sqrt{a+b} B \cot (c+d x) \Pi \left (\frac{a+b}{a};\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{d}+\frac{2 b (5 b B+8 a C) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 b C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [B] time = 25.0387, size = 7124, normalized size = 16.12 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
________________________________________________________________________________________
Maple [B] time = 0.781, size = 3285, normalized size = 7.4 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)
[Out]
2/15/d*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(23*C*cos(d*x+c)^3*a^3+35*B*cos(
d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elliptic
E((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-45*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1
/2))*a^2*b-9*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)
+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+23*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*a^3+9*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-9*C*sin(d*x+c)*cos(d*x
+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-5*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-5*B*cos(
d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elliptic
F((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+35*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*a*b^2-35*B*cos(d*x+c)^3*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+
1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+35*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c
),((a-b)/(a+b))^(1/2))*a^2*b+35*B*cos(d*x+c)^2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(
d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-35*B*cos(d*x+c)^
2*sin(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+c
os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+23*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^
2*b+9*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1
/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-23*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-
b)/(a+b))^(1/2))*a^2*b-17*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c)
)/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+23*C*sin(d*x+c)*cos(d*
x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x
+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+9*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a
+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-23*
C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El
lipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-17*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+
b))^(1/2))*a*b^2-45*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos
(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-11*C*cos(d*x+c)^4*a^2*b-9*C*
cos(d*x+c)^4*a*b^2-23*C*cos(d*x+c)^3*a^2*b-5*C*cos(d*x+c)^3*a*b^2+14*C*cos(d*x+c)*a*b^2-35*B*cos(d*x+c)^3*a*b^
2+40*B*cos(d*x+c)^2*a*b^2-35*B*cos(d*x+c)^4*a^2*b-5*B*cos(d*x+c)^4*a*b^2+35*B*cos(d*x+c)^3*a^2*b+34*C*cos(d*x+
c)^2*a^2*b+15*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c
)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-30*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*
x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+
c),-1,((a-b)/(a+b))^(1/2))*a^3-15*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*co
s(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+15*B*sin(d*x+c)*
cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+c
os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-30*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*
a^3-23*C*cos(d*x+c)^4*a^3-9*C*cos(d*x+c)^3*b^3+6*C*cos(d*x+c)^2*b^3-5*B*cos(d*x+c)^3*b^3+5*B*cos(d*x+c)*b^3+23
*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*E
llipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+9*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+
c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b)
)^(1/2))*b^3+3*C*b^3-15*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/
(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3)/(b+a*cos(d*x+c))/cos(d*x+
c)^2/sin(d*x+c)^5
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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 )\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{4} + B a^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right ) +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} +{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*b^2*cos(d*x + c)*sec(d*x + c)^4 + B*a^2*cos(d*x + c)*sec(d*x + c) + (2*C*a*b + B*b^2)*cos(d*x + c)
*sec(d*x + c)^3 + (C*a^2 + 2*B*a*b)*cos(d*x + c)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
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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)*(a+b*sec(d*x+c))**(5/2)*(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 )\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(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))*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c), x)