3.261 \(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=250 \[ -\frac {(15 A-11 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac {(147 A-95 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \sec (c+d x)+a}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

[Out]

-1/2*(A-B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2)-1/4*(15*A-11*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))/a^(3/2)/d*2^(1/2)+1/10*(9*A-5*B)*sin(d*x+c)/a/d/sec(d*x+c)^(
3/2)/(a+a*sec(d*x+c))^(1/2)-1/30*(39*A-35*B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/30*(147*
A-95*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.73, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4020, 4022, 4013, 3808, 206} \[ -\frac {(15 A-11 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac {(147 A-95 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \sec (c+d x)+a}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

-((15*A - 11*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt
[2]*a^(3/2)*d) - ((A - B)*Sin[c + d*x])/(2*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((9*A - 5*B)*Sin
[c + d*x])/(10*a*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((39*A - 35*B)*Sin[c + d*x])/(30*a*d*Sqrt[Se
c[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((147*A - 95*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(30*a*d*Sqrt[a + a*Se
c[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]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (9 A-5 B)-3 a (A-B) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {-\frac {1}{4} a^2 (39 A-35 B)+a^2 (9 A-5 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\frac {1}{8} a^3 (147 A-95 B)-\frac {1}{4} a^3 (39 A-35 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-11 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}+\frac {(15 A-11 B) \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 )}{2 a d}\\ &=-\frac {(15 A-11 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 )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.50, size = 171, normalized size = 0.68 \[ \frac {\sec (c+d x) \left (\frac {15 \sqrt {2} (15 A-11 B) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \tan (c+d x) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {\sec (c+d x)}}{\sqrt {1-\sec (c+d x)}}\right )}{\sqrt {1-\sec (c+d x)}}+\sin (c+d x) \sqrt {\sec (c+d x)} (3 (39 A-20 B) \cos (c+d x)+(10 B-6 A) \cos (2 (c+d x))+3 A \cos (3 (c+d x))+141 A-85 B)\right )}{30 d (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

(Sec[c + d*x]*((141*A - 85*B + 3*(39*A - 20*B)*Cos[c + d*x] + (-6*A + 10*B)*Cos[2*(c + d*x)] + 3*A*Cos[3*(c +
d*x)])*Sqrt[Sec[c + d*x]]*Sin[c + d*x] + (15*Sqrt[2]*(15*A - 11*B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1
- Sec[c + d*x]]]*Cos[(c + d*x)/2]^2*Tan[c + d*x])/Sqrt[1 - Sec[c + d*x]]))/(30*d*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [A]  time = 0.49, size = 500, normalized size = 2.00 \[ \left [-\frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, 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 ) - \frac {4 \, {\left (12 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (147 \, A - 95 \, B\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 )}}}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A - 11 \, B\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, 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 ) + \frac {2 \, {\left (12 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (147 \, A - 95 \, B\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 )}}}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/120*(15*sqrt(2)*((15*A - 11*B)*cos(d*x + c)^2 + 2*(15*A - 11*B)*cos(d*x + c) + 15*A - 11*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*(12*A*cos(d*x + c)^4 - 4*(3*A - 5*B)*cos(d
*x + c)^3 + 12*(9*A - 5*B)*cos(d*x + c)^2 + (147*A - 95*B)*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^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/60*(15*sqrt(2)*((
15*A - 11*B)*cos(d*x + c)^2 + 2*(15*A - 11*B)*cos(d*x + c) + 15*A - 11*B)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqr
t((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(12*A*cos(d*x + c)^4 - 4*(3*A -
5*B)*cos(d*x + c)^3 + 12*(9*A - 5*B)*cos(d*x + c)^2 + (147*A - 95*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/c
os(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), x)

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maple [A]  time = 2.80, size = 339, normalized size = 1.36 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (225 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-24 A \left (\cos ^{4}\left (d x +c \right )\right )-165 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+225 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \sin \left (d x +c \right )+48 A \left (\cos ^{3}\left (d x +c \right )\right )-165 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, B \sin \left (d x +c \right )-40 B \left (\cos ^{3}\left (d x +c \right )\right )-240 A \left (\cos ^{2}\left (d x +c \right )\right )+160 B \left (\cos ^{2}\left (d x +c \right )\right )-78 A \cos \left (d x +c \right )+70 B \cos \left (d x +c \right )+294 A -190 B \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{60 d \sin \left (d x +c \right )^{3} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

-1/60/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(225*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)-24*A*cos(d*x+c)^4-165*B*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)+225*arctan(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)+48*A*cos(d*x+c)^3-165*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))
^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*B*sin(d*x+c)-40*B*cos(d*x+c)^3-240*A*cos(d*x+c)^2+160*B*cos(d*x+c)^2-78*A*co
s(d*x+c)+70*B*cos(d*x+c)+294*A-190*B)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)^3/a^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^(3/2)*(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+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Timed out

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