\(\int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [193]

Optimal result

Integrand size = 23, antiderivative size = 167 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {35 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{128 \sqrt {2} a^{5/2} d}-\frac {\sec (c+d x)}{6 d (a+a \sin (c+d x))^{5/2}}-\frac {35 \cos (c+d x)}{128 a d (a+a \sin (c+d x))^{3/2}}-\frac {7 \sec (c+d x)}{48 a d (a+a \sin (c+d x))^{3/2}}+\frac {35 \sec (c+d x)}{96 a^2 d \sqrt {a+a \sin (c+d x)}} \] Output:

-35/256*arctanh(1/2*a^(1/2)*cos(d*x+c)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^( 
1/2)/a^(5/2)/d-1/6*sec(d*x+c)/d/(a+a*sin(d*x+c))^(5/2)-35/128*cos(d*x+c)/a 
/d/(a+a*sin(d*x+c))^(3/2)-7/48*sec(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)+35/96 
*sec(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.31 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},4,\frac {1}{2},\frac {1}{2} (1-\sin (c+d x))\right ) \sec (c+d x) \sqrt {a (1+\sin (c+d x))}}{8 a^3 d} \] Input:

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

Output:

(Hypergeometric2F1[-1/2, 4, 1/2, (1 - Sin[c + d*x])/2]*Sec[c + d*x]*Sqrt[a 
*(1 + Sin[c + d*x])])/(8*a^3*d)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3160, 3042, 3160, 3042, 3166, 3042, 3129, 3042, 3128, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\sec ^2(c+d x)}{(a \sin (c+d x)+a)^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^2 (a \sin (c+d x)+a)^{5/2}}dx\)

\(\Big \downarrow \) 3160

\(\displaystyle \frac {7 \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^{3/2}}dx}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^{3/2}}dx}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3160

\(\displaystyle \frac {7 \left (\frac {5 \int \frac {\sec ^2(c+d x)}{\sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \left (\frac {5 \int \frac {1}{\cos (c+d x)^2 \sqrt {\sin (c+d x) a+a}}dx}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3166

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \int \frac {1}{(\sin (c+d x) a+a)^{3/2}}dx+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3129

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \left (\frac {\int \frac {1}{\sqrt {\sin (c+d x) a+a}}dx}{4 a}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 3128

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\int \frac {1}{2 a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{2 a d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {7 \left (\frac {5 \left (\frac {3}{2} a \left (-\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {\cos (c+d x)}{2 d (a \sin (c+d x)+a)^{3/2}}\right )+\frac {\sec (c+d x)}{d \sqrt {a \sin (c+d x)+a}}\right )}{8 a}-\frac {\sec (c+d x)}{4 d (a \sin (c+d x)+a)^{3/2}}\right )}{12 a}-\frac {\sec (c+d x)}{6 d (a \sin (c+d x)+a)^{5/2}}\)

Input:

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

Output:

-1/6*Sec[c + d*x]/(d*(a + a*Sin[c + d*x])^(5/2)) + (7*(-1/4*Sec[c + d*x]/( 
d*(a + a*Sin[c + d*x])^(3/2)) + (5*(Sec[c + d*x]/(d*Sqrt[a + a*Sin[c + d*x 
]]) + (3*a*(-1/2*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c 
+ d*x]])]/(Sqrt[2]*a^(3/2)*d) - Cos[c + d*x]/(2*d*(a + a*Sin[c + d*x])^(3/ 
2))))/2))/(8*a)))/(12*a)
 

Defintions of rubi rules used

rule 219

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3128

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

rule 3129

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c 
+ d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n 
+ 1))   Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] 
&& EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
 

rule 3160

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^m/(a*f*g*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1))   Int[ 
(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, 
f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & 
& IntegersQ[2*m, 2*p]
 

rule 3166

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. 
)*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S 
qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1)))   Int[(g 
*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e 
, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(287\) vs. \(2(140)=280\).

Time = 0.18 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.72

method	result	size
default	\(\frac {-210 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) a^{\frac {7}{2}}-490 \cos \left (d x +c \right )^{2} a^{\frac {7}{2}}+105 \sqrt {2}\, \sqrt {a -a \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )+448 \sin \left (d x +c \right ) a^{\frac {7}{2}}+315 \sqrt {2}\, \sqrt {a -a \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \cos \left (d x +c \right )^{2}+320 a^{\frac {7}{2}}-420 \sqrt {a -a \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, \sin \left (d x +c \right ) a^{3}-420 \sqrt {a -a \sin \left (d x +c \right )}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{3}}{768 a^{\frac {11}{2}} \left (1+\sin \left (d x +c \right )\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(288\)

Input:

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

Output:

1/768*(-210*cos(d*x+c)^2*sin(d*x+c)*a^(7/2)-490*cos(d*x+c)^2*a^(7/2)+105*2 
^(1/2)*(a-a*sin(d*x+c))^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a 
^(1/2))*a^3*cos(d*x+c)^2*sin(d*x+c)+448*sin(d*x+c)*a^(7/2)+315*2^(1/2)*(a- 
a*sin(d*x+c))^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^ 
3*cos(d*x+c)^2+320*a^(7/2)-420*(a-a*sin(d*x+c))^(1/2)*arctanh(1/2*(a-a*sin 
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*sin(d*x+c)*a^3-420*(a-a*sin(d*x+c) 
)^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^3)/a 
^(11/2)/(1+sin(d*x+c))^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.68 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {105 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )\right )} \sin \left (d x + c\right ) - 4 \, \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (245 \, \cos \left (d x + c\right )^{2} + 7 \, {\left (15 \, \cos \left (d x + c\right )^{2} - 32\right )} \sin \left (d x + c\right ) - 160\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1536 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:

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

Output:

1/1536*(105*sqrt(2)*(3*cos(d*x + c)^3 + (cos(d*x + c)^3 - 4*cos(d*x + c))* 
sin(d*x + c) - 4*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)* 
sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*c 
os(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c) + 2*a)/(cos(d*x + c)^2 - 
 (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*(245*cos(d*x + c 
)^2 + 7*(15*cos(d*x + c)^2 - 32)*sin(d*x + c) - 160)*sqrt(a*sin(d*x + c) + 
 a))/(3*a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^ 
3 - 4*a^3*d*cos(d*x + c))*sin(d*x + c))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a} {\left (\frac {105 \, \sqrt {2} \log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {105 \, \sqrt {2} \log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {96 \, \sqrt {2}}{a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} - \frac {2 \, {\left (57 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 136 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 87 \, \sqrt {2} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{1536 \, d} \] Input:

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

Output:

1/1536*sqrt(a)*(105*sqrt(2)*log(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1)/(a^3*s 
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 105*sqrt(2)*log(-sin(-1/4*pi + 1/2*d 
*x + 1/2*c) + 1)/(a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 96*sqrt(2)/(a 
^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)) - 2 
*(57*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 136*sqrt(2)*sin(-1/4*pi + 
1/2*d*x + 1/2*c)^3 + 87*sqrt(2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((sin(-1/4 
*pi + 1/2*d*x + 1/2*c)^2 - 1)^3*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))))/ 
d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1}d x \right )}{a^{3}} \] Input:

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

Output:

(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*sec(c + d*x)**2)/(sin(c + d*x)**3 + 3 
*sin(c + d*x)**2 + 3*sin(c + d*x) + 1),x))/a**3