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

Optimal result

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

[Out]

-2/9/d/e/(e*cos(d*x+c))^(3/2)/(a+a*sin(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)
-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)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\frac {32 (a \sin (c+d x)+a)^{3/2}}{45 a^3 d e (e \cos (c+d x))^{3/2}}-\frac {16 \sqrt {a \sin (c+d x)+a}}{15 a^2 d e (e \cos (c+d x))^{3/2}}-\frac {4}{15 a d e \sqrt {a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}-\frac {2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}} \]

[In]

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

[Out]

-2/(9*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^(3/2)) - 4/(15*a*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*S
in[c + d*x]]) - (16*Sqrt[a + a*Sin[c + d*x]])/(15*a^2*d*e*(e*Cos[c + d*x])^(3/2)) + (32*(a + a*Sin[c + d*x])^(
3/2))/(45*a^3*d*e*(e*Cos[c + d*x])^(3/2))

Rule 2750

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2751

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[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[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}+\frac {2 \int \frac {1}{(e \cos (c+d x))^{5/2} \sqrt {a+a \sin (c+d x)}} \, dx}{3 a} \\ & = -\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 {8 \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx}{15 a^2} \\ & = -\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 {16 \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{5/2}} \, dx}{15 a^3} \\ & = -\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}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 (7+12 \cos (2 (c+d x))-6 \sin (c+d x)+4 \sin (3 (c+d x)))}{45 d e (e \cos (c+d x))^{3/2} (a (1+\sin (c+d x)))^{3/2}} \]

[In]

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

[Out]

(-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 + Si
n[c + d*x]))^(3/2))

Maple [A] (verified)

Time = 2.67 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.54

method	result	size
default	\(-\frac {2 \left (16 \cos \left (d x +c \right ) \sin \left (d x +c \right )+24 \cos \left (d x +c \right )-10 \tan \left (d x +c \right )-5 \sec \left (d x +c \right )\right )}{45 d a \,e^{2} \left (1+\sin \left (d x +c \right )\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \sqrt {e \cos \left (d x +c \right )}}\)	\(83\)

[In]

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

[Out]

-2/45/d/a/e^2/(1+sin(d*x+c))/(a*(1+sin(d*x+c)))^(1/2)/(e*cos(d*x+c))^(1/2)*(16*cos(d*x+c)*sin(d*x+c)+24*cos(d*
x+c)-10*tan(d*x+c)-5*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (24 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (8 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 5\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{45 \, {\left (a^{2} d e^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

2/45*sqrt(e*cos(d*x + c))*(24*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^2 - 5)*sin(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. \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

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 = 0.34 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.42 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (19 \, \sqrt {a} \sqrt {e} + \frac {12 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {58 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {116 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {116 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {58 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {12 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {19 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{45 \, {\left (a^{2} e^{3} + \frac {4 \, a^{2} e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{2} e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{2} e^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{2} e^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {11}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}}} \]

[In]

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

[Out]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 10.55 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.49 \[ \int \frac {1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx=-\frac {14\,\sqrt {a+a\,\sin \left (c+d\,x\right )}-12\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+24\,\cos \left (2\,c+2\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}+8\,\sin \left (3\,c+3\,d\,x\right )\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{\frac {225\,a^2\,d\,e^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}-\frac {45\,a^2\,d\,e^2\,\cos \left (3\,c+3\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}{4}+45\,a^2\,d\,e^2\,\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {e\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {e\,{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}} \]

[In]

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

[Out]

-(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 + 45*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))