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]

-((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]])

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Rubi [A]  time = 0.734469, antiderivative size = 250, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 1.40641, 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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Maple [A]  time = 0.324, size = 339, normalized size = 1.4 \begin{align*}{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{60\,d{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}} \left ( 24\,A \left ( \cos \left ( dx+c \right ) \right ) ^{4}-225\,A\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}+165\,B\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) \arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}-48\,A \left ( \cos \left ( dx+c \right ) \right ) ^{3}-225\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}A\sin \left ( dx+c \right ) +40\,B \left ( \cos \left ( dx+c \right ) \right ) ^{3}+165\,\arctan \left ( 1/2\,\sin \left ( dx+c \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}} \right ) \sqrt{-2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}B\sin \left ( dx+c \right ) +240\,A \left ( \cos \left ( dx+c \right ) \right ) ^{2}-160\,B \left ( \cos \left ( dx+c \right ) \right ) ^{2}+78\,A\cos \left ( dx+c \right ) -70\,B\cos \left ( dx+c \right ) -294\,A+190\,B \right ) \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) ^{{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

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

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Fricas [A]  time = 0.543179, size = 1328, normalized size = 5.31 \begin{align*} \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 ] \end{align*}

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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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))/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),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{3}{2}} \sec \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))/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)