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*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)

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Rubi [A]  time = 0.232092, antiderivative size = 135, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 0.540617, size = 120, normalized size = 0.89 \[ -\frac{2 a^3 \left (15 \sin ^2(c+d x)-50 \sin (c+d x)+67\right ) \sqrt{a (\sin (c+d x)+1)}+15 \sqrt{2} a^{7/2} \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^6 \tanh ^{-1}\left (\frac{\sqrt{a (\sin (c+d x)+1)}}{\sqrt{2} \sqrt{a}}\right )}{384 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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))/(384*d*(-1 + Sin[c + d*x])
^3)

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Maple [A]  time = 0.218, size = 144, normalized size = 1.1 \begin{align*} 2\,{\frac{{a}^{7}}{d} \left ( -1/12\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a \left ( a\sin \left ( dx+c \right ) -a \right ) ^{3}}}-{\frac{5}{12\,a} \left ( -1/8\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a \left ( a\sin \left ( dx+c \right ) -a \right ) ^{2}}}-3/8\,{\frac{1}{a} \left ( -1/4\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }}{a \left ( a\sin \left ( dx+c \right ) -a \right ) }}+1/8\,{\frac{\sqrt{2}}{{a}^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) } \right ) } \right ) } \right ) } \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.80116, size = 501, normalized size = 3.71 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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