3.39 \(\int (a+a \sec (c+d x))^3 \sin ^7(c+d x) \, dx\)

Optimal. Leaf size=131 \[ \frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^6(c+d x)}{2 d}-\frac {2 a^3 \cos ^4(c+d x)}{d}-\frac {2 a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^2(c+d x)}{d}+\frac {8 a^3 \cos (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 131, 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 = {3872, 2836, 12, 88} \[ \frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {a^3 \cos ^6(c+d x)}{2 d}-\frac {2 a^3 \cos ^4(c+d x)}{d}-\frac {2 a^3 \cos ^3(c+d x)}{d}+\frac {3 a^3 \cos ^2(c+d x)}{d}+\frac {8 a^3 \cos (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}+\frac {3 a^3 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^7,x]

[Out]

(8*a^3*Cos[c + d*x])/d + (3*a^3*Cos[c + d*x]^2)/d - (2*a^3*Cos[c + d*x]^3)/d - (2*a^3*Cos[c + d*x]^4)/d + (a^3
*Cos[c + d*x]^6)/(2*d) + (a^3*Cos[c + d*x]^7)/(7*d) + (3*a^3*Sec[c + d*x])/d + (a^3*Sec[c + d*x]^2)/(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

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*x)/b
)^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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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 (a+a \sec (c+d x))^3 \sin ^7(c+d x) \, dx &=-\int (-a-a \cos (c+d x))^3 \sin ^4(c+d x) \tan ^3(c+d x) \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {a^3 (-a-x)^3 (-a+x)^6}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^7 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(-a-x)^3 (-a+x)^6}{x^3} \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-8 a^6-\frac {a^9}{x^3}+\frac {3 a^8}{x^2}+6 a^5 x+6 a^4 x^2-8 a^3 x^3+3 a x^5-x^6\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=\frac {8 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos ^2(c+d x)}{d}-\frac {2 a^3 \cos ^3(c+d x)}{d}-\frac {2 a^3 \cos ^4(c+d x)}{d}+\frac {a^3 \cos ^6(c+d x)}{2 d}+\frac {a^3 \cos ^7(c+d x)}{7 d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 106, normalized size = 0.81 \[ \frac {a^3 (14014 \cos (c+d x)-210 \cos (2 (c+d x))+2548 \cos (3 (c+d x))+196 \cos (4 (c+d x))-188 \cos (5 (c+d x))-56 \cos (6 (c+d x))+9 \cos (7 (c+d x))+7 \cos (8 (c+d x))+\cos (9 (c+d x))+427) \sec ^2(c+d x)}{1792 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Sin[c + d*x]^7,x]

[Out]

(a^3*(427 + 14014*Cos[c + d*x] - 210*Cos[2*(c + d*x)] + 2548*Cos[3*(c + d*x)] + 196*Cos[4*(c + d*x)] - 188*Cos
[5*(c + d*x)] - 56*Cos[6*(c + d*x)] + 9*Cos[7*(c + d*x)] + 7*Cos[8*(c + d*x)] + Cos[9*(c + d*x)])*Sec[c + d*x]
^2)/(1792*d)

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fricas [A]  time = 0.77, size = 121, normalized size = 0.92 \[ \frac {32 \, a^{3} \cos \left (d x + c\right )^{9} + 112 \, a^{3} \cos \left (d x + c\right )^{8} - 448 \, a^{3} \cos \left (d x + c\right )^{6} - 448 \, a^{3} \cos \left (d x + c\right )^{5} + 672 \, a^{3} \cos \left (d x + c\right )^{4} + 1792 \, a^{3} \cos \left (d x + c\right )^{3} - 203 \, a^{3} \cos \left (d x + c\right )^{2} + 672 \, a^{3} \cos \left (d x + c\right ) + 112 \, a^{3}}{224 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^7,x, algorithm="fricas")

[Out]

1/224*(32*a^3*cos(d*x + c)^9 + 112*a^3*cos(d*x + c)^8 - 448*a^3*cos(d*x + c)^6 - 448*a^3*cos(d*x + c)^5 + 672*
a^3*cos(d*x + c)^4 + 1792*a^3*cos(d*x + c)^3 - 203*a^3*cos(d*x + c)^2 + 672*a^3*cos(d*x + c) + 112*a^3)/(d*cos
(d*x + c)^2)

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giac [A]  time = 1.11, size = 239, normalized size = 1.82 \[ \frac {2 \, {\left (\frac {7 \, {\left (3 \, a^{3} + \frac {2 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}} - \frac {43 \, a^{3} - \frac {273 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {672 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {630 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {343 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {105 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {14 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{7}}\right )}}{7 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^7,x, algorithm="giac")

[Out]

2/7*(7*(3*a^3 + 2*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2 - (
43*a^3 - 273*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 672*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 6
30*a^3*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 + 343*a^3*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 105*a^3
*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5 + 14*a^3*(cos(d*x + c) - 1)^6/(cos(d*x + c) + 1)^6)/((cos(d*x + c)
- 1)/(cos(d*x + c) + 1) - 1)^7)/d

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maple [A]  time = 0.71, size = 130, normalized size = 0.99 \[ \frac {64 a^{3} \cos \left (d x +c \right )}{7 d}+\frac {20 a^{3} \cos \left (d x +c \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{7 d}+\frac {24 a^{3} \cos \left (d x +c \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{7 d}+\frac {32 a^{3} \cos \left (d x +c \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{7 d}+\frac {3 a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )}+\frac {a^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*sin(d*x+c)^7,x)

[Out]

64/7*a^3*cos(d*x+c)/d+20/7/d*a^3*cos(d*x+c)*sin(d*x+c)^6+24/7/d*a^3*cos(d*x+c)*sin(d*x+c)^4+32/7/d*a^3*cos(d*x
+c)*sin(d*x+c)^2+3/d*a^3*sin(d*x+c)^8/cos(d*x+c)+1/2/d*a^3*sin(d*x+c)^8/cos(d*x+c)^2

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maxima [A]  time = 0.32, size = 107, normalized size = 0.82 \[ \frac {2 \, a^{3} \cos \left (d x + c\right )^{7} + 7 \, a^{3} \cos \left (d x + c\right )^{6} - 28 \, a^{3} \cos \left (d x + c\right )^{4} - 28 \, a^{3} \cos \left (d x + c\right )^{3} + 42 \, a^{3} \cos \left (d x + c\right )^{2} + 112 \, a^{3} \cos \left (d x + c\right ) + \frac {7 \, {\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )}}{\cos \left (d x + c\right )^{2}}}{14 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*sin(d*x+c)^7,x, algorithm="maxima")

[Out]

1/14*(2*a^3*cos(d*x + c)^7 + 7*a^3*cos(d*x + c)^6 - 28*a^3*cos(d*x + c)^4 - 28*a^3*cos(d*x + c)^3 + 42*a^3*cos
(d*x + c)^2 + 112*a^3*cos(d*x + c) + 7*(6*a^3*cos(d*x + c) + a^3)/cos(d*x + c)^2)/d

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mupad [B]  time = 0.90, size = 107, normalized size = 0.82 \[ \frac {\frac {3\,a^3\,\cos \left (c+d\,x\right )+\frac {a^3}{2}}{{\cos \left (c+d\,x\right )}^2}+8\,a^3\,\cos \left (c+d\,x\right )+3\,a^3\,{\cos \left (c+d\,x\right )}^2-2\,a^3\,{\cos \left (c+d\,x\right )}^3-2\,a^3\,{\cos \left (c+d\,x\right )}^4+\frac {a^3\,{\cos \left (c+d\,x\right )}^6}{2}+\frac {a^3\,{\cos \left (c+d\,x\right )}^7}{7}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^7*(a + a/cos(c + d*x))^3,x)

[Out]

((3*a^3*cos(c + d*x) + a^3/2)/cos(c + d*x)^2 + 8*a^3*cos(c + d*x) + 3*a^3*cos(c + d*x)^2 - 2*a^3*cos(c + d*x)^
3 - 2*a^3*cos(c + d*x)^4 + (a^3*cos(c + d*x)^6)/2 + (a^3*cos(c + d*x)^7)/7)/d

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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((a+a*sec(d*x+c))**3*sin(d*x+c)**7,x)

[Out]

Timed out

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