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

Optimal. Leaf size=250 \[ -\frac{(299 A-147 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

[Out]

((163*A - 75*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqr
t[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)) - ((17*A - 9*B)*S
in[c + d*x])/(16*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((95*A - 39*B)*Sin[c + d*x])/(48*a^2*d*S
qrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((299*A - 147*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(48*a^2*d*Sqrt
[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.761036, 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} \[ -\frac{(299 A-147 B) \sin (c+d x) \sqrt{\sec (c+d x)}}{48 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a \sec (c+d x)+a}}+\frac{(163 A-75 B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((163*A - 75*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqr
t[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(5/2)) - ((17*A - 9*B)*S
in[c + d*x])/(16*a*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)) + ((95*A - 39*B)*Sin[c + d*x])/(48*a^2*d*S
qrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((299*A - 147*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(48*a^2*d*Sqrt
[a + a*Sec[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{3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\frac{1}{2} a (11 A-3 B)-3 a (A-B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\frac{1}{4} a^2 (95 A-39 B)-a^2 (17 A-9 B) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-\frac{1}{8} a^3 (299 A-147 B)+\frac{1}{4} a^3 (95 A-39 B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(163 A-75 B) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(163 A-75 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 )}{16 a^2 d}\\ &=\frac{(163 A-75 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 )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B) \sin (c+d x)}{4 d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{5/2}}-\frac{(17 A-9 B) \sin (c+d x)}{16 a d \sqrt{\sec (c+d x)} (a+a \sec (c+d x))^{3/2}}+\frac{(95 A-39 B) \sin (c+d x)}{48 a^2 d \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(299 A-147 B) \sqrt{\sec (c+d x)} \sin (c+d x)}{48 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 1.73511, size = 193, normalized size = 0.77 \[ \frac{2 \tan (c+d x) \sqrt{1-\sec (c+d x)} \sec ^2(c+d x) ((255 B-479 A) \cos (c+d x)+(48 B-80 A) \cos (2 (c+d x))+8 A \cos (3 (c+d x))-379 A+195 B)-12 \sqrt{2} (163 A-75 B) \sin (c+d x) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^{\frac{7}{2}}(c+d x) \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{\sec (c+d x)}}{\sqrt{1-\sec (c+d x)}}\right )}{96 d \sqrt{-(\sec (c+d x)-1) \sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-12*Sqrt[2]*(163*A - 75*B)*ArcTan[(Sqrt[2]*Sqrt[Sec[c + d*x]])/Sqrt[1 - Sec[c + d*x]]]*Cos[(c + d*x)/2]^4*Sec
[c + d*x]^(7/2)*Sin[c + d*x] + 2*(-379*A + 195*B + (-479*A + 255*B)*Cos[c + d*x] + (-80*A + 48*B)*Cos[2*(c + d
*x)] + 8*A*Cos[3*(c + d*x)])*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]^2*Tan[c + d*x])/(96*d*Sqrt[-((-1 + Sec[c + d*
x])*Sec[c + d*x])]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.323, size = 449, normalized size = 1.8 \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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/96/d/a^3*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(489*A*sin(d*x+c)*cos(d*x+c)^2*arctan(1/2*si
n(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-225*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))*(-2/(cos(d*x+c)+1))^(1/2)+978*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)+64*A*cos(d*x+c)^4-450*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)+489*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)-384*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)*B*sin(d*x+c)+192*B*cos(d*x+c)^3-686*A*cos(d*x+c)^2+318*B*cos(d*x+c)^2+408*A*cos
(d*x+c)-216*B*cos(d*x+c)+598*A-294*B)*cos(d*x+c)^2*(1/cos(d*x+c))^(3/2)/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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.543725, size = 1503, normalized size = 6.01 \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))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/192*(3*sqrt(2)*((163*A - 75*B)*cos(d*x + c)^3 + 3*(163*A - 75*B)*cos(d*x + c)^2 + 3*(163*A - 75*B)*cos(d*x
+ c) + 163*A - 75*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*(32*A*
cos(d*x + c)^4 - 32*(5*A - 3*B)*cos(d*x + c)^3 - (503*A - 255*B)*cos(d*x + c)^2 - (299*A - 147*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^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), -1/96*(3*sqrt(2)*((163*A - 75*B)*cos(d*x + c)^3 + 3*(163*A - 75*B
)*cos(d*x + c)^2 + 3*(163*A - 75*B)*cos(d*x + c) + 163*A - 75*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*(32*A*cos(d*x + c)^4 - 32*(5*A - 3*B)*cos
(d*x + c)^3 - (503*A - 255*B)*cos(d*x + c)^2 - (299*A - 147*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^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))/sec(d*x+c)**(3/2)/(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}} \sec \left (d x + c\right )^{\frac{3}{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)^(3/2)/(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)*sec(d*x + c)^(3/2)), x)