\(\int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx\) [278]

Optimal result

Integrand size = 27, antiderivative size = 115 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {16 (a+a \sin (c+d x))^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}} \]

[Out]

8/3*(a+a*sin(d*x+c))^(3/2)/a/d/e/(e*cos(d*x+c))^(5/2)-16/15*(a+a*sin(d*x+c))^(5/2)/a^2/d/e/(e*cos(d*x+c))^(5/2
)-2/3*(a+a*sin(d*x+c))^(1/2)/d/e/(e*cos(d*x+c))^(5/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2751, 2750} \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=-\frac {16 (a \sin (c+d x)+a)^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{3 d e (e \cos (c+d x))^{5/2}} \]

[In]

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

[Out]

(-2*Sqrt[a + a*Sin[c + d*x]])/(3*d*e*(e*Cos[c + d*x])^(5/2)) + (8*(a + a*Sin[c + d*x])^(3/2))/(3*a*d*e*(e*Cos[
c + d*x])^(5/2)) - (16*(a + a*Sin[c + d*x])^(5/2))/(15*a^2*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 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {4 \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx}{3 a} \\ & = -\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {8 \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{7/2}} \, dx}{3 a^2} \\ & = -\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {16 (a+a \sin (c+d x))^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \sqrt {a (1+\sin (c+d x))} (3+4 \cos (2 (c+d x))+4 \sin (c+d x))}{15 d e (e \cos (c+d x))^{5/2}} \]

[In]

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

[Out]

(2*Sqrt[a*(1 + Sin[c + d*x])]*(3 + 4*Cos[2*(c + d*x)] + 4*Sin[c + d*x]))/(15*d*e*(e*Cos[c + d*x])^(5/2))

Maple [A] (verified)

Time = 2.57 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {2 \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \left (8+4 \sec \left (d x +c \right ) \tan \left (d x +c \right )-\left (\sec ^{2}\left (d x +c \right )\right )\right )}{15 d \sqrt {e \cos \left (d x +c \right )}\, e^{3}}\)	\(57\)

[In]

int((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/15/d*(a*(1+sin(d*x+c)))^(1/2)/(e*cos(d*x+c))^(1/2)/e^3*(8+4*sec(d*x+c)*tan(d*x+c)-sec(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, d e^{4} \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/15*sqrt(e*cos(d*x + c))*(8*cos(d*x + c)^2 + 4*sin(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a)/(d*e^4*cos(d*x + c)
^3)

Sympy [F(-1)]

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

[In]

integrate((a+a*sin(d*x+c))**(1/2)/(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. 282 vs. \(2 (97) = 194\).

Time = 0.32 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.45 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {2 \, {\left (7 \, \sqrt {a} \sqrt {e} + \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, {\left (e^{4} + \frac {3 \, e^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, e^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {e^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}}} \]

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/15*(7*sqrt(a)*sqrt(e) + 8*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(a)*sqrt(e)*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + 25*sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 8*sqrt(a)*sqrt(e)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 1)^3/((e^4 + 3*e^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*e^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ e^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(
d*x + c) + 1) + 1)^(7/2))

Giac [F(-1)]

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

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.82 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx=\frac {8\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (2\,\sin \left (c+d\,x\right )+7\,\cos \left (2\,c+2\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+2\,\sin \left (3\,c+3\,d\,x\right )+5\right )}{15\,d\,e^3\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )} \]

[In]

int((a + a*sin(c + d*x))^(1/2)/(e*cos(c + d*x))^(7/2),x)

[Out]

(8*(a*(sin(c + d*x) + 1))^(1/2)*(2*sin(c + d*x) + 7*cos(2*c + 2*d*x) + 2*cos(4*c + 4*d*x) + 2*sin(3*c + 3*d*x)
+ 5))/(15*d*e^3*(e*cos(c + d*x))^(1/2)*(4*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) + 3))