\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [153]

Optimal result

Integrand size = 33, antiderivative size = 171 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d} \]

[Out]

-1/4*(7*A-11*B)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/2*(A-B)*sec(
d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/3*(9*A-13*B)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/6*(3*A-7*B
)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {4104, 4095, 4086, 3880, 209} \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A-11 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(3 A-7 B) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{6 a^2 d}+\frac {(A-B) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a \sec (c+d x)+a}} \]

[In]

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

[Out]

-1/2*((7*A - 11*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) + ((
A - B)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((9*A - 13*B)*Tan[c + d*x])/(3*a*d*Sqrt
[a + a*Sec[c + d*x]]) - ((3*A - 7*B)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

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 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 4086

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 4095

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)),
Int[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] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4104

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*} \text {integral}& = \frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^2(c+d x) \left (2 a (A-B)-\frac {1}{2} a (3 A-7 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2} \\ & = \frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac {\int \frac {\sec (c+d x) \left (-\frac {1}{4} a^2 (3 A-7 B)+\frac {1}{2} a^2 (9 A-13 B) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{3 a^3} \\ & = \frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac {(7 A-11 B) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = \frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac {(7 A-11 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 )}{2 a d} \\ & = -\frac {(7 A-11 B) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(A-B) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-13 B) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}-\frac {(3 A-7 B) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-3 \sqrt {2} (7 A-11 B) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (15 A-19 B+12 (A-B) \sec (c+d x)+4 B \sec ^2(c+d x)\right )\right ) \tan (c+d x)}{6 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

[In]

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

[Out]

((-3*Sqrt[2]*(7*A - 11*B)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 - S
ec[c + d*x]]*(15*A - 19*B + 12*(A - B)*Sec[c + d*x] + 4*B*Sec[c + d*x]^2))*Tan[c + d*x])/(6*d*Sqrt[1 - Sec[c +
d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(330\) vs. \(2(148)=296\).

Time = 4.10 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.94

method	result	size
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-3 B \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+21 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-33 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-30 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+46 B \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-27 B \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{12 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\)	\(331\)
parts	\(-\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+7 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}-9 \csc \left (d x +c \right )+9 \cot \left (d x +c \right )\right )}{4 d \,a^{2}}+\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (3 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+33 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}}-46 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+27 \csc \left (d x +c \right )-27 \cot \left (d x +c \right )\right )}{12 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )}\)	\(340\)

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/12/a^2/d*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(3*A*(1-cos(d*x+c))^5*csc(d*x+c)^5-3*B*(1-cos(d*x+c
))^5*csc(d*x+c)^5+21*A*ln(csc(d*x+c)-cot(d*x+c)+((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)^(3/2)-33*B*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*((1-cos(d*x+c))^2*cs
c(d*x+c)^2-1)^(3/2)-30*A*(1-cos(d*x+c))^3*csc(d*x+c)^3+46*B*(1-cos(d*x+c))^3*csc(d*x+c)^3+27*A*(-cot(d*x+c)+cs
c(d*x+c))-27*B*(-cot(d*x+c)+csc(d*x+c)))/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 459, normalized size of antiderivative = 2.68 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, \frac {3 \, \sqrt {2} {\left ({\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, A - 11 \, B\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (15 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{2} + 12 \, {\left (A - B\right )} \cos \left (d x + c\right ) + 4 \, B\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/24*(3*sqrt(2)*((7*A - 11*B)*cos(d*x + c)^3 + 2*(7*A - 11*B)*cos(d*x + c)^2 + (7*A - 11*B)*cos(d*x + c))*sqr
t(-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*((15*A - 19*B)*cos(d*x + c)^2 + 12*(
A - B)*cos(d*x + c) + 4*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2
*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c)), 1/12*(3*sqrt(2)*((7*A - 11*B)*cos(d*x + c)^3 + 2*(7*A - 11*B)*cos(d*x
+ c)^2 + (7*A - 11*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*((15*A - 19*B)*cos(d*x + c)^2 + 12*(A - B)*cos(d*x + c) + 4*B)*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*sec(c + d*x))*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)

Maxima [F]

\[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/(a*sec(d*x + c) + a)^(3/2), x)

Giac [A] (verification not implemented)

none

Time = 1.63 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} - \frac {2 \, {\left (15 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 23 \, \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {27 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - \sqrt {2} B a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, {\left (7 \, \sqrt {2} A - 11 \, \sqrt {2} B\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{12 \, d} \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-1/12*(((3*(sqrt(2)*A*a*sgn(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a - 2*(15*sq
rt(2)*A*a*sgn(cos(d*x + c)) - 23*sqrt(2)*B*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)^2 + 27*(sqrt(2)*A*a*sg
n(cos(d*x + c)) - sqrt(2)*B*a*sgn(cos(d*x + c)))/a)*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*(7*sqrt(2)*A - 11*sqrt(2)*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*sgn(cos(d*x + c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x))/(cos(c + d*x)^3*(a + a/cos(c + d*x))^(3/2)), x)