3.151 \(\int \sec ^6(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=30 \[ \frac{2 a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{5 d} \]

[Out]

(2*a*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2))/(5*d)

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Rubi [A]  time = 0.056813, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {2673} \[ \frac{2 a \sec ^5(c+d x) (a \sin (c+d x)+a)^{5/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(2*a*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(5/2))/(5*d)

Rule 2673

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 - 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]

Rubi steps

\begin{align*} \int \sec ^6(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{2 a \sec ^5(c+d x) (a+a \sin (c+d x))^{5/2}}{5 d}\\ \end{align*}

Mathematica [B]  time = 5.27195, size = 69, normalized size = 2.3 \[ \frac{2 (a (\sin (c+d x)+1))^{7/2}}{5 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^7} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(2*(a*(1 + Sin[c + d*x]))^(7/2))/(5*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5*(Cos[(c + d*x)/2] + Sin[(c + d*x
)/2])^7)

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Maple [A]  time = 0.106, size = 47, normalized size = 1.6 \begin{align*}{\frac{2\,{a}^{4} \left ( 1+\sin \left ( dx+c \right ) \right ) }{5\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{2}\cos \left ( dx+c \right ) d}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6*(a+a*sin(d*x+c))^(7/2),x)

[Out]

2/5*a^4*(1+sin(d*x+c))/(sin(d*x+c)-1)^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [B]  time = 1.67032, size = 365, normalized size = 12.17 \begin{align*} -\frac{2 \,{\left (a^{\frac{7}{2}} + \frac{6 \, a^{\frac{7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{15 \, a^{\frac{7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{20 \, a^{\frac{7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{15 \, a^{\frac{7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac{6 \, a^{\frac{7}{2}} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac{a^{\frac{7}{2}} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )}}{5 \, d{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2/5*(a^(7/2) + 6*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ 20*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6*a^(7/2)
*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^(7/2)*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)/(d*(5*sin(d*x + c)/(co
s(d*x + c) + 1) - 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 5*sin(d*x
+ c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
1)^(7/2))

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Fricas [B]  time = 1.62921, size = 142, normalized size = 4.73 \begin{align*} -\frac{2 \, \sqrt{a \sin \left (d x + c\right ) + a} a^{3}}{5 \,{\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-2/5*sqrt(a*sin(d*x + c) + a)*a^3/(d*cos(d*x + c)^3 + 2*d*cos(d*x + c)*sin(d*x + c) - 2*d*cos(d*x + c))

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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)**6*(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)^6*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out