\(\int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\) [148]

Optimal result

Integrand size = 23, antiderivative size = 91 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {3 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d} \]

[Out]

a*sec(d*x+c)^2*(a+a*sin(d*x+c))^(5/2)/d-3*a^(7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/
d+3*a^3*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2755, 2746, 52, 65, 212} \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {3 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a \sin (c+d x)+a}}{d}+\frac {a \sec ^2(c+d x) (a \sin (c+d x)+a)^{5/2}}{d} \]

[In]

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

[Out]

(-3*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + (3*a^3*Sqrt[a + a*Sin[c + d*x]])/
d + (a*Sec[c + d*x]^2*(a + a*Sin[c + d*x])^(5/2))/d

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2746

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2755

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[-2*b*(
g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(p + 1))), x] + Dist[b^2*((2*m + p - 1)/(g^2*(p + 1
))), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2
- b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && IntegersQ[2*m, 2*p]

Rubi steps \begin{align*} \text {integral}& = \frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {1}{2} \left (3 a^2\right ) \int \sec (c+d x) (a+a \sin (c+d x))^{3/2} \, dx \\ & = \frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (3 a^3\right ) \text {Subst}\left (\int \frac {\sqrt {a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{2 d} \\ & = \frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (3 a^4\right ) \text {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{d} \\ & = \frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d}-\frac {\left (6 a^4\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{d} \\ & = -\frac {3 \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {3 a^3 \sqrt {a+a \sin (c+d x)}}{d}+\frac {a \sec ^2(c+d x) (a+a \sin (c+d x))^{5/2}}{d} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.46 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {a \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},\frac {1}{2} (1+\sin (c+d x))\right ) (a+a \sin (c+d x))^{5/2}}{10 d} \]

[In]

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

[Out]

(a*Hypergeometric2F1[2, 5/2, 7/2, (1 + Sin[c + d*x])/2]*(a + a*Sin[c + d*x])^(5/2))/(10*d)

Maple [A] (verified)

Time = 29.89 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {2 a^{3} \left (\sqrt {a +a \sin \left (d x +c \right )}+4 a \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 \left (a \sin \left (d x +c \right )-a \right )}-\frac {3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}\right )\right )}{d}\)	\(83\)

[In]

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

[Out]

2*a^3*((a+a*sin(d*x+c))^(1/2)+4*a*(-1/4*(a+a*sin(d*x+c))^(1/2)/(a*sin(d*x+c)-a)-3/8*2^(1/2)/a^(1/2)*arctanh(1/
2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.27 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {3 \, \sqrt {2} {\left (a^{3} \sin \left (d x + c\right ) - a^{3}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

[In]

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

[Out]

1/2*(3*sqrt(2)*(a^3*sin(d*x + c) - a^3)*sqrt(a)*log(-(a*sin(d*x + c) - 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt
(a) + 3*a)/(sin(d*x + c) - 1)) + 4*(a^3*sin(d*x + c) - 2*a^3)*sqrt(a*sin(d*x + c) + a))/(d*sin(d*x + c) - d)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=\frac {3 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + 4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4} - \frac {4 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{5}}{a \sin \left (d x + c\right ) - a}}{2 \, a d} \]

[In]

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

[Out]

1/2*(3*sqrt(2)*a^(9/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c
) + a))) + 4*sqrt(a*sin(d*x + c) + a)*a^4 - 4*sqrt(a*sin(d*x + c) + a)*a^5/(a*sin(d*x + c) - a))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.18 \[ \int \sec ^3(c+d x) (a+a \sin (c+d x))^{7/2} \, dx=-\frac {\sqrt {2} a^{\frac {7}{2}} {\left (\frac {2 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 4 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 3 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*sqrt(2)*a^(7/2)*(2*cos(-1/4*pi + 1/2*d*x + 1/2*c)/(cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1) - 4*cos(-1/4*pi
+ 1/2*d*x + 1/2*c) + 3*log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) - 3*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*s
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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