3.619 \(\int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=254 \[ \frac {(3 A-43 B+115 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
[Out]
(2*B-5*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/4*(A-B+C)*sec(d*x+c)^(7/2)*sin(d*x+c)
/d/(a+a*sec(d*x+c))^(5/2)+1/16*(A+7*B-15*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)+1/32*(3*A-4
3*B+115*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
/16*(3*A-11*B+35*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)
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Rubi [A] time = 0.82, antiderivative size = 254, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.178, Rules used =
{4084, 4019, 4021, 4023, 3808, 206, 3801, 215} \[ \frac {(3 A-11 B+35 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(3 A-43 B+115 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
((2*B - 5*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) + ((3*A - 43*B + 115*C)*Arc
Tanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - (
(A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((A + 7*B - 15*C)*Sec[c + d*x]
^(5/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((3*A - 11*B + 35*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x
])/(16*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 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 4019
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(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 - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b,
d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 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 4084
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*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Rubi steps
\begin {align*} \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (\frac {1}{2} a (3 A+5 B-5 C)+a (A-B+5 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\frac {3}{4} a^2 (A+7 B-15 C)+\frac {1}{2} a^2 (3 A-11 B+35 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sqrt {\sec (c+d x)} \left (\frac {1}{4} a^3 (3 A-11 B+35 C)+4 a^3 (2 B-5 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(2 B-5 C) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx}{2 a^3}+\frac {(3 A-43 B+115 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(2 B-5 C) \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 )}{a^3 d}-\frac {(3 A-43 B+115 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 {(2 B-5 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac {(3 A-43 B+115 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-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 3.65, size = 222, normalized size = 0.87 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((6 A-86 B+230 C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \tan \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^3\left (\frac {1}{2} (c+d x)\right ) (2 (7 A-15 B+55 C) \cos (c+d x)+(3 A-11 B+35 C) \cos (2 (c+d x))+3 A-11 B+67 C)+32 \sqrt {2} (2 B-5 C) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 d (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
(Cos[(c + d*x)/2]^5*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((6*A - 86*B + 230*C)*ArcTanh[S
in[(c + d*x)/2]] + 32*Sqrt[2]*(2*B - 5*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]] + ((3*A - 11*B + 67*C + 2*(7*A - 1
5*B + 55*C)*Cos[c + d*x] + (3*A - 11*B + 35*C)*Cos[2*(c + d*x)])*Sec[(c + d*x)/2]^3*Sec[c + d*x]*Tan[(c + d*x)
/2])/2))/(4*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(5/2))
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fricas [A] time = 0.74, size = 837, normalized size = 3.30 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[1/64*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + 3*(3*A - 43*B +
115*C)*cos(d*x + c) + 3*A - 43*B + 115*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)) - 16*((2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 + 3*(2*B - 5*C)*cos(d*x + c) + 2*B -
5*C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4
*((3*A - 11*B + 35*C)*cos(d*x + c)^2 + (7*A - 15*B + 55*C)*cos(d*x + c) + 16*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), -1/32*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + 3*(
3*A - 43*B + 115*C)*cos(d*x + c) + 3*A - 43*B + 115*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))) - 16*((2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*
x + c)^2 + 3*(2*B - 5*C)*cos(d*x + c) + 2*B - 5*C)*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)) - 2*((3*A - 11*B + 35*C)*co
s(d*x + c)^2 + (7*A - 15*B + 55*C)*cos(d*x + c) + 16*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/s
qrt(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*sec(d*x + c) + a)^(5/2), x)
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maple [B] time = 3.19, size = 982, normalized size = 3.87 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
[Out]
-1/16/d*(1/cos(d*x+c))^(5/2)*cos(d*x+c)^2*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-16*B*arctan(
1/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1+sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+16*B*arctan(1
/4*(-2/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)+1-sin(d*x+c))*2^(1/2))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+40*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)-40*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)+3*A*cos(d*x+c)^3
*(-2/(1+cos(d*x+c)))^(1/2)-3*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))-11*B*(
-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3+43*B*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos(d*x
+c)^2-16*B*cos(d*x+c)*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))*2^(1/
2)+16*B*cos(d*x+c)*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))*2^(1/2)+
35*C*cos(d*x+c)^3*(-2/(1+cos(d*x+c)))^(1/2)-115*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))+40*C*sin(d*x+c)*2^(1/2)*cos(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))-40*C*sin(d*x+c)*2^(1/2)*cos(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))+4*A*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-3*A*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d
*x+c)))^(1/2))-4*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+43*B*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2
/(1+cos(d*x+c)))^(1/2))+20*C*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)-115*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin
(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))-7*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+15*B*(-2/(1+cos(d*x+c)))^(1/2)*cos
(d*x+c)-39*C*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-16*C*(-2/(1+cos(d*x+c)))^(1/2))/(-2/(1+cos(d*x+c)))^(1/2)/si
n(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(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((1/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(5/2),x)
[Out]
int(((1/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/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(sec(d*x+c)**(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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