3.1166 \(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=285 \[ -\frac {(5 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(8 A+19 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-1/2*(A+C)*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(3/2)+1/4*(8*A+19*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a
*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d-1/4*(5*A+13*C)*arctanh(1/2*sin(d*x+c)*a^(1/2)*
sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(3/2)/d*2^(1/2)+1/2*(A+2*
C)*sin(d*x+c)/a/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(1/2)-1/4*(2*A+7*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)/(a+a*s
ec(d*x+c))^(1/2)

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Rubi [A]  time = 0.91, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {4265, 4085, 4021, 4023, 3808, 206, 3801, 215} \[ -\frac {(5 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(8 A+19 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

((8*A + 19*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/
(4*a^(3/2)*d) - ((5*A + 13*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*
x]])]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(2*Sqrt[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Cos[c + d*x]^
(7/2)*(a + a*Sec[c + d*x])^(3/2)) + ((A + 2*C)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*x]
]) - ((2*A + 7*C)*Sin[c + d*x])/(4*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 3801

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]

Rule 3808

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]

Rule 4021

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b
^2, 0] && GtQ[n, 1]

Rule 4023

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Dist[B
/b, Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A
*b - a*B, 0] && EqQ[a^2 - b^2, 0]

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(a*f*(
2*m + 1)), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b*(
2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x]
&& EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {1}{2} a (A+5 C)-2 a (A+2 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (-3 a^2 (A+2 C)+a^2 (2 A+7 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{2} a^3 (2 A+7 C)-\frac {1}{2} a^3 (8 A+19 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left ((5 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}+\frac {\left ((8 A+19 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx}{8 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\left ((5 A+13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}-\frac {\left ((8 A+19 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^2 d}\\ &=\frac {(8 A+19 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{4 a^{3/2} d}-\frac {(5 A+13 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(A+2 C) \sin (c+d x)}{2 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(2 A+7 C) \sin (c+d x)}{4 a d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.28, size = 213, normalized size = 0.75 \[ -\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (A \cos ^2(c+d x)+C\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right ) ((2 A+7 C) \cos (2 (c+d x))+2 A+6 C \cos (c+d x)+3 C)+(5 A+13 C) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {(8 A+19 C) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {3}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {2}}\right )}{4 a d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a (\sec (c+d x)+1)} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

-1/4*((C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]*((5*A + 13*C)*ArcTanh[Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[
(3*(c + d*x))/2])^2 - ((8*A + 19*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Cos[(3*(c + d*x))/2]
)^2)/Sqrt[2] + (2*A + 3*C + 6*C*Cos[c + d*x] + (2*A + 7*C)*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(a*d*Cos[c + d
*x]^(5/2)*(A + 2*C + A*Cos[2*(c + d*x)])*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.62, size = 754, normalized size = 2.65 \[ \left [\frac {2 \, \sqrt {2} {\left ({\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 4 \, \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} {\left (\cos \left (d x + c\right ) - 2\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right )}{16 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {2 \, \sqrt {2} {\left ({\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (5 \, A + 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (2 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) - 2 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left ({\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (8 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/16*(2*sqrt(2)*((5*A + 13*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c)^3 + (5*A + 13*C)*cos(d*x + c)^2)*s
qrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*s
in(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 4*((2*A + 7*C)*cos(d*x + c)^2 +
3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*
C)*cos(d*x + c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 19*C)*cos(d*x + c)^2)*sqrt(a)*log((a*cos(d*x + c)^3
- 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(cos(d*x + c))*sin(d*x + c) - 7*a*
cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2
*d*cos(d*x + c)^2), 1/8*(2*sqrt(2)*((5*A + 13*C)*cos(d*x + c)^4 + 2*(5*A + 13*C)*cos(d*x + c)^3 + (5*A + 13*C)
*cos(d*x + c)^2)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(
a*sin(d*x + c))) - 2*((2*A + 7*C)*cos(d*x + c)^2 + 3*C*cos(d*x + c) - 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((8*A + 19*C)*cos(d*x + c)^4 + 2*(8*A + 19*C)*cos(d*x + c)^3 + (8*A + 1
9*C)*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin
(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*c
os(d*x + c)^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2)), x)

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maple [B]  time = 2.55, size = 508, normalized size = 1.78 \[ -\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (8 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-8 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )+19 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}-19 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right )+1-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}+4 A \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-20 A \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )+14 C \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-52 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )-4 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )-8 C \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-10 C \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+4 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right )}{8 d \,a^{2} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/8/d*(-1+cos(d*x+c))*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(8*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*arctan(1/4*(-2
/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))-8*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*arctan(1/4*(-2/(
1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))+19*C*sin(d*x+c)*cos(d*x+c)^2*arctan(1/4*(-2/(1+cos(d*x
+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*2^(1/2)-19*C*sin(d*x+c)*cos(d*x+c)^2*arctan(1/4*(-2/(1+cos(d*x+
c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*2^(1/2)+4*A*cos(d*x+c)^3*(-2/(1+cos(d*x+c)))^(1/2)-20*A*sin(d*x+
c)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))+14*C*cos(d*x+c)^3*(-2/(1+cos(d*x+c)))^(1/2)-5
2*C*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))-4*A*cos(d*x+c)^2*(-2/(1+cos(d*x+c
)))^(1/2)-8*C*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-10*C*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+4*C*(-2/(1+cos(
d*x+c)))^(1/2))/a^2/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^(3/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^(5/2)*(a + a/cos(c + d*x))^(3/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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