3.1.76 \(\int \frac {\sin ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [76]

Optimal. Leaf size=73 \[ -\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^6(c+d x)}{3 a^2 d}+\frac {\cos ^7(c+d x)}{7 a^2 d} \]

[Out]

-1/3*cos(d*x+c)^3/a^2/d+1/2*cos(d*x+c)^4/a^2/d-1/3*cos(d*x+c)^6/a^2/d+1/7*cos(d*x+c)^7/a^2/d

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Rubi [A]
time = 0.11, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3957, 2915, 12, 76} \begin {gather*} \frac {\cos ^7(c+d x)}{7 a^2 d}-\frac {\cos ^6(c+d x)}{3 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^3(c+d x)}{3 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Cos[c + d*x]^3/(a^2*d) + Cos[c + d*x]^4/(2*a^2*d) - Cos[c + d*x]^6/(3*a^2*d) + Cos[c + d*x]^7/(7*a^2*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin ^7(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin ^7(c+d x)}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {(-a-x)^3 x^2 (-a+x)}{a^2} \, dx,x,-a \cos (c+d x)\right )}{a^7 d}\\ &=\frac {\text {Subst}\left (\int (-a-x)^3 x^2 (-a+x) \, dx,x,-a \cos (c+d x)\right )}{a^9 d}\\ &=\frac {\text {Subst}\left (\int \left (a^4 x^2+2 a^3 x^3-2 a x^5-x^6\right ) \, dx,x,-a \cos (c+d x)\right )}{a^9 d}\\ &=-\frac {\cos ^3(c+d x)}{3 a^2 d}+\frac {\cos ^4(c+d x)}{2 a^2 d}-\frac {\cos ^6(c+d x)}{3 a^2 d}+\frac {\cos ^7(c+d x)}{7 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 1.19, size = 53, normalized size = 0.73 \begin {gather*} \frac {4 (17 \cos (c+d x)+10 \cos (2 (c+d x))+3 (4+\cos (3 (c+d x)))) \sin ^8\left (\frac {1}{2} (c+d x)\right )}{21 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(17*Cos[c + d*x] + 10*Cos[2*(c + d*x)] + 3*(4 + Cos[3*(c + d*x)]))*Sin[(c + d*x)/2]^8)/(21*a^2*d)

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Maple [A]
time = 0.14, size = 50, normalized size = 0.68

method	result	size
derivativedivides	\(-\frac {\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{4}}-\frac {1}{7 \sec \left (d x +c \right )^{7}}+\frac {1}{3 \sec \left (d x +c \right )^{6}}}{d \,a^{2}}\)	\(50\)
default	\(-\frac {\frac {1}{3 \sec \left (d x +c \right )^{3}}-\frac {1}{2 \sec \left (d x +c \right )^{4}}-\frac {1}{7 \sec \left (d x +c \right )^{7}}+\frac {1}{3 \sec \left (d x +c \right )^{6}}}{d \,a^{2}}\)	\(50\)
risch	\(-\frac {11 \cos \left (d x +c \right )}{64 a^{2} d}+\frac {\cos \left (7 d x +7 c \right )}{448 d \,a^{2}}-\frac {\cos \left (6 d x +6 c \right )}{96 d \,a^{2}}+\frac {\cos \left (5 d x +5 c \right )}{64 d \,a^{2}}-\frac {7 \cos \left (3 d x +3 c \right )}{192 d \,a^{2}}+\frac {3 \cos \left (2 d x +2 c \right )}{32 d \,a^{2}}\)	\(101\)
norman	\(\frac {-\frac {16 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {8}{21 a d}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {40 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {16 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} a}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/a^2*(1/3/sec(d*x+c)^3-1/2/sec(d*x+c)^4-1/7/sec(d*x+c)^7+1/3/sec(d*x+c)^6)

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Maxima [A]
time = 0.26, size = 49, normalized size = 0.67 \begin {gather*} \frac {6 \, \cos \left (d x + c\right )^{7} - 14 \, \cos \left (d x + c\right )^{6} + 21 \, \cos \left (d x + c\right )^{4} - 14 \, \cos \left (d x + c\right )^{3}}{42 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*(6*cos(d*x + c)^7 - 14*cos(d*x + c)^6 + 21*cos(d*x + c)^4 - 14*cos(d*x + c)^3)/(a^2*d)

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Fricas [A]
time = 3.78, size = 49, normalized size = 0.67 \begin {gather*} \frac {6 \, \cos \left (d x + c\right )^{7} - 14 \, \cos \left (d x + c\right )^{6} + 21 \, \cos \left (d x + c\right )^{4} - 14 \, \cos \left (d x + c\right )^{3}}{42 \, a^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/42*(6*cos(d*x + c)^7 - 14*cos(d*x + c)^6 + 21*cos(d*x + c)^4 - 14*cos(d*x + c)^3)/(a^2*d)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (65) = 130\).
time = 0.46, size = 141, normalized size = 1.93 \begin {gather*} -\frac {8 \, {\left (\frac {7 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {21 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {35 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {42 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - 1\right )}}{21 \, a^{2} d {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/21*(7*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 21*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 35*(cos(d*x +
c) - 1)^3/(cos(d*x + c) + 1)^3 - 14*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 + 42*(cos(d*x + c) - 1)^5/(cos(d
*x + c) + 1)^5 - 1)/(a^2*d*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^7)

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Mupad [B]
time = 0.92, size = 58, normalized size = 0.79 \begin {gather*} -\frac {\frac {{\cos \left (c+d\,x\right )}^3}{3\,a^2}-\frac {{\cos \left (c+d\,x\right )}^4}{2\,a^2}+\frac {{\cos \left (c+d\,x\right )}^6}{3\,a^2}-\frac {{\cos \left (c+d\,x\right )}^7}{7\,a^2}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(c + d*x)^3/(3*a^2) - cos(c + d*x)^4/(2*a^2) + cos(c + d*x)^6/(3*a^2) - cos(c + d*x)^7/(7*a^2))/d

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