3.554 \(\int \frac{A+B \sec (c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=287 \[ \frac{(9 A-13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(12 A-19 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}+\frac{(6 A-7 B) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-((12*A - 19*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]]
)/(4*a^(3/2)*d) + ((9*A - 13*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]])/(2*Sqrt[2]*a^(3/2)*d) + ((A - B)*Sin[c + d*x])/(2*d*Cos[c + d*x
]^(7/2)*(a + a*Sec[c + d*x])^(3/2)) - ((A - 2*B)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*
x]]) + ((6*A - 7*B)*Sin[c + d*x])/(4*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.94735, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {2955, 4019, 4021, 4023, 3808, 206, 3801, 215} \[ \frac{(9 A-13 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(12 A-19 B) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}+\frac{(6 A-7 B) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

-((12*A - 19*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]]
)/(4*a^(3/2)*d) + ((9*A - 13*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]])/(2*Sqrt[2]*a^(3/2)*d) + ((A - B)*Sin[c + d*x])/(2*d*Cos[c + d*x
]^(7/2)*(a + a*Sec[c + d*x])^(3/2)) - ((A - 2*B)*Sin[c + d*x])/(2*a*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Sec[c + d*
x]]) + ((6*A - 7*B)*Sin[c + d*x])/(4*a*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

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 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 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{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{7}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{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{5}{2}}(c+d x) \left (\frac{5}{2} a (A-B)-2 a (A-2 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{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{\sec ^{\frac{3}{2}}(c+d x) \left (-3 a^2 (A-2 B)+a^2 (6 A-7 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{(6 A-7 B) \sin (c+d x)}{4 a 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}{2} a^3 (6 A-7 B)-\frac{1}{2} a^3 (12 A-19 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{(6 A-7 B) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left ((12 A-19 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}{8 a^2}+\frac{\left ((9 A-13 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}{4 a}\\ &=\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{(6 A-7 B) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{\left ((12 A-19 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 )}{4 a^2 d}-\frac{\left ((9 A-13 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 )}{2 a d}\\ &=-\frac{(12 A-19 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)}}{4 a^{3/2} d}+\frac{(9 A-13 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)}}{2 \sqrt{2} a^{3/2} d}+\frac{(A-B) \sin (c+d x)}{2 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{(A-2 B) \sin (c+d x)}{2 a d \cos ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{(6 A-7 B) \sin (c+d x)}{4 a d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.30096, size = 328, normalized size = 1.14 \[ \frac{\sin (c+d x) \sec ^{\frac{3}{2}}(c+d x) \left (2 (6 A-7 B) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sin ^{-1}\left (\sqrt{1-\sec (c+d x)}\right )-2 \sqrt{2} (9 A-13 B) \cos ^2\left (\frac{1}{2} (c+d x)\right ) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )+4 A \sqrt{-(\sec (c+d x)-1) \sec (c+d x)}+6 A \cos (c+d x) \sqrt{(\cos (c+d x)-1) \sec ^2(c+d x)}+36 A \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )+2 B \sqrt{1-\sec (c+d x)} \sec ^{\frac{3}{2}}(c+d x)-3 B \sqrt{-(\sec (c+d x)-1) \sec (c+d x)}-7 B \cos (c+d x) \sqrt{(\cos (c+d x)-1) \sec ^2(c+d x)}-52 B \cos ^2\left (\frac{1}{2} (c+d x)\right ) \sin ^{-1}\left (\sqrt{\sec (c+d x)}\right )\right )}{4 d \sqrt{\cos (c+d x)-1} (a (\sec (c+d x)+1))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(7/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

(Sec[c + d*x]^(3/2)*(2*(6*A - 7*B)*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2 + 36*A*ArcSin[Sqrt[Sec[c
+ d*x]]]*Cos[(c + d*x)/2]^2 - 52*B*ArcSin[Sqrt[Sec[c + d*x]]]*Cos[(c + d*x)/2]^2 - 2*Sqrt[2]*(9*A - 13*B)*ArcT
an[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^2 + 2*B*Sqrt[1 - Sec[c + d*x]]*Sec[c
+ d*x]^(3/2) + 4*A*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] - 3*B*Sqrt[-((-1 + Sec[c + d*x])*Sec[c + d*x])] +
6*A*Cos[c + d*x]*Sqrt[(-1 + Cos[c + d*x])*Sec[c + d*x]^2] - 7*B*Cos[c + d*x]*Sqrt[(-1 + Cos[c + d*x])*Sec[c +
d*x]^2])*Sin[c + d*x])/(4*d*Sqrt[-1 + Cos[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B]  time = 0.291, size = 531, normalized size = 1.9 \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)^(7/2)/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/8/d*(-1+cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(12*A*sin(d*x+c)*cos(d*x+c)^2*arctan(1/4*2^(1/2)*(-2
/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)-12*A*sin(d*x+c)*cos(d*x+c)^2*arctan(1/4*2^(1/2)*(-2/
(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*2^(1/2)-19*B*sin(d*x+c)*cos(d*x+c)^2*arctan(1/4*2^(1/2)*(-2/(
cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*2^(1/2)+19*B*sin(d*x+c)*cos(d*x+c)^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)))*2^(1/2)-36*A*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*(-2/
(cos(d*x+c)+1))^(1/2))+12*A*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1/2)+52*B*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*sin
(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-14*B*cos(d*x+c)^3*(-2/(cos(d*x+c)+1))^(1/2)-4*A*cos(d*x+c)^2*(-2/(cos(d*x+c
)+1))^(1/2)+8*B*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)-8*A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+10*B*cos(d*x+c
)*(-2/(cos(d*x+c)+1))^(1/2)-4*B*(-2/(cos(d*x+c)+1))^(1/2))/a^2/(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^3/cos(d*x+
c)^(3/2)

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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)^(7/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.862492, size = 1989, normalized size = 6.93 \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)^(7/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(2*sqrt(2)*((9*A - 13*B)*cos(d*x + c)^4 + 2*(9*A - 13*B)*cos(d*x + c)^3 + (9*A - 13*B)*cos(d*x + c)^2)*
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*((6*A - 7*B)*cos(d*x + c)^2
+ (4*A - 3*B)*cos(d*x + c) + 2*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((
12*A - 19*B)*cos(d*x + c)^4 + 2*(12*A - 19*B)*cos(d*x + c)^3 + (12*A - 19*B)*cos(d*x + c)^2)*sqrt(a)*log((a*co
s(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^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x
+ c)^3 + a^2*d*cos(d*x + c)^2), -1/8*(2*sqrt(2)*((9*A - 13*B)*cos(d*x + c)^4 + 2*(9*A - 13*B)*cos(d*x + c)^3 +
(9*A - 13*B)*cos(d*x + c)^2)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(co
s(d*x + c))/(a*sin(d*x + c))) - 2*((6*A - 7*B)*cos(d*x + c)^2 + (4*A - 3*B)*cos(d*x + c) + 2*B)*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) + ((12*A - 19*B)*cos(d*x + c)^4 + 2*(12*A - 19*B)*co
s(d*x + c)^3 + (12*A - 19*B)*cos(d*x + c)^2)*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^2*d*cos(d*x + c)^4 + 2*a^2*d*
cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

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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)**(7/2)/(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}} \cos \left (d x + c\right )^{\frac{7}{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)^(7/2)/(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)*cos(d*x + c)^(7/2)), x)