\(\int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [164]

Optimal result

Integrand size = 25, antiderivative size = 164 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-3 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

[Out]

2*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/32*(43*A-3*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)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(11*A-3*B
)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4007, 4005, 3859, 209, 3880} \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {(43 A-3 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d) - ((43*A - 3*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)*Tan[c + d*x])/(4*d*(a + a*Sec
[c + d*x])^(5/2)) - ((11*A - 3*B)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))

Rule 209

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

Rule 3859

Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(a + x^2), x], x, b*(C
ot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 3880

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 4005

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

Rule 4007

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

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {\int \frac {-4 a A+\frac {3}{2} a (A-B) \sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {8 a^2 A-\frac {1}{4} a^2 (11 A-3 B) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {A \int \sqrt {a+a \sec (c+d x)} \, dx}{a^3}-\frac {(43 A-3 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2} \\ & = -\frac {(A-B) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(2 A) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^2 d}+\frac {(43 A-3 B) \text {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 {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-3 B) \arctan \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) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-3 B) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(343\) vs. \(2(164)=328\).

Time = 7.50 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.09 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\left ((-43 A+3 B) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+32 \sqrt {2} A \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sec ^{\frac {3}{2}}(c+d x) \sqrt {1+\sec (c+d x)} (A+B \sec (c+d x))}{4 d (B+A \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a (1+\sec (c+d x)))^{5/2}}+\frac {\cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) (A+B \sec (c+d x)) \left (\frac {1}{2} (-15 A+7 B) \sin \left (\frac {1}{2} (c+d x)\right )+\frac {1}{4} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (19 A \sin \left (\frac {1}{2} (c+d x)\right )-11 B \sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1}{2} \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (-A \sin \left (\frac {1}{2} (c+d x)\right )+B \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )}{d (B+A \cos (c+d x)) (a (1+\sec (c+d x)))^{5/2}} \]

[In]

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

[Out]

(((-43*A + 3*B)*ArcSin[Tan[(c + d*x)/2]] + 32*Sqrt[2]*A*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c +
d*x])]])*Cos[(c + d*x)/2]^4*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sec[c + d*x]^(3/2)*Sqrt[1 + Sec[c + d*x]]*(
A + B*Sec[c + d*x]))/(4*d*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*(a*(1 + Sec[c + d*x]))^(5/2)) + (Cos[(
c + d*x)/2]^5*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*(((-15*A + 7*B)*Sin[(c + d*x)/2])/2 + (Sec[(c + d*x)/2]^2*(1
9*A*Sin[(c + d*x)/2] - 11*B*Sin[(c + d*x)/2]))/4 + (Sec[(c + d*x)/2]^4*(-(A*Sin[(c + d*x)/2]) + B*Sin[(c + d*x
)/2]))/2))/(d*(B + A*Cos[c + d*x])*(a*(1 + Sec[c + d*x]))^(5/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(139)=278\).

Time = 3.46 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.29

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-2 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+2 B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+32 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+11 A \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 B \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-43 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+3 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 a^{3} d}\)	\(376\)
parts	\(\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-2 \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}+32 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+13 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-43 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}+\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\)	\(441\)

[In]

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

[Out]

1/32/a^3/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-2*A*((1-co
s(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d*x+c)+csc(d*x+c))+2*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d
*x+c)+csc(d*x+c))+32*A*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c)
))+11*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-3*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)
^(1/2)*(-cot(d*x+c)+csc(d*x+c))-43*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))+3*B*ln(
csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (139) = 278\).

Time = 3.60 (sec) , antiderivative size = 670, normalized size of antiderivative = 4.09 \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [\frac {\sqrt {2} {\left ({\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 43 \, A - 3 \, B\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 ) - 64 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} + 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 ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (15 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 3 \, 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 )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, \frac {\sqrt {2} {\left ({\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (43 \, A - 3 \, B\right )} \cos \left (d x + c\right ) + 43 \, A - 3 \, B\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 ) - 64 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\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 (15 \, A - 7 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (11 \, A - 3 \, 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 )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

[In]

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

[Out]

[1/64*(sqrt(2)*((43*A - 3*B)*cos(d*x + c)^3 + 3*(43*A - 3*B)*cos(d*x + c)^2 + 3*(43*A - 3*B)*cos(d*x + c) + 43
*A - 3*B)*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)) - 64*(A*cos(d*x + c)^3 + 3*
A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) +
a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*((15*A - 7*B)*cos(d*x
+ c)^2 + (11*A - 3*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(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), 1/32*(sqrt(2)*((43*A - 3*B)*cos(d*x + c)^3 + 3*(4
3*A - 3*B)*cos(d*x + c)^2 + 3*(43*A - 3*B)*cos(d*x + c) + 43*A - 3*B)*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))) - 64*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*
cos(d*x + c) + A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))
- 2*((15*A - 7*B)*cos(d*x + c)^2 + (11*A - 3*B)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(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)]

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m i_lex_is_greater Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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