\(\int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx\) [311]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [B] (verification not implemented)
Giac [F(-1)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 154 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}-\frac {4}{15 a d e (e \cos (c+d x))^{3/2} \sqrt {a+a \sin (c+d x)}}-\frac {16 \sqrt {a+a \sin (c+d x)}}{15 a^2 d e (e \cos (c+d x))^{3/2}}+\frac {32 (a+a \sin (c+d x))^{3/2}}{45 a^3 d e (e \cos (c+d x))^{3/2}} \] Output:

-2/9/d/e/(e*cos(d*x+c))^(3/2)/(a+a*sin(d*x+c))^(3/2)-4/15/a/d/e/(e*cos(d*x 
+c))^(3/2)/(a+a*sin(d*x+c))^(1/2)-16/15*(a+a*sin(d*x+c))^(1/2)/a^2/d/e/(e* 
cos(d*x+c))^(3/2)+32/45*(a+a*sin(d*x+c))^(3/2)/a^3/d/e/(e*cos(d*x+c))^(3/2 
)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (7+12 \cos (2 (c+d x))-6 \sin (c+d x)+4 \sin (3 (c+d x)))}{45 d e (e \cos (c+d x))^{3/2} (a (1+\sin (c+d x)))^{3/2}} \] Input:

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^(3/2)),x]
 

Output:

(-2*(7 + 12*Cos[2*(c + d*x)] - 6*Sin[c + d*x] + 4*Sin[3*(c + d*x)]))/(45*d 
*e*(e*Cos[c + d*x])^(3/2)*(a*(1 + Sin[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3151, 3042, 3151, 3042, 3151, 3042, 3150}

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.

\(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {2 \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {\sin (c+d x) a+a}}dx}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {\sin (c+d x) a+a}}dx}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {2 \left (\frac {4 \int \frac {\sqrt {\sin (c+d x) a+a}}{(e \cos (c+d x))^{5/2}}dx}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\right )}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {4 \int \frac {\sqrt {\sin (c+d x) a+a}}{(e \cos (c+d x))^{5/2}}dx}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\right )}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3151

\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{(e \cos (c+d x))^{5/2}}dx}{a}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\right )}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {2 \int \frac {(\sin (c+d x) a+a)^{3/2}}{(e \cos (c+d x))^{5/2}}dx}{a}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\right )}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3150

\(\displaystyle \frac {2 \left (\frac {4 \left (\frac {4 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{3/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{d e (e \cos (c+d x))^{3/2}}\right )}{5 a}-\frac {2}{5 d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}\right )}{3 a}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}}\)

Input:

Int[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^(3/2)),x]
 

Output:

-2/(9*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^(3/2)) + (2*(-2/(5*d 
*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*Sin[c + d*x]]) + (4*((-2*Sqrt[a + a*S 
in[c + d*x]])/(d*e*(e*Cos[c + d*x])^(3/2)) + (4*(a + a*Sin[c + d*x])^(3/2) 
)/(3*a*d*e*(e*Cos[c + d*x])^(3/2))))/(5*a)))/(3*a)
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3150
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*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] 
 && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]
 

rule 3151
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*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl 
ify[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] && ILtQ[Simpli 
fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]
 
Maple [A] (verified)

Time = 9.09 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.23

method result size
default \(-\frac {2 \left (128 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-128 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-96 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )+19\right )}{45 d \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e^{2} a \sqrt {\left (1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}\, \sqrt {e \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}}\) \(190\)

Input:

int(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-2/45/d*(128*sin(1/2*d*x+1/2*c)*cos(1/2*d*x+1/2*c)^5+96*cos(1/2*d*x+1/2*c) 
^4-128*sin(1/2*d*x+1/2*c)*cos(1/2*d*x+1/2*c)^3-96*cos(1/2*d*x+1/2*c)^2+12* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)+19)/(2*cos(1/2*d*x+1/2*c)^2-1)/(1+2* 
cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c))/e^2/a/((1+2*cos(1/2*d*x+1/2*c)*sin( 
1/2*d*x+1/2*c))*a)^(1/2)/(e*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (24 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45 \, {\left (a^{2} d e^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2}\right )}} \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fric 
as")
 

Output:

2/45*sqrt(e*cos(d*x + c))*(24*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^2 - 5)*si 
n(d*x + c) - 5)*sqrt(a*sin(d*x + c) + a)/(a^2*d*e^3*cos(d*x + c)^4 - 2*a^2 
*d*e^3*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*e^3*cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cos(d*x+c))**(5/2)/(a+a*sin(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (130) = 260\).

Time = 0.17 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.42 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (19 \, \sqrt {a} \sqrt {e} + \frac {12 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {58 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {116 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {116 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {58 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {12 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {19 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{45 \, {\left (a^{2} e^{3} + \frac {4 \, a^{2} e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} e^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} e^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxi 
ma")
 

Output:

-2/45*(19*sqrt(a)*sqrt(e) + 12*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) 
+ 1) - 58*sqrt(a)*sqrt(e)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 116*sqrt(a 
)*sqrt(e)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 116*sqrt(a)*sqrt(e)*sin(d* 
x + c)^5/(cos(d*x + c) + 1)^5 + 58*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x 
 + c) + 1)^6 - 12*sqrt(a)*sqrt(e)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 19 
*sqrt(a)*sqrt(e)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*(sin(d*x + c)^2/(cos 
(d*x + c) + 1)^2 + 1)^4/((a^2*e^3 + 4*a^2*e^3*sin(d*x + c)^2/(cos(d*x + c) 
 + 1)^2 + 6*a^2*e^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^2*e^3*sin(d* 
x + c)^6/(cos(d*x + c) + 1)^6 + a^2*e^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^ 
8)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x 
+ c) + 1) + 1)^(5/2))
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac 
")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 20.15 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {14\,\sqrt {a+a\,\sin \left (c+d\,x\right )}-12\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+24\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+8\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\frac {225\,a^2\,d\,e^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}-\frac {45\,a^2\,d\,e^2\,\cos \left (3\,c+3\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}+45\,a^2\,d\,e^2\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}} \] Input:

int(1/((e*cos(c + d*x))^(5/2)*(a + a*sin(c + d*x))^(3/2)),x)
 

Output:

-(14*(a + a*sin(c + d*x))^(1/2) - 12*sin(c + d*x)*(a + a*sin(c + d*x))^(1/ 
2) + 24*cos(2*c + 2*d*x)*(a + a*sin(c + d*x))^(1/2) + 8*sin(3*c + 3*d*x)*( 
a + a*sin(c + d*x))^(1/2))/((225*a^2*d*e^2*cos(c + d*x)*((e*exp(- c*1i - d 
*x*1i))/2 + (e*exp(c*1i + d*x*1i))/2)^(1/2))/4 - (45*a^2*d*e^2*cos(3*c + 3 
*d*x)*((e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i + d*x*1i))/2)^(1/2))/4 + 4 
5*a^2*d*e^2*sin(2*c + 2*d*x)*((e*exp(- c*1i - d*x*1i))/2 + (e*exp(c*1i + d 
*x*1i))/2)^(1/2))
 

Reduce [F]

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

int(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x)
 

Output:

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