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]

-1/32*(75*A-163*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/4*(A-B)*s
ec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(9*A-17*B)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(
3/2)+1/24*(93*A-197*B)*tan(d*x+c)/a^2/d/(a+a*sec(d*x+c))^(1/2)-1/48*(39*A-95*B)*(a+a*sec(d*x+c))^(1/2)*tan(d*x
+c)/a^3/d

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Rubi [A]  time = 0.65, antiderivative size = 216, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 2.71, 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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fricas [A]  time = 0.49, size = 557, normalized size = 2.58 \[ \left [\frac {3 \, \sqrt {2} {\left ({\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (147 \, A - 299 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (255 \, A - 503 \, B\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 32 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}, \frac {3 \, \sqrt {2} {\left ({\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (75 \, A - 163 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (147 \, A - 299 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (255 \, A - 503 \, B\right )} \cos \left (d x + c\right )^{2} + 32 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 32 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{96 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d \cos \left (d x + c\right )\right )}}\right ] \]

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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giac [A]  time = 3.40, size = 311, normalized size = 1.44 \[ \frac {\frac {{\left ({\left (3 \, {\left (\frac {2 \, \sqrt {2} {\left (A a^{5} - B a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {\sqrt {2} {\left (15 \, A a^{5} - 23 \, B a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {4 \, \sqrt {2} {\left (75 \, A a^{5} - 167 \, B a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {3 \, \sqrt {2} {\left (83 \, A a^{5} - 155 \, B a^{5}\right )}}{a^{6} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, \sqrt {2} {\left (75 \, A - 163 \, B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}{96 \, d} \]

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

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maple [B]  time = 1.75, size = 795, normalized size = 3.68 \[ \text {result too large to display} \]

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*(-1+cos(d*x+c))^2*(-225*A*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+489*B*cos(d*x+c)^3*sin(d*x+c)*ln(-(-(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-675*A*l
n(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(3/2)*sin(d*x+c)*cos(d*x+c)^2+1467*B*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d
*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)*cos(d*x+c)^2-675*A*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+1467*B*
cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(3/2)-225*A*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x
+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*sin(d*x+c)+489*B*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*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*cos(
d*x+c)-128*B)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^5/cos(d*x+c)/a^3

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maxima [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(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

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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