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

Optimal. Leaf size=295 \[ -\frac {(283 A+75 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(2671 A+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2)-1/16*(21*A+5*C)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2
)/(a+a*sec(d*x+c))^(3/2)-1/32*(283*A+75*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))/a^(5/2)/d*2^(1/2)+1/80*(157*A+45*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)-1/24
0*(787*A+195*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/240*(2671*A+735*C)*sin(d*x+c)*sec(d
*x+c)^(1/2)/a^2/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.91, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {4085, 4020, 4022, 4013, 3808, 206} \[ \frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(2671 A+735 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{240 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}-\frac {(283 A+75 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((283*A + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sq
rt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)) - ((21*A + 5*C)*
Sin[c + d*x])/(16*a*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((157*A + 45*C)*Sin[c + d*x])/(80*a^2*d
*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((787*A + 195*C)*Sin[c + d*x])/(240*a^2*d*Sqrt[Sec[c + d*x]]*S
qrt[a + a*Sec[c + d*x]]) + ((2671*A + 735*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(240*a^2*d*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 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 4013

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

Rule 4020

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

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*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] && LtQ[n, 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)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a (13 A+5 C)+4 a A \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {-\frac {1}{4} a^2 (157 A+45 C)+\frac {3}{2} a^2 (21 A+5 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\frac {1}{8} a^3 (787 A+195 C)-\frac {1}{2} a^3 (157 A+45 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {-\frac {1}{16} a^4 (2671 A+735 C)+\frac {1}{8} a^4 (787 A+195 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(283 A+75 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(283 A+75 C) \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 )}{16 a^2 d}\\ &=-\frac {(283 A+75 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 )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \sin (c+d x)}{4 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac {(21 A+5 C) \sin (c+d x)}{16 a d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(157 A+45 C) \sin (c+d x)}{80 a^2 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(787 A+195 C) \sin (c+d x)}{240 a^2 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(2671 A+735 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 4.25, size = 349, normalized size = 1.18 \[ \frac {(\sec (c+d x)+1)^{5/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {\left (\sin \left (\frac {3}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)+1} (5 (887 A+255 C) \cos (c+d x)+16 (52 A+15 C) \cos (2 (c+d x))-40 A \cos (3 (c+d x))+12 A \cos (4 (c+d x))+3491 A+975 C)}{\sec ^{\frac {3}{2}}(c+d x)}-15 \sqrt {2} (283 A+75 C) \cos ^2(c+d x) \sqrt {\tan ^2(c+d x)} \cot (c+d x) \left (\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)-2 \sqrt {2} \sqrt {\tan ^2(c+d x)} \sqrt {\sec (c+d x)+1} \sqrt {\sec (c+d x)}+1\right )-\log \left (-3 \sec ^2(c+d x)-2 \sec (c+d x)+2 \sqrt {2} \sqrt {\tan ^2(c+d x)} \sqrt {\sec (c+d x)+1} \sqrt {\sec (c+d x)}+1\right )\right )\right )}{960 d (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + Sec[c + d*x])^(5/2)*(A + C*Sec[c + d*x]^2)*(((3491*A + 975*C + 5*(887*A + 255*C)*Cos[c + d*x] + 16*(52*A
+ 15*C)*Cos[2*(c + d*x)] - 40*A*Cos[3*(c + d*x)] + 12*A*Cos[4*(c + d*x)])*Sec[(c + d*x)/2]^5*Sqrt[1 + Sec[c +
d*x]]*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2]))/Sec[c + d*x]^(3/2) - 15*Sqrt[2]*(283*A + 75*C)*Cos[c + d*x]
^2*Cot[c + d*x]*(Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x
]]*Sqrt[Tan[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + S
ec[c + d*x]]*Sqrt[Tan[c + d*x]^2]])*Sqrt[Tan[c + d*x]^2]))/(960*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c
+ d*x]))^(5/2))

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fricas [A]  time = 0.50, size = 584, normalized size = 1.98 \[ \left [\frac {15 \, \sqrt {2} {\left ({\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right ) + 283 \, A + 75 \, C\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 ) + \frac {4 \, {\left (96 \, A \cos \left (d x + c\right )^{5} - 160 \, A \cos \left (d x + c\right )^{4} + 32 \, {\left (49 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (911 \, A + 255 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (2671 \, A + 735 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{960 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (283 \, A + 75 \, C\right )} \cos \left (d x + c\right ) + 283 \, A + 75 \, C\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 ) + \frac {2 \, {\left (96 \, A \cos \left (d x + c\right )^{5} - 160 \, A \cos \left (d x + c\right )^{4} + 32 \, {\left (49 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (911 \, A + 255 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (2671 \, A + 735 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{480 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/960*(15*sqrt(2)*((283*A + 75*C)*cos(d*x + c)^3 + 3*(283*A + 75*C)*cos(d*x + c)^2 + 3*(283*A + 75*C)*cos(d*x
+ c) + 283*A + 75*C)*sqrt(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))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(96*A*
cos(d*x + c)^5 - 160*A*cos(d*x + c)^4 + 32*(49*A + 15*C)*cos(d*x + c)^3 + 5*(911*A + 255*C)*cos(d*x + c)^2 + (
2671*A + 735*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*
cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/480*(15*sqrt(2)*((283*A + 75*C)*cos
(d*x + c)^3 + 3*(283*A + 75*C)*cos(d*x + c)^2 + 3*(283*A + 75*C)*cos(d*x + c) + 283*A + 75*C)*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*(96*A*cos(d*
x + c)^5 - 160*A*cos(d*x + c)^4 + 32*(49*A + 15*C)*cos(d*x + c)^3 + 5*(911*A + 255*C)*cos(d*x + c)^2 + (2671*A
+ 735*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3*d*cos(d*
x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.19, size = 460, normalized size = 1.56 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right )^{2} \left (192 A \left (\cos ^{5}\left (d x +c \right )\right )-4245 A \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 ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-1125 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 ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-512 A \left (\cos ^{4}\left (d x +c \right )\right )-8490 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-2250 C \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+3456 A \left (\cos ^{3}\left (d x +c \right )\right )-4245 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \sin \left (d x +c \right )+960 C \left (\cos ^{3}\left (d x +c \right )\right )-1125 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+5974 A \left (\cos ^{2}\left (d x +c \right )\right )+1590 C \left (\cos ^{2}\left (d x +c \right )\right )-3768 A \cos \left (d x +c \right )-1080 C \cos \left (d x +c \right )-5342 A -1470 C \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{480 d \sin \left (d x +c \right )^{5} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(192*A*cos(d*x+c)^5-4245*A*cos(d*x+c)^2*arctan(
1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-1125*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))*(-2/(1+cos(d*x+c)))^(1/2)-512*A*cos(d*x+c)^4-8490*A*arctan(1/
2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*sin(d*x+c)-2250*C*sin(d*x+c)*arct
an(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+3456*A*cos(d*x+c)^3-4245*arc
tan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*A*sin(d*x+c)+960*C*cos(d*x+c)^3-1125*C
*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+5974*A*cos(d*x+c)^2+159
0*C*cos(d*x+c)^2-3768*A*cos(d*x+c)-1080*C*cos(d*x+c)-5342*A-1470*C)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+
c)^5/a^3

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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)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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

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))^(5/2)*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(5/2)), 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)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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