3.10.53 \(\int \cos (c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [953]
Optimal. Leaf size=505 \[ -\frac {(a-b) \sqrt {a+b} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}+\frac {\sqrt {a+b} \left (a^2 b (15 A+90 B-46 C)+30 a^3 C-2 b^3 (15 A-5 B+9 C)+2 a b^2 (45 A-35 B+17 C)\right ) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {a \sqrt {a+b} (5 A b+2 a B) \cot (c+d x) \Pi \left (\frac {a+b}{a};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \]
[Out]
A*(a+b*sec(d*x+c))^(5/2)*sin(d*x+c)/d-1/15*(a-b)*(70*a*b*B-a^2*(15*A-46*C)+6*b^2*(5*A+3*C))*cot(d*x+c)*Ellipti
cE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+s
ec(d*x+c))/(a-b))^(1/2)/b/d+1/15*(a^2*b*(15*A+90*B-46*C)+30*a^3*C-2*b^3*(15*A-5*B+9*C)+2*a*b^2*(45*A-35*B+17*C
))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/
(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b/d-a*(5*A*b+2*B*a)*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/
(a+b)^(1/2),(a+b)/a,((a+b)/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^
(1/2)/d-1/5*b*(5*A-2*C)*(a+b*sec(d*x+c))^(3/2)*tan(d*x+c)/d-1/15*b*(15*A*a-10*B*b-16*C*a)*(a+b*sec(d*x+c))^(1/
2)*tan(d*x+c)/d
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Rubi [A]
time = 0.66, antiderivative size = 505, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4179, 4141,
4143, 4006, 3869, 3917, 4089} \begin {gather*} -\frac {(a-b) \sqrt {a+b} \cot (c+d x) \left (-\left (a^2 (15 A-46 C)\right )+70 a b B+6 b^2 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}+\frac {\sqrt {a+b} \cot (c+d x) \left (30 a^3 C+a^2 b (15 A+90 B-46 C)+2 a b^2 (45 A-35 B+17 C)-2 b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b d}-\frac {a \sqrt {a+b} (2 a B+5 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};\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{d}-\frac {b \tan (c+d x) (15 a A-16 a C-10 b B) \sqrt {a+b \sec (c+d x)}}{15 d}-\frac {b (5 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}+\frac {A \sin (c+d x) (a+b \sec (c+d x))^{5/2}}{d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
-1/15*((a - b)*Sqrt[a + b]*(70*a*b*B - a^2*(15*A - 46*C) + 6*b^2*(5*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sq
rt[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))])/(b*d) + (Sqrt[a + b]*(a^2*b*(15*A + 90*B - 46*C) + 30*a^3*C - 2*b^3*(15*A - 5*B + 9*C) +
2*a*b^2*(45*A - 35*B + 17*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) - (a*Sqrt[a + b]
*(5*A*b + 2*a*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 + (A*(a + b*Sec[c + d*x])
^(5/2)*Sin[c + d*x])/d - (b*(15*a*A - 10*b*B - 16*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) - (b*(5*A
- 2*C)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d)
Rule 3869
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*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)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Rule 3917
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(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)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4006
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 4089
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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rule 4141
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), I
nt[(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 4143
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 4179
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]
Rubi steps
\begin {align*} \int \cos (c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+2 a B)+(b B+a C) \sec (c+d x)-\frac {1}{2} b (5 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (5 A-2 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}{4} a (5 A b+2 a B)+\frac {1}{2} \left (5 A b^2+10 a b B+5 a^2 C+3 b^2 C\right ) \sec (c+d x)-\frac {1}{4} b (15 a A-10 b B-16 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15}{8} a^2 (5 A b+2 a B)+\frac {1}{4} \left (45 a^2 b B+5 b^3 B+15 a^3 C+a b^2 (45 A+17 C)\right ) \sec (c+d x)+\frac {1}{8} b \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {4}{15} \int \frac {\frac {15}{8} a^2 (5 A b+2 a B)+\left (-\frac {1}{8} b \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right )+\frac {1}{4} \left (45 a^2 b B+5 b^3 B+15 a^3 C+a b^2 (45 A+17 C)\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{30} \left (b \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 C)\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {(a-b) \sqrt {a+b} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 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 {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac {1}{2} \left (a^2 (5 A b+2 a B)\right ) \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx+\frac {1}{30} \left (a^2 b (15 A+90 B-46 C)+30 a^3 C-2 b^3 (15 A-5 B+9 C)+2 a b^2 (45 A-35 B+17 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=-\frac {(a-b) \sqrt {a+b} \left (70 a b B-a^2 (15 A-46 C)+6 b^2 (5 A+3 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 {\sqrt {a+b} \left (a^2 b (15 A+90 B-46 C)+30 a^3 C-2 b^3 (15 A-5 B+9 C)+2 a b^2 (45 A-35 B+17 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 {a \sqrt {a+b} (5 A b+2 a 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 {A (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{d}-\frac {b (15 a A-10 b B-16 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}-\frac {b (5 A-2 C) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1490\) vs. \(2(505)=1010\).
time = 20.97, size = 1490, normalized size = 2.95 \begin {gather*} \frac {2 (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (15 a^3 A \tan \left (\frac {1}{2} (c+d x)\right )+15 a^2 A b \tan \left (\frac {1}{2} (c+d x)\right )-30 a A b^2 \tan \left (\frac {1}{2} (c+d x)\right )-30 A b^3 \tan \left (\frac {1}{2} (c+d x)\right )-70 a^2 b B \tan \left (\frac {1}{2} (c+d x)\right )-70 a b^2 B \tan \left (\frac {1}{2} (c+d x)\right )-46 a^3 C \tan \left (\frac {1}{2} (c+d x)\right )-46 a^2 b C \tan \left (\frac {1}{2} (c+d x)\right )-18 a b^2 C \tan \left (\frac {1}{2} (c+d x)\right )-18 b^3 C \tan \left (\frac {1}{2} (c+d x)\right )-30 a^3 A \tan ^3\left (\frac {1}{2} (c+d x)\right )+60 a A b^2 \tan ^3\left (\frac {1}{2} (c+d x)\right )+140 a^2 b B \tan ^3\left (\frac {1}{2} (c+d x)\right )+92 a^3 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+36 a b^2 C \tan ^3\left (\frac {1}{2} (c+d x)\right )+15 a^3 A \tan ^5\left (\frac {1}{2} (c+d x)\right )-15 a^2 A b \tan ^5\left (\frac {1}{2} (c+d x)\right )-30 a A b^2 \tan ^5\left (\frac {1}{2} (c+d x)\right )+30 A b^3 \tan ^5\left (\frac {1}{2} (c+d x)\right )-70 a^2 b B \tan ^5\left (\frac {1}{2} (c+d x)\right )+70 a b^2 B \tan ^5\left (\frac {1}{2} (c+d x)\right )-46 a^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+46 a^2 b C \tan ^5\left (\frac {1}{2} (c+d x)\right )-18 a b^2 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+18 b^3 C \tan ^5\left (\frac {1}{2} (c+d x)\right )+150 a^2 A b \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+60 a^3 B \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+150 a^2 A b \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+60 a^3 B \Pi \left (-1;\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+(a+b) \left (-70 a b B+a^2 (15 A-46 C)-6 b^2 (5 A+3 C)\right ) E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-2 \left (a^2 b (45 A-45 B-23 C)+15 a^3 (B-C)-b^3 (15 A+5 B+9 C)-a b^2 (45 A+35 B+17 C)\right ) F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}\right )}{15 d (b+a \cos (c+d x))^{5/2} (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sec ^{\frac {9}{2}}(c+d x) \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {\cos ^4(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4}{15} \left (15 A b^2+35 a b B+23 a^2 C+9 b^2 C\right ) \sin (c+d x)+\frac {4}{15} \sec (c+d x) \left (5 b^2 B \sin (c+d x)+11 a b C \sin (c+d x)\right )+\frac {4}{5} b^2 C \sec (c+d x) \tan (c+d x)\right )}{d (b+a \cos (c+d x))^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x))} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
[Out]
(2*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(15*
a^3*A*Tan[(c + d*x)/2] + 15*a^2*A*b*Tan[(c + d*x)/2] - 30*a*A*b^2*Tan[(c + d*x)/2] - 30*A*b^3*Tan[(c + d*x)/2]
- 70*a^2*b*B*Tan[(c + d*x)/2] - 70*a*b^2*B*Tan[(c + d*x)/2] - 46*a^3*C*Tan[(c + d*x)/2] - 46*a^2*b*C*Tan[(c +
d*x)/2] - 18*a*b^2*C*Tan[(c + d*x)/2] - 18*b^3*C*Tan[(c + d*x)/2] - 30*a^3*A*Tan[(c + d*x)/2]^3 + 60*a*A*b^2*
Tan[(c + d*x)/2]^3 + 140*a^2*b*B*Tan[(c + d*x)/2]^3 + 92*a^3*C*Tan[(c + d*x)/2]^3 + 36*a*b^2*C*Tan[(c + d*x)/2
]^3 + 15*a^3*A*Tan[(c + d*x)/2]^5 - 15*a^2*A*b*Tan[(c + d*x)/2]^5 - 30*a*A*b^2*Tan[(c + d*x)/2]^5 + 30*A*b^3*T
an[(c + d*x)/2]^5 - 70*a^2*b*B*Tan[(c + d*x)/2]^5 + 70*a*b^2*B*Tan[(c + d*x)/2]^5 - 46*a^3*C*Tan[(c + d*x)/2]^
5 + 46*a^2*b*C*Tan[(c + d*x)/2]^5 - 18*a*b^2*C*Tan[(c + d*x)/2]^5 + 18*b^3*C*Tan[(c + d*x)/2]^5 + 150*a^2*A*b*
EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c
+ d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 60*a^3*B*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a +
b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 150*a^2
*A*b*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]
*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 60*a^3*B*EllipticPi[-1, ArcSin[Tan[(c +
d*x)/2]], (a - b)/(a + b)]*Tan[(c + d*x)/2]^2*Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2
+ b*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*(-70*a*b*B + a^2*(15*A - 46*C) - 6*b^2*(5*A + 3*C))*EllipticE[ArcS
in[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*T
an[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] - 2*(a^2*b*(45*A - 45*B - 23*C) + 15*a^3*(B - C) - b^3*(15*
A + 5*B + 9*C) - a*b^2*(45*A + 35*B + 17*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan
[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)])
)/(15*d*(b + a*Cos[c + d*x])^(5/2)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sec[c + d*x]^(9/2)*(1 + T
an[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)])
+ (Cos[c + d*x]^4*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(15*A*b^2 + 35*a*b*B
+ 23*a^2*C + 9*b^2*C)*Sin[c + d*x])/15 + (4*Sec[c + d*x]*(5*b^2*B*Sin[c + d*x] + 11*a*b*C*Sin[c + d*x]))/15 +
(4*b^2*C*Sec[c + d*x]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x])^2*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2
*d*x]))
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(4980\) vs.
\(2(464)=928\).
time = 0.57, size = 4981, normalized size = 9.86
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\(\text {Expression too large to display}\) |
\(4981\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
[Out]
1/15/d*(1+cos(d*x+c))^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(30*A*sin(d*x+c)*cos(d*x+c)^3*(c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*a*b^2+15*A*cos(d*x+c)^3*a^2*b+80*B*cos(d*x+c)^2*a*b^2-70*B*cos(d*x+c)^4*a^2*b-10*B*c
os(d*x+c)^4*a*b^2+70*B*cos(d*x+c)^3*a^2*b-70*B*cos(d*x+c)^3*a*b^2-15*A*cos(d*x+c)^5*a^3+6*C*b^3+90*A*sin(d*x+c
)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+c
os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-150*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))
*a^2*b+90*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^
(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-150*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1
,((a-b)/(a+b))^(1/2))*a^2*b-90*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b-90*B*cos(d*x+c)^3*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b+30*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b
+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-30*A*
sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipt
icF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+46*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*a^3+18*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^
(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-18*C*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)
/(a+b))^(1/2))*b^3+30*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-30*A*sin(d*x+c)*cos(d*x+c)^2*(
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*b^3+46*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c)
)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+18*C*sin(d*x+c)*co
s(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d
*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-18*C*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-15*A*c
os(d*x+c)^4*a^2*b-10*B*cos(d*x+c)^3*b^3+10*B*cos(d*x+c)*b^3-46*C*cos(d*x+c)^4*a^3-30*A*cos(d*x+c)^3*b^3+46*C*c
os(d*x+c)^3*a^3-18*C*cos(d*x+c)^3*b^3+30*A*cos(d*x+c)^2*b^3+12*C*cos(d*x+c)^2*b^3-15*A*cos(d*x+c)^3*(cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b-15*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+co
s(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b-15*A*cos(d*x
+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3+30*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos
(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-
60*B*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((
-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-30*C*cos(d*x+c)^3*(cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*sin(d*x+c)*a^3-15*A*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1
/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3+30*B*cos(d*x+c)^2*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(
a+b))^(1/2))*sin(d*x+c)*a^3-60*B*cos(d*x+c)^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))/sin(d*x+c),-1,((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3-30*C*cos(d*x+c)^2
*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*...
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+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) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c), x)
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(A+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 + (2*C*a*b + B*b^2)*cos(d*x + c)*sec(d*x + c)^3 + A*a^2*cos(d*x +
c) + (C*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)*sec(d*x + c))*sqrt
(b*sec(d*x + c) + a), x)
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^(5/2)*(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)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)
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