\(\int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [138]

Optimal result

Integrand size = 21, antiderivative size = 107 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {3 \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

[Out]

3/32*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*tan(d*x+c)/d/(a+a*sec
(d*x+c))^(5/2)+3/16*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3881, 3880, 209} \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {3 \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac {\tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[In]

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

[Out]

(3*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Tan[c + d*x]/(4
*d*(a + a*Sec[c + d*x])^(5/2)) + (3*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 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 3881

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.49 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \tan (c+d x)}{4 a^2 d \sqrt {a (1+\sec (c+d x))}} \]

[In]

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

[Out]

(Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c + d*x])/2]*Tan[c + d*x])/(4*a^2*d*Sqrt[a*(1 + Sec[c + d*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(88)=176\).

Time = 0.96 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\)	\(192\)

[In]

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

[Out]

1/32/d/a^3*(-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*((1-cos(d
*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d*x+c)+csc(d*x+c))-3*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)
+csc(d*x+c))+3*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. 177 vs. \(2 (88) = 176\).

Time = 0.33 (sec) , antiderivative size = 403, normalized size of antiderivative = 3.77 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {3 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - 2 \, {\left (7 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

[In]

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

[Out]

[-1/64*(3*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sq
rt((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*(7*cos(d*x + c)^2 + 3*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*(3
*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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*(7*cos(d*x + c)^2 + 3*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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.86 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {5 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {3 \, \sqrt {2} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \]

[In]

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

[Out]

-1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c)^2/(a^3*sgn(cos(d*x + c))) - 5*sqrt(
2)/(a^3*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + 3*sqrt(2)*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(cos(d*x + c))))/d

Mupad [F(-1)]

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

[In]

int(1/(cos(c + d*x)*(a + a/cos(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)*(a + a/cos(c + d*x))^(5/2)), x)