\(\int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [147]

Optimal result

Integrand size = 23, antiderivative size = 89 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {64 a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {16 a^2 \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{5/2}}{3 d} \]

[Out]

-16/3*a^2*sec(d*x+c)*(a+a*sin(d*x+c))^(3/2)/d-2/3*a*sec(d*x+c)*(a+a*sin(d*x+c))^(5/2)/d+64/3*a^3*sec(d*x+c)*(a
+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2753, 2752} \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {64 a^3 \sec (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {16 a^2 \sec (c+d x) (a \sin (c+d x)+a)^{3/2}}{3 d}-\frac {2 a \sec (c+d x) (a \sin (c+d x)+a)^{5/2}}{3 d} \]

[In]

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

[Out]

(64*a^3*Sec[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d) - (16*a^2*Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2))/(3*d)
- (2*a*Sec[c + d*x]*(a + a*Sin[c + d*x])^(5/2))/(3*d)

Rule 2752

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 - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2753

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 - 1)/(f*g*(m + p))), x] + Dist[a*((2*m + p - 1)/(m + p)), 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] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{5/2}}{3 d}+\frac {1}{3} (8 a) \int \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2} \, dx \\ & = -\frac {16 a^2 \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{5/2}}{3 d}+\frac {1}{3} \left (32 a^2\right ) \int \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = \frac {64 a^3 \sec (c+d x) \sqrt {a+a \sin (c+d x)}}{3 d}-\frac {16 a^2 \sec (c+d x) (a+a \sin (c+d x))^{3/2}}{3 d}-\frac {2 a \sec (c+d x) (a+a \sin (c+d x))^{5/2}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.54 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {a^3 \sec (c+d x) (45+\cos (2 (c+d x))-20 \sin (c+d x)) \sqrt {a (1+\sin (c+d x))}}{3 d} \]

[In]

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

[Out]

(a^3*Sec[c + d*x]*(45 + Cos[2*(c + d*x)] - 20*Sin[c + d*x])*Sqrt[a*(1 + Sin[c + d*x])])/(3*d)

Maple [A] (verified)

Time = 4.86 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62

method	result	size
default	\(-\frac {2 a^{4} \left (1+\sin \left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )+10 \sin \left (d x +c \right )-23\right )}{3 \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(55\)

[In]

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

[Out]

-2/3*a^4*(1+sin(d*x+c))*(sin(d*x+c)^2+10*sin(d*x+c)-23)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.61 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {2 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - 10 \, a^{3} \sin \left (d x + c\right ) + 22 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{3 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

2/3*(a^3*cos(d*x + c)^2 - 10*a^3*sin(d*x + c) + 22*a^3)*sqrt(a*sin(d*x + c) + a)/(d*cos(d*x + c))

Sympy [F(-1)]

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

[In]

integrate(sec(d*x+c)**2*(a+a*sin(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. 237 vs. \(2 (77) = 154\).

Time = 0.31 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.66 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {2 \, {\left (23 \, a^{\frac {7}{2}} - \frac {20 \, a^{\frac {7}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {88 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {60 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {130 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {60 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {88 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {20 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {23 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )}}{3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/3*(23*a^(7/2) - 20*a^(7/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 88*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2
- 60*a^(7/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 130*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 60*a^(7/
2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 88*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 20*a^(7/2)*sin(d*x +
c)^7/(cos(d*x + c) + 1)^7 + 23*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)/(d*(sin(d*x + c)/(cos(d*x + c) +
1) - 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^(7/2))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18 \[ \int \sec ^2(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {4 \, \sqrt {2} {\left (a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {3 \, a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}\right )} \sqrt {a}}{3 \, d} \]

[In]

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

[Out]

4/3*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)^3 - 6*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) - 3*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))*sqrt(a)/d

Mupad [F(-1)]

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

[In]

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

[Out]

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