3.160 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=216 \[ -\frac{(75 A-163 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(39 A-95 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}+\frac{(93 A-197 B) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(9 A-17 B) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

-((75*A - 163*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (
(A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((9*A - 17*B)*Sec[c + d*x]^2*Tan[c + d
*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((93*A - 197*B)*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) -
((39*A - 95*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(48*a^3*d)

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Rubi [A]  time = 0.654887, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {4019, 4010, 4001, 3795, 203} \[ -\frac{(75 A-163 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(39 A-95 B) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{48 a^3 d}+\frac{(93 A-197 B) \tan (c+d x)}{24 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(A-B) \tan (c+d x) \sec ^3(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}+\frac{(9 A-17 B) \tan (c+d x) \sec ^2(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

-((75*A - 163*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (
(A - B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((9*A - 17*B)*Sec[c + d*x]^2*Tan[c + d
*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + ((93*A - 197*B)*Tan[c + d*x])/(24*a^2*d*Sqrt[a + a*Sec[c + d*x]]) -
((39*A - 95*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(48*a^3*d)

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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*B*m + A*b*(m + 1))/(b*(
m + 1)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B, e, f, m}, x] && NeQ[A*b - a*B,
0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] &&  !LtQ[m, -2^(-1)]

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\sec ^3(c+d x) \left (3 a (A-B)-\frac{1}{2} a (3 A-11 B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 A-17 B) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^2(c+d x) \left (a^2 (9 A-17 B)-\frac{1}{4} a^2 (39 A-95 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 A-17 B) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(39 A-95 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{\int \frac{\sec (c+d x) \left (-\frac{1}{8} a^3 (39 A-95 B)+\frac{1}{4} a^3 (93 A-197 B) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{12 a^5}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 A-17 B) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 A-197 B) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-95 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}-\frac{(75 A-163 B) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 A-17 B) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 A-197 B) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-95 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}+\frac{(75 A-163 B) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac{(75 A-163 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{(A-B) \sec ^3(c+d x) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{(9 A-17 B) \sec ^2(c+d x) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(93 A-197 B) \tan (c+d x)}{24 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(39 A-95 B) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{48 a^3 d}\\ \end{align*}

Mathematica [A]  time = 2.57115, size = 161, normalized size = 0.75 \[ \frac{\tan (c+d x) \left (\sqrt{1-\sec (c+d x)} \left (32 (3 A-5 B) \sec ^2(c+d x)+(255 A-503 B) \sec (c+d x)+147 A+32 B \sec ^3(c+d x)-299 B\right )-6 \sqrt{2} (75 A-163 B) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) \tanh ^{-1}\left (\frac{\sqrt{1-\sec (c+d x)}}{\sqrt{2}}\right )\right )}{48 d \sqrt{1-\sec (c+d x)} (a (\sec (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((-6*Sqrt[2]*(75*A - 163*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^4*Sec[c + d*x]^2 + Sqrt[1
- Sec[c + d*x]]*(147*A - 299*B + (255*A - 503*B)*Sec[c + d*x] + 32*(3*A - 5*B)*Sec[c + d*x]^2 + 32*B*Sec[c +
d*x]^3))*Tan[c + d*x])/(48*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(5/2))

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Maple [B]  time = 0.275, size = 795, normalized size = 3.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/192/d/a^3*(-1+cos(d*x+c))^2*(225*A*sin(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-489*B*sin(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+675*A
*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d
*x+c)+cos(d*x+c)-1)/sin(d*x+c))-1467*B*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))+675*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-1467*
B*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*
x+c)+cos(d*x+c)-1)/sin(d*x+c))+225*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-489*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*ln(-(-(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-588*A*cos(d*x+c)^4+1196*B*cos(d
*x+c)^4-432*A*cos(d*x+c)^3+816*B*cos(d*x+c)^3+636*A*cos(d*x+c)^2-1372*B*cos(d*x+c)^2+384*A*cos(d*x+c)-768*B*co
s(d*x+c)+128*B)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)

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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(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.613782, size = 1474, normalized size = 6.82 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/192*(3*sqrt(2)*((75*A - 163*B)*cos(d*x + c)^4 + 3*(75*A - 163*B)*cos(d*x + c)^3 + 3*(75*A - 163*B)*cos(d*x
+ c)^2 + (75*A - 163*B)*cos(d*x + c))*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))
*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1))
+ 4*((147*A - 299*B)*cos(d*x + c)^3 + (255*A - 503*B)*cos(d*x + c)^2 + 32*(3*A - 5*B)*cos(d*x + c) + 32*B)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*c
os(d*x + c)^2 + a^3*d*cos(d*x + c)), 1/96*(3*sqrt(2)*((75*A - 163*B)*cos(d*x + c)^4 + 3*(75*A - 163*B)*cos(d*x
+ c)^3 + 3*(75*A - 163*B)*cos(d*x + c)^2 + (75*A - 163*B)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*
x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((147*A - 299*B)*cos(d*x + c)^3 + (255*A -
503*B)*cos(d*x + c)^2 + 32*(3*A - 5*B)*cos(d*x + c) + 32*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x +
c))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(5/2), x)

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Giac [A]  time = 10.1536, size = 420, normalized size = 1.94 \begin{align*} \frac{\frac{{\left ({\left (3 \,{\left (\frac{2 \, \sqrt{2}{\left (A a^{5} - B a^{5}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{\sqrt{2}{\left (15 \, A a^{5} - 23 \, B a^{5}\right )}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - \frac{4 \, \sqrt{2}{\left (75 \, A a^{5} - 167 \, B a^{5}\right )}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{3 \, \sqrt{2}{\left (83 \, A a^{5} - 155 \, B a^{5}\right )}}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{3 \, \sqrt{2}{\left (75 \, A - 163 \, B\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/96*(((3*(2*sqrt(2)*(A*a^5 - B*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + sqrt(2)*(1
5*A*a^5 - 23*B*a^5)/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 - 4*sqrt(2)*(75*A*a^5 - 167*
B*a^5)/(a^6*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2 + 3*sqrt(2)*(83*A*a^5 - 155*B*a^5)/(a^6*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-a*tan(1/2*d*x + 1/
2*c)^2 + a)) - 3*sqrt(2)*(75*A - 163*B)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^
2 + a)))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d