3.1291 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=294 \[ \frac {(3 A-43 B+115 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-1/4*(A-B+C)*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(A+7*B-15*C)*sin(d*x+c)/a/d/cos(d*x+c)^
(5/2)/(a+a*sec(d*x+c))^(3/2)+(2*B-5*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^(5/2)/d+1/32*(3*A-43*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)+1/16*(3*A-11*B+35*C)*sin(d*x+c)/a^2/d/cos(
d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 1.00, antiderivative size = 294, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4265, 4084, 4019, 4021, 4023, 3808, 206, 3801, 215} \[ \frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {(3 A-43 B+115 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Cos[c + d*x]^(5/2)*(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]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(
a^(5/2)*d) + ((3*A - 43*B + 115*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]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c + d*x])/(4*d*Cos
[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(5/2)) + ((A + 7*B - 15*C)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(5/2)*(a +
a*Sec[c + d*x])^(3/2)) + ((3*A - 11*B + 35*C)*Sin[c + d*x])/(16*a^2*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 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)]

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+B \sec (c+d x)+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+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) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\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 (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) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{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 {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) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 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}{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) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\left ((2 B-5 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}{2 a^3}+\frac {\left ((3 A-43 B+115 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}{32 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {\left ((2 B-5 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 )}{a^3 d}-\frac {\left ((3 A-43 B+115 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 )}{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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B+C) \sin (c+d x)}{4 d \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(A+7 B-15 C) \sin (c+d x)}{16 a d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(3 A-11 B+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 3.94, size = 222, normalized size = 0.76 \[ \frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left ((6 A-86 B+230 C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \tan \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^3\left (\frac {1}{2} (c+d x)\right ) (2 (7 A-15 B+55 C) \cos (c+d x)+(3 A-11 B+35 C) \cos (2 (c+d x))+3 A-11 B+67 C)+32 \sqrt {2} (2 B-5 C) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{4 d \sqrt {\cos (c+d x)} (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((6*A - 86*B + 230*C)*ArcTanh[Sin[(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 - 15*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*Sqrt[
Cos[c + d*x]]*(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 = 0.90, size = 884, normalized size = 3.01 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(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)^4 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B +
115*C)*cos(d*x + c)^2 + (3*A - 43*B + 115*C)*cos(d*x + 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*((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))*sqrt(cos(d*x + c))*sin(d*x + c) - 16*((2*B - 5*C)*cos(d*x + c)^4 + 3*(
2*B - 5*C)*cos(d*x + c)^3 + 3*(2*B - 5*C)*cos(d*x + c)^2 + (2*B - 5*C)*cos(d*x + c))*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^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 +
3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/32*(sqrt(2)*((3*A - 43*B + 115*C)*cos(d*x + c)^4 + 3*(3*A -
43*B + 115*C)*cos(d*x + c)^3 + 3*(3*A - 43*B + 115*C)*cos(d*x + c)^2 + (3*A - 43*B + 115*C)*cos(d*x + 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*
((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))*sqrt(cos(d*x + c))*sin(d*x + c) - 16*((2*B - 5*C)*cos(d*x + c)^4 + 3*(2*B - 5*C)*cos(d*x + c)^3 + 3*(
2*B - 5*C)*cos(d*x + c)^2 + (2*B - 5*C)*cos(d*x + 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)))/(a^3*d*cos(d*x + c)^4 +
3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 2.34, size = 972, normalized size = 3.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/d*(-1+cos(d*x+c))^2*(a*(1+cos(d*x+c))/cos(d*x+c))^(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)+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+c
os(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+co
s(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)*a
rctan(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)*arct
an(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))/a^3/(-2/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^5/cos(d*x+c)^(1/2)

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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+B*sec(d*x+c)+C*sec(d*x+c)^2)/cos(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 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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