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

Optimal. Leaf size=135 \[ \frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{64 d}+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{7/2}}{6 d}+\frac {5 a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{48 d} \]

[Out]

5/64*a^2*sec(d*x+c)^2*(a+a*sin(d*x+c))^(3/2)/d+5/48*a*sec(d*x+c)^4*(a+a*sin(d*x+c))^(5/2)/d+1/6*sec(d*x+c)^6*(
a+a*sin(d*x+c))^(7/2)/d+5/128*a^(7/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d

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Rubi [A]  time = 0.23, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2675, 2667, 63, 206} \[ \frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a \sin (c+d x)+a}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a \sin (c+d x)+a)^{3/2}}{64 d}+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{7/2}}{6 d}+\frac {5 a \sec ^4(c+d x) (a \sin (c+d x)+a)^{5/2}}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^(7/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(64*Sqrt[2]*d) + (5*a^2*Sec[c + d*x]^2*(a + a*
Sin[c + d*x])^(3/2))/(64*d) + (5*a*Sec[c + d*x]^4*(a + a*Sin[c + d*x])^(5/2))/(48*d) + (Sec[c + d*x]^6*(a + a*
Sin[c + d*x])^(7/2))/(6*d)

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 ^7(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{12} (5 a) \int \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{32} \left (5 a^2\right ) \int \sec ^3(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {1}{128} \left (5 a^3\right ) \int \sec (c+d x) \sqrt {a+a \sin (c+d x)} \, dx\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,a \sin (c+d x)\right )}{128 d}\\ &=\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}+\frac {\left (5 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+a \sin (c+d x)}\right )}{64 d}\\ &=\frac {5 a^{7/2} \tanh ^{-1}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{64 \sqrt {2} d}+\frac {5 a^2 \sec ^2(c+d x) (a+a \sin (c+d x))^{3/2}}{64 d}+\frac {5 a \sec ^4(c+d x) (a+a \sin (c+d x))^{5/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{7/2}}{6 d}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 120, normalized size = 0.89 \[ -\frac {15 \sqrt {2} a^{7/2} \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6 \tanh ^{-1}\left (\frac {\sqrt {a (\sin (c+d x)+1)}}{\sqrt {2} \sqrt {a}}\right )+2 a^3 \left (15 \sin ^2(c+d x)-50 \sin (c+d x)+67\right ) \sqrt {a (\sin (c+d x)+1)}}{384 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(15*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a*(1 + Sin[c + d*x])]/(Sqrt[2]*Sqrt[a])]*(Cos[(c + d*x)/2] - Sin[(c +
d*x)/2])^6 + 2*a^3*Sqrt[a*(1 + Sin[c + d*x])]*(67 - 50*Sin[c + d*x] + 15*Sin[c + d*x]^2))/(d*(-1 + Sin[c + d*x
])^3)

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fricas [A]  time = 0.64, size = 193, normalized size = 1.43 \[ \frac {15 \, {\left (3 \, \sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - 4 \, \sqrt {2} a^{3} - {\left (\sqrt {2} a^{3} \cos \left (d x + c\right )^{2} - 4 \, \sqrt {2} a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) + 4 \, {\left (15 \, a^{3} \cos \left (d x + c\right )^{2} + 50 \, a^{3} \sin \left (d x + c\right ) - 82 \, a^{3}\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(15*(3*sqrt(2)*a^3*cos(d*x + c)^2 - 4*sqrt(2)*a^3 - (sqrt(2)*a^3*cos(d*x + c)^2 - 4*sqrt(2)*a^3)*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*(15*a^3*cos(d*x + c)^2 + 50*a^3*sin(d*x + c) - 82*a^3)*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^2 - (d*co
s(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)

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

[Out]

Timed out

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maple [A]  time = 0.35, size = 144, normalized size = 1.07 \[ \frac {2 a^{7} \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{12 a \left (a \sin \left (d x +c \right )-a \right )^{3}}-\frac {5 \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{8 a \left (a \sin \left (d x +c \right )-a \right )^{2}}-\frac {3 \left (-\frac {\sqrt {a +a \sin \left (d x +c \right )}}{4 a \left (a \sin \left (d x +c \right )-a \right )}+\frac {\sqrt {2}\, \arctanh \left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{8 a}\right )}{12 a}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^7*(-1/12*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a)^3-5/12/a*(-1/8*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a
)^2-3/8/a*(-1/4*(a+a*sin(d*x+c))^(1/2)/a/(a*sin(d*x+c)-a)+1/8/a^(3/2)*2^(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 = 0.46, size = 168, normalized size = 1.24 \[ -\frac {15 \, \sqrt {2} a^{\frac {9}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{5} - 80 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{6} + 132 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{7}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{3} - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a + 12 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{2} - 8 \, a^{3}}}{768 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(15*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*(15*(a*sin(d*x + c) + a)^(5/2)*a^5 - 80*(a*sin(d*x + c) + a)^(3/2)*a^6 + 132*sqrt(a*sin(d*x +
c) + a)*a^7)/((a*sin(d*x + c) + a)^3 - 6*(a*sin(d*x + c) + a)^2*a + 12*(a*sin(d*x + c) + a)*a^2 - 8*a^3))/(a*
d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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