3.11.89 \(\int \frac {(a+a \sec (c+d x)) (A+C \sec ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\) [1089]

Optimal. Leaf size=165 \[ -\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}} \]

[Out]

-2/5*a*(5*A+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*
a*(7*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/7*a*C*si
n(d*x+c)/d/cos(d*x+c)^(7/2)+2/5*a*C*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/21*a*(7*A+5*C)*sin(d*x+c)/d/cos(d*x+c)^(3/
2)+2/5*a*(5*A+3*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

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Rubi [A]
time = 0.18, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4199, 3111, 3100, 2827, 2716, 2720, 2719} \begin {gather*} \frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}-\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

(-2*a*(5*A + 3*C)*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*a*(7*A + 5*C)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*a
*C*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (2*a*C*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*a*(7*A + 5*C)*S
in[c + d*x])/(21*d*Cos[c + d*x]^(3/2)) + (2*a*(5*A + 3*C)*Sin[c + d*x])/(5*d*Sqrt[Cos[c + d*x]])

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3111

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e
_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m +
1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp
[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e +
f*x] + b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4199

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sec[(e_.)
+ (f_.)*(x_)]^2), x_Symbol] :> Dist[d^(m + 2), Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A
*Cos[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\int \frac {(a+a \cos (c+d x)) \left (C+A \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {\frac {7 a C}{2}+\frac {1}{2} a (7 A+5 C) \cos (c+d x)+\frac {7}{2} a A \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {\frac {5}{4} a (7 A+5 C)+\frac {7}{4} a (5 A+3 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} (a (5 A+3 C)) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{7} (a (7 A+5 C)) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}-\frac {1}{5} (a (5 A+3 C)) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{21} (a (7 A+5 C)) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=-\frac {2 a (5 A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 a (7 A+5 C) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 a C \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 a C \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 a (7 A+5 C) \sin (c+d x)}{21 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a (5 A+3 C) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.56, size = 895, normalized size = 5.42 \begin {gather*} a \left (\sqrt {\cos (c+d x)} (1+\cos (c+d x)) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {(5 A+3 C) \csc (c) \sec (c)}{5 d}+\frac {C \sec (c) \sec ^4(c+d x) \sin (d x)}{7 d}+\frac {\sec (c) \sec ^3(c+d x) (5 C \sin (c)+7 C \sin (d x))}{35 d}+\frac {\sec (c) \sec ^2(c+d x) (21 C \sin (c)+35 A \sin (d x)+25 C \sin (d x))}{105 d}+\frac {\sec (c) \sec (c+d x) (35 A \sin (c)+25 C \sin (c)+105 A \sin (d x)+63 C \sin (d x))}{105 d}\right )-\frac {A (1+\cos (c+d x)) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}-\frac {5 C (1+\cos (c+d x)) \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\text {ArcTan}(\cot (c)))\right ) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{21 d \sqrt {1+\cot ^2(c)}}+\frac {A (1+\cos (c+d x)) \csc (c) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d}+\frac {3 C (1+\cos (c+d x)) \csc (c) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{10 d}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[((a + a*Sec[c + d*x])*(A + C*Sec[c + d*x]^2))/Cos[c + d*x]^(3/2),x]

[Out]

a*(Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])*Sec[c/2 + (d*x)/2]^2*(((5*A + 3*C)*Csc[c]*Sec[c])/(5*d) + (C*Sec[c]*S
ec[c + d*x]^4*Sin[d*x])/(7*d) + (Sec[c]*Sec[c + d*x]^3*(5*C*Sin[c] + 7*C*Sin[d*x]))/(35*d) + (Sec[c]*Sec[c + d
*x]^2*(21*C*Sin[c] + 35*A*Sin[d*x] + 25*C*Sin[d*x]))/(105*d) + (Sec[c]*Sec[c + d*x]*(35*A*Sin[c] + 25*C*Sin[c]
+ 105*A*Sin[d*x] + 63*C*Sin[d*x]))/(105*d)) - (A*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4
}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c
]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sq
rt[1 + Cot[c]^2]) - (5*C*(1 + Cos[c + d*x])*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c
]]]^2]*Sec[c/2 + (d*x)/2]^2*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[
c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(21*d*Sqrt[1 + Cot[c]^2]) + (A*(
1 + Cos[c + d*x])*Csc[c]*Sec[c/2 + (d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]
]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]
]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]
*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))
/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d) + (3*C*(1 + Cos[c + d*x])*Csc[c]*Sec[c/2 +
(d*x)/2]^2*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan
[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan
[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Co
s[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c
]]]*Sqrt[1 + Tan[c]^2]]))/(10*d))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(810\) vs. \(2(197)=394\).
time = 0.26, size = 811, normalized size = 4.92

method	result	size
default	\(\text {Expression too large to display}\)	\(811\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-4*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a*(1/10*C/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+
1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*(2*sin(
1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c
)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*
x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+1/2*C*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+1/2*A/sin(1/2*d*x+1/2
*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*
cos(1/2*d*x+1/2*c)-(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),
2^(1/2)))+1/2*A*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2
*c)^2-1/2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1
/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(
1/2)/d

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Maxima [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((a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.49, size = 243, normalized size = 1.47 \begin {gather*} \frac {-5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (7 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (5 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, {\left (5 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 5 \, {\left (7 \, A + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 21 \, C a \cos \left (d x + c\right ) + 15 \, C a\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/105*(-5*I*sqrt(2)*(7*A + 5*C)*a*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5
*I*sqrt(2)*(7*A + 5*C)*a*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(
2)*(5*A + 3*C)*a*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c
))) + 21*I*sqrt(2)*(5*A + 3*C)*a*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c)
- I*sin(d*x + c))) + 2*(21*(5*A + 3*C)*a*cos(d*x + c)^3 + 5*(7*A + 5*C)*a*cos(d*x + c)^2 + 21*C*a*cos(d*x + c
) + 15*C*a)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a \left (\int \frac {A}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A \sec {\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{3}{\left (c + d x \right )}}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2)/cos(d*x+c)**(3/2),x)

[Out]

a*(Integral(A/cos(c + d*x)**(3/2), x) + Integral(A*sec(c + d*x)/cos(c + d*x)**(3/2), x) + Integral(C*sec(c + d
*x)**2/cos(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**3/cos(c + d*x)**(3/2), x))

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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((a+a*sec(d*x+c))*(A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(a*sec(d*x + c) + a)/cos(d*x + c)^(3/2), x)

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Mupad [B]
time = 6.21, size = 177, normalized size = 1.07 \begin {gather*} \frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,A\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x)))/cos(c + d*x)^(3/2),x)

[Out]

(2*A*a*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))
+ (2*A*a*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(
1/2)) + (2*C*a*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x
)^2)^(1/2)) + (2*C*a*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))/(7*d*cos(c + d*x)^(7/2)*(sin(c
+ d*x)^2)^(1/2))

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