\(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\) [560]

Optimal result

Integrand size = 35, antiderivative size = 234 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 B \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{5/2} d}+\frac {(3 A-43 B) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \]

[Out]

1/4*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(3*A-11*B)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)
/(a+a*sec(d*x+c))^(3/2)+2*B*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)*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))*c
os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3034, 4104, 4108, 3893, 212, 3886, 221} \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {(3 A-43 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \text {arcsinh}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(2*B*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)*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)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(5/2)*(a +
a*Sec[c + d*x])^(5/2)) + ((3*A - 11*B)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2))

Rule 212

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

Rule 221

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 3034

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] &&  !IntegerQ[p] &&  !(IntegerQ[m] && IntegerQ[n])

Rule 3886

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], 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 3893

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 4104

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 4108

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]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx \\ & = \frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{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 {3}{2}}(c+d x) \left (\frac {3}{2} a (A-B)+4 a B \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{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 {\sqrt {\sec (c+d x)} \left (\frac {1}{4} a^2 (3 A-11 B)+8 a^2 B \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4} \\ & = \frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {\left ((3 A-43 B) \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 {\left (B \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}{a^3} \\ & = \frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac {\left ((3 A-43 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 {\left (2 B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \text {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 {2 B \text {arcsinh}\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) \text {arctanh}\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) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(772\) vs. \(2(234)=468\).

Time = 6.21 (sec) , antiderivative size = 772, normalized size of antiderivative = 3.30 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\frac {A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} (1+\sec (c+d x))^{5/2} \left (-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{4 d (1+\sec (c+d x))^{5/2}}-\frac {\sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{16 d (1+\sec (c+d x))^{3/2}}+\frac {3 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 d \sqrt {1+\sec (c+d x)}}+\frac {\sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 d \sqrt {1+\sec (c+d x)}}+\frac {3 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {3 \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {3 \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{16 \sqrt {2} d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{(a (1+\sec (c+d x)))^{5/2}}+\frac {B \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} (1+\sec (c+d x))^{5/2} \left (-\frac {\sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{4 d (1+\sec (c+d x))^{5/2}}+\frac {3 \sec ^{\frac {9}{2}}(c+d x) \sin (c+d x)}{16 d (1+\sec (c+d x))^{3/2}}-\frac {11 \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 d \sqrt {1+\sec (c+d x)}}+\frac {7 \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{16 d \sqrt {1+\sec (c+d x)}}-\frac {3 \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{16 d \sqrt {1+\sec (c+d x)}}-\frac {11 \arcsin \left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}-\frac {43 \arcsin \left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)}{16 d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}+\frac {43 \arctan \left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right ) \tan (c+d x)}{16 \sqrt {2} d \sqrt {1-\sec (c+d x)} \sqrt {1+\sec (c+d x)}}\right )}{(a (1+\sec (c+d x)))^{5/2}} \]

[In]

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

[Out]

(A*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])^(5/2)*(-1/4*(Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(
1 + Sec[c + d*x])^(5/2)) - (Sec[c + d*x]^(7/2)*Sin[c + d*x])/(16*d*(1 + Sec[c + d*x])^(3/2)) + (3*Sec[c + d*x]
^(3/2)*Sin[c + d*x])/(16*d*Sqrt[1 + Sec[c + d*x]]) + (Sec[c + d*x]^(5/2)*Sin[c + d*x])/(16*d*Sqrt[1 + Sec[c +
d*x]]) + (3*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])
+ (3*ArcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]) - (3*ArcTan
[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(16*Sqrt[2]*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[
1 + Sec[c + d*x]])))/(a*(1 + Sec[c + d*x]))^(5/2) + (B*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x]
)^(5/2)*(-1/4*(Sec[c + d*x]^(9/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])^(5/2)) + (3*Sec[c + d*x]^(9/2)*Sin[c + d
*x])/(16*d*(1 + Sec[c + d*x])^(3/2)) - (11*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(16*d*Sqrt[1 + Sec[c + d*x]]) + (7
*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(16*d*Sqrt[1 + Sec[c + d*x]]) - (3*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(16*d*Sq
rt[1 + Sec[c + d*x]]) - (11*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 +
Sec[c + d*x]]) - (43*ArcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[1 + Sec[c + d
*x]]) + (43*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x])/(16*Sqrt[2]*d*Sqrt[1 - S
ec[c + d*x]]*Sqrt[1 + Sec[c + d*x]])))/(a*(1 + Sec[c + d*x]))^(5/2)

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(689\) vs. \(2(195)=390\).

Time = 8.22 (sec) , antiderivative size = 690, normalized size of antiderivative = 2.95

method	result	size
default	\(\frac {-\frac {B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2} \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+16 \sqrt {2}\, \arctan \left (\frac {\left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+16 \sqrt {2}\, \arctan \left (\frac {\left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )-1\right ) \sqrt {2}}{2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+13 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-43 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 \left (-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}\right )^{\frac {3}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1\right ) a \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}-\frac {A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (2 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+5 \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \arctan \left (\frac {-\cot \left (d x +c \right )+\csc \left (d x +c \right )}{\sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )\right )}{32 \sqrt {-\frac {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}+1}}\, a \sqrt {-\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}}{a^{2} d}\)	\(690\)

[In]

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

[Out]

1/a^2/d*(-1/32*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^2*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(2*(-(1-co
s(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3+16*2^(1/2)*arctan(1/2*(csc(d*x+c)-cot(d*x+c)+1
)*2^(1/2)/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+16*2^(1/2)*arctan(1/2*(-cot(d*x+c)+csc(d*x+c)-1)*2^(1/2)/(
-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+13*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-
43*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))))/(-((1-cos(d*x+c))^2*csc(d*x+c)
^2-1)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))^(3/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1)/a/(-(1-cos(d*x+c))^2*csc(d*x+
c)^2-1)^(1/2)-1/32*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(2*(-(1-
cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(1-cos(d*x+c))^3*csc(d*x+c)^3+5*(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(
-cot(d*x+c)+csc(d*x+c))-3*arctan(1/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))))/(-((1-c
os(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^2*csc(d*x+c)^2+1))^(1/2)/a/(-(1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/
2))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 720, normalized size of antiderivative = 3.08 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B\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 (3 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 15 \, B\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 ) - 32 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\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 )}{64 \, {\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 {\sqrt {2} {\left ({\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 43 \, B\right )} \cos \left (d x + c\right ) + 3 \, A - 43 \, B\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 (3 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 7 \, A - 15 \, B\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 ) - 32 \, {\left (B \cos \left (d x + c\right )^{3} + 3 \, B \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + B\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 )}{32 \, {\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 ] \]

[In]

integrate((A+B*sec(d*x+c))/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)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c)^2 + 3*(3*A - 43*B)*cos(d*x + c) + 3
*A - 43*B)*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)*cos
(d*x + c) + 7*A - 15*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 32*(B*cos(d*
x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*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)^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)*cos(d*x + c)^3 + 3*(3*A - 43*B)*cos(d*x + c)^2 + 3*(3*A - 43*B)*cos(d*x + c) + 3
*A - 43*B)*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)*cos(d*x + c) + 7*A - 15*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x +
c))*sin(d*x + c) - 32*(B*cos(d*x + c)^3 + 3*B*cos(d*x + c)^2 + 3*B*cos(d*x + c) + B)*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)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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