\(\int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx\) [286]

Optimal result

Integrand size = 27, antiderivative size = 74 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 (a+a \sin (c+d x))^{3/2}}{d e (e \cos (c+d x))^{5/2}}-\frac {4 (a+a \sin (c+d x))^{5/2}}{5 a d e (e \cos (c+d x))^{5/2}} \]

[Out]

2*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(5/2)-4/5*(a+a*sin(d*x+c))^(5/2)/a/d/e/(e*cos(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 (a \sin (c+d x)+a)^{3/2}}{d e (e \cos (c+d x))^{5/2}}-\frac {4 (a \sin (c+d x)+a)^{5/2}}{5 a d e (e \cos (c+d x))^{5/2}} \]

[In]

Int[(a + a*Sin[c + d*x])^(3/2)/(e*Cos[c + d*x])^(7/2),x]

[Out]

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

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 (a+a \sin (c+d x))^{3/2}}{d e (e \cos (c+d x))^{5/2}}-\frac {2 \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{7/2}} \, dx}{a} \\ & = \frac {2 (a+a \sin (c+d x))^{3/2}}{d e (e \cos (c+d x))^{5/2}}-\frac {4 (a+a \sin (c+d x))^{5/2}}{5 a d e (e \cos (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {2 a \sqrt {a (1+\sin (c+d x))} (-3+2 \sin (c+d x))}{5 d e^3 \sqrt {e \cos (c+d x)} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2} \]

[In]

Integrate[(a + a*Sin[c + d*x])^(3/2)/(e*Cos[c + d*x])^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 \left (2 \sin \left (d x +c \right )-3\right ) a \sqrt {a \left (1+\sin \left (d x +c \right )\right )}}{5 d \left (\sin \left (d x +c \right )-1\right ) \sqrt {e \cos \left (d x +c \right )}\, e^{3}}\)	\(52\)

[In]

int((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (2 \, a \sin \left (d x + c\right ) - 3 \, a\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{5 \, {\left (d e^{4} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d e^{4} \cos \left (d x + c\right )\right )}} \]

[In]

integrate((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/5*sqrt(e*cos(d*x + c))*(2*a*sin(d*x + c) - 3*a)*sqrt(a*sin(d*x + c) + a)/(d*e^4*cos(d*x + c)*sin(d*x + c) -
d*e^4*cos(d*x + c))

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))**(3/2)/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (64) = 128\).

Time = 0.32 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.80 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \, {\left (3 \, a^{\frac {3}{2}} \sqrt {e} - \frac {4 \, a^{\frac {3}{2}} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {4 \, a^{\frac {3}{2}} \sqrt {e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {3 \, a^{\frac {3}{2}} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{5 \, {\left (e^{4} + \frac {2 \, e^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {e^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}}} \]

[In]

integrate((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/5*(3*a^(3/2)*sqrt(e) - 4*a^(3/2)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^(3/2)*sqrt(e)*sin(d*x + c)^3/
(cos(d*x + c) + 1)^3 - 3*a^(3/2)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*(sin(d*x + c)^2/(cos(d*x + c) +
1)^2 + 1)^2/((e^4 + 2*e^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + e^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d*sqr
t(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2))

Giac [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(d*x+c))^(3/2)/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.94 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {4\,a\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (5\,\sin \left (c+d\,x\right )+\cos \left (2\,c+2\,d\,x\right )-4\right )}{5\,d\,e^3\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (4\,\sin \left (c+d\,x\right )+\cos \left (2\,c+2\,d\,x\right )-3\right )} \]

[In]

int((a + a*sin(c + d*x))^(3/2)/(e*cos(c + d*x))^(7/2),x)

[Out]

(4*a*(a*(sin(c + d*x) + 1))^(1/2)*(5*sin(c + d*x) + cos(2*c + 2*d*x) - 4))/(5*d*e^3*(e*cos(c + d*x))^(1/2)*(4*
sin(c + d*x) + cos(2*c + 2*d*x) - 3))