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]

2/5*a*sec(d*x+c)^5*(a+a*sin(d*x+c))^(5/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.00, 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*}

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Mathematica [B]  time = 5.27, size = 69, normalized size = 2.30 \[ \frac {2 (a (\sin (c+d x)+1))^{7/2}}{5 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^7} \]

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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fricas [B]  time = 0.68, size = 54, normalized size = 1.80 \[ -\frac {2 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{3}}{5 \, {\left (d \cos \left (d x + c\right )^{3} + 2 \, d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )\right )}} \]

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

[Out]

Timed out

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maple [A]  time = 0.18, size = 47, normalized size = 1.57 \[ \frac {2 a^{4} \left (1+\sin \left (d x +c \right )\right )}{5 \left (\sin \left (d x +c \right )-1\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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 = 0.53, size = 270, normalized size = 9.00 \[ -\frac {2 \, {\left (a^{\frac {7}{2}} + \frac {6 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{\frac {7}{2}} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )}}{5 \, d {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{\frac {7}{2}}} \]

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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mupad [B]  time = 8.43, size = 86, normalized size = 2.87 \[ -\frac {16\,a^3\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {a+a\,\left (\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}}{5\,d\,{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}-\mathrm {i}\right )}^5\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(16*a^3*exp(c*3i + d*x*3i)*(a + a*((exp(- c*1i - d*x*1i)*1i)/2 - (exp(c*1i + d*x*1i)*1i)/2))^(1/2))/(5*d*(exp
(c*1i + d*x*1i) - 1i)^5*(exp(c*1i + d*x*1i) + 1i))

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

[Out]

Timed out

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