3.154 \(\int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=191 \[ \frac {315 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{2048 \sqrt {2} d}-\frac {315 a^4}{2048 d \sqrt {a \sin (c+d x)+a}}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{256 d}+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}+\frac {3 a \sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{32 d} \]

[Out]

21/256*a^2*sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2)/d+3/32*a*sec(d*x+c)^6*(a+a*sin(d*x+c))^(5/2)/d+1/8*sec(d*x+c)^8
*(a+a*sin(d*x+c))^(7/2)/d+315/4096*a^(7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-315/2
048*a^4/d/(a+a*sin(d*x+c))^(1/2)+105/1024*a^3*sec(d*x+c)^2*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.31, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2675, 2667, 51, 63, 206} \[ -\frac {315 a^4}{2048 d \sqrt {a \sin (c+d x)+a}}+\frac {315 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{2048 \sqrt {2} d}+\frac {21 a^2 \sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{256 d}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{1024 d}+\frac {\sec ^8(c+d x) (a \sin (c+d x)+a)^{7/2}}{8 d}+\frac {3 a \sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{32 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(315*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(2048*Sqrt[2]*d) - (315*a^4)/(2048*d*Sqrt[a
+ a*Sin[c + d*x]]) + (105*a^3*Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(1024*d) + (21*a^2*Sec[c + d*x]^4*(a +
a*Sin[c + d*x])^(3/2))/(256*d) + (3*a*Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2))/(32*d) + (Sec[c + d*x]^8*(a +
a*Sin[c + d*x])^(7/2))/(8*d)

Rule 51

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

Rule 63

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 206

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

Rule 2667

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 2675

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)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rubi steps

\begin {align*} \int \sec ^9(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {1}{16} (9 a) \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {1}{64} \left (21 a^2\right ) \int \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {1}{512} \left (105 a^3\right ) \int \sec ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {\left (315 a^4\right ) \int \frac {\sec (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{2048}\\ &=\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {\left (315 a^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,a \sin (c+d x)\right )}{2048 d}\\ &=-\frac {315 a^4}{2048 d \sqrt {a+a \sin (c+d x)}}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {\left (315 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{4096 d}\\ &=-\frac {315 a^4}{2048 d \sqrt {a+a \sin (c+d x)}}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}+\frac {\left (315 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{2048 d}\\ &=\frac {315 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2048 \sqrt {2} d}-\frac {315 a^4}{2048 d \sqrt {a+a \sin (c+d x)}}+\frac {105 a^3 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{1024 d}+\frac {21 a^2 \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{256 d}+\frac {3 a \sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{32 d}+\frac {\sec ^8(c+d x) (a+a \sin (c+d x))^{7/2}}{8 d}\\ \end {align*}

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Mathematica [C]  time = 0.11, size = 44, normalized size = 0.23 \[ -\frac {a^4 \, _2F_1\left (-\frac {1}{2},5;\frac {1}{2};\frac {1}{2} (\sin (c+d x)+1)\right )}{16 d \sqrt {a \sin (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

-1/16*(a^4*Hypergeometric2F1[-1/2, 5, 1/2, (1 + Sin[c + d*x])/2])/(d*Sqrt[a + a*Sin[c + d*x]])

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fricas [A]  time = 0.56, size = 254, normalized size = 1.33 \[ \frac {315 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{4} - 4 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (315 \, a^{3} \cos \left (d x + c\right )^{4} - 1722 \, a^{3} \cos \left (d x + c\right )^{2} + 896 \, a^{3} + 6 \, {\left (175 \, a^{3} \cos \left (d x + c\right )^{2} - 192 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{8192 \, {\left (3 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/8192*(315*(3*sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqrt(2)*a^3*cos(d*x + c)^2 - (sqrt(2)*a^3*cos(d*x + c)^4 - 4*sqr
t(2)*a^3*cos(d*x + c)^2)*sin(d*x + c))*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*(315*a^3*cos(d*x + c)^4 - 1722*a^3*cos(d*x + c)^2 + 896*a^3 + 6*(175*a^3*cos
(d*x + c)^2 - 192*a^3)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 - (d*c
os(d*x + c)^4 - 4*d*cos(d*x + c)^2)*sin(d*x + c))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.42, size = 129, normalized size = 0.68 \[ -\frac {2 a^{9} \left (\frac {1}{32 a^{5} \sqrt {a +a \sin \left (d x +c \right )}}+\frac {-\frac {\sqrt {a +a \sin \left (d x +c \right )}\, a^{3} \left (187 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-725 \left (\cos ^{2}\left (d x +c \right )\right )-1236 \sin \left (d x +c \right )+1364\right )}{128 \left (a \sin \left (d x +c \right )-a \right )^{4}}-\frac {315 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{256 \sqrt {a}}}{32 a^{5}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x)

[Out]

-2*a^9*(1/32/a^5/(a+a*sin(d*x+c))^(1/2)+1/32/a^5*(-1/128*(a+a*sin(d*x+c))^(1/2)*a^3*(187*cos(d*x+c)^2*sin(d*x+
c)-725*cos(d*x+c)^2-1236*sin(d*x+c)+1364)/(a*sin(d*x+c)-a)^4-315/256*2^(1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+
c))^(1/2)*2^(1/2)/a^(1/2))))/d

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maxima [A]  time = 0.45, size = 219, normalized size = 1.15 \[ -\frac {315 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{4} a^{5} - 2310 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{6} + 6132 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{7} - 6696 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{8} + 2048 \, a^{9}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 8 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 24 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 32 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3} + 16 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{4}}}{8192 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-1/8192*(315*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*(315*(a*sin(d*x + c) + a)^4*a^5 - 2310*(a*sin(d*x + c) + a)^3*a^6 + 6132*(a*sin(d*x + c) +
a)^2*a^7 - 6696*(a*sin(d*x + c) + a)*a^8 + 2048*a^9)/((a*sin(d*x + c) + a)^(9/2) - 8*(a*sin(d*x + c) + a)^(7/2
)*a + 24*(a*sin(d*x + c) + a)^(5/2)*a^2 - 32*(a*sin(d*x + c) + a)^(3/2)*a^3 + 16*sqrt(a*sin(d*x + c) + a)*a^4)
)/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{7/2}}{{\cos \left (c+d\,x\right )}^9} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^9,x)

[Out]

int((a + a*sin(c + d*x))^(7/2)/cos(c + d*x)^9, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**9*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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