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

Optimal. Leaf size=234 \[ \frac{(3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[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))

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Rubi [A]  time = 0.713895, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2955, 4019, 4023, 3808, 206, 3801, 215} \[ \frac{(3 A-43 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{2 B \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(3 A-11 B) \sin (c+d x)}{16 a d \cos ^{\frac{3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac{(A-B) \sin (c+d x)}{4 d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[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 2955

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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \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 (\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 ) \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{\left (2 B \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{2 B \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) \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) \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]  time = 6.15611, size = 965, normalized size = 4.12 \[ \frac{3 A \sin ^{-1}\left (\sqrt{1-\sec (c+d x)}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}-\frac{11 B \sin ^{-1}\left (\sqrt{1-\sec (c+d x)}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac{3 A \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}-\frac{43 B \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}-\frac{3 A \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 \sqrt{2} d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac{43 B \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x) (\sec (c+d x)+1)^2}{16 \sqrt{2} d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}}+\frac{3 A \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \cos ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}-\frac{11 B \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \cos ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}+\frac{A \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \cos ^{\frac{5}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}+\frac{7 B \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \cos ^{\frac{5}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}-\frac{3 B \sin (c+d x) (\sec (c+d x)+1)^2}{16 d \cos ^{\frac{7}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}-\frac{A \sin (c+d x) (\sec (c+d x)+1)}{16 d \cos ^{\frac{7}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}+\frac{3 B \sin (c+d x) (\sec (c+d x)+1)}{16 d \cos ^{\frac{9}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}-\frac{A \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}}-\frac{B \sin (c+d x)}{4 d \cos ^{\frac{9}{2}}(c+d x) (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(B*Sin[c + d*x])/(4*d*Cos[c + d*x]^(9/2)*(a*(1 + Sec[c + d*x]))^(5/2)) - (A*Sin[c + d*x])/(4*d*Cos[c + d*x]^(
7/2)*(a*(1 + Sec[c + d*x]))^(5/2)) + (3*B*(1 + Sec[c + d*x])*Sin[c + d*x])/(16*d*Cos[c + d*x]^(9/2)*(a*(1 + Se
c[c + d*x]))^(5/2)) - (A*(1 + Sec[c + d*x])*Sin[c + d*x])/(16*d*Cos[c + d*x]^(7/2)*(a*(1 + Sec[c + d*x]))^(5/2
)) - (3*B*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*Cos[c + d*x]^(7/2)*(a*(1 + Sec[c + d*x]))^(5/2)) + (A*(1 +
Sec[c + d*x])^2*Sin[c + d*x])/(16*d*Cos[c + d*x]^(5/2)*(a*(1 + Sec[c + d*x]))^(5/2)) + (7*B*(1 + Sec[c + d*x])
^2*Sin[c + d*x])/(16*d*Cos[c + d*x]^(5/2)*(a*(1 + Sec[c + d*x]))^(5/2)) + (3*A*(1 + Sec[c + d*x])^2*Sin[c + d*
x])/(16*d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2)) - (11*B*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*Co
s[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(5/2)) + (3*A*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sec[c
+ d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) - (
11*B*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(
16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) + (3*A*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]
*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/
2)) - (43*B*ArcSin[Sqrt[Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x]
)/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) - (3*A*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1
- Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*Sqrt[2]*d*Sqrt[
1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2)) + (43*B*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d
*x]]]*Sqrt[Cos[c + d*x]]*Sec[c + d*x]^(3/2)*(1 + Sec[c + d*x])^2*Sin[c + d*x])/(16*Sqrt[2]*d*Sqrt[1 - Sec[c +
d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.321, size = 540, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 0.688488, size = 1906, normalized size = 8.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)