3.71 \(\int \sec ^9(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=174 \[ \frac {a^3 \tan (c+d x)}{d}+\frac {b \left (3 a^2+2 b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {a \left (a^2+6 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (6 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a \left (2 a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {3 a b^2 \tan ^7(c+d x)}{7 d}+\frac {b^3 \tan ^8(c+d x)}{8 d} \]

[Out]

a^3*tan(d*x+c)/d+3/2*a^2*b*tan(d*x+c)^2/d+1/3*a*(2*a^2+3*b^2)*tan(d*x+c)^3/d+1/4*b*(6*a^2+b^2)*tan(d*x+c)^4/d+
1/5*a*(a^2+6*b^2)*tan(d*x+c)^5/d+1/6*b*(3*a^2+2*b^2)*tan(d*x+c)^6/d+3/7*a*b^2*tan(d*x+c)^7/d+1/8*b^3*tan(d*x+c
)^8/d

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Rubi [A]  time = 0.14, antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {3088, 948} \[ \frac {b \left (3 a^2+2 b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {a \left (a^2+6 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (6 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a \left (2 a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {3 a b^2 \tan ^7(c+d x)}{7 d}+\frac {b^3 \tan ^8(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(a^3*Tan[c + d*x])/d + (3*a^2*b*Tan[c + d*x]^2)/(2*d) + (a*(2*a^2 + 3*b^2)*Tan[c + d*x]^3)/(3*d) + (b*(6*a^2 +
b^2)*Tan[c + d*x]^4)/(4*d) + (a*(a^2 + 6*b^2)*Tan[c + d*x]^5)/(5*d) + (b*(3*a^2 + 2*b^2)*Tan[c + d*x]^6)/(6*d
) + (3*a*b^2*Tan[c + d*x]^7)/(7*d) + (b^3*Tan[c + d*x]^8)/(8*d)

Rule 948

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && (IGtQ[m, 0] || (EqQ[m, -2] && EqQ[p, 1] && EqQ[d, 0]))

Rule 3088

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> -Dist[d^(-1), Subst[Int[(x^m*(b + a*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

\begin {align*} \int \sec ^9(c+d x) (a \cos (c+d x)+b \sin (c+d x))^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^3 \left (1+x^2\right )^2}{x^9} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {b^3}{x^9}+\frac {3 a b^2}{x^8}+\frac {3 a^2 b+2 b^3}{x^7}+\frac {a^3+6 a b^2}{x^6}+\frac {6 a^2 b+b^3}{x^5}+\frac {2 a^3+3 a b^2}{x^4}+\frac {3 a^2 b}{x^3}+\frac {a^3}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {a^3 \tan (c+d x)}{d}+\frac {3 a^2 b \tan ^2(c+d x)}{2 d}+\frac {a \left (2 a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac {b \left (6 a^2+b^2\right ) \tan ^4(c+d x)}{4 d}+\frac {a \left (a^2+6 b^2\right ) \tan ^5(c+d x)}{5 d}+\frac {b \left (3 a^2+2 b^2\right ) \tan ^6(c+d x)}{6 d}+\frac {3 a b^2 \tan ^7(c+d x)}{7 d}+\frac {b^3 \tan ^8(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 115, normalized size = 0.66 \[ \frac {\frac {1}{3} \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^6-\frac {4}{5} a \left (a^2+b^2\right ) (a+b \tan (c+d x))^5+\frac {1}{4} \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^4+\frac {1}{8} (a+b \tan (c+d x))^8-\frac {4}{7} a (a+b \tan (c+d x))^7}{b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^9*(a*Cos[c + d*x] + b*Sin[c + d*x])^3,x]

[Out]

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^4)/4 - (4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^5)/5 + ((3*a^2 + b^2)*(a + b
*Tan[c + d*x])^6)/3 - (4*a*(a + b*Tan[c + d*x])^7)/7 + (a + b*Tan[c + d*x])^8/8)/(b^5*d)

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fricas [A]  time = 0.81, size = 128, normalized size = 0.74 \[ \frac {105 \, b^{3} + 140 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (8 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{7} + 4 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} + 45 \, a b^{2} \cos \left (d x + c\right ) + 3 \, {\left (7 \, a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/840*(105*b^3 + 140*(3*a^2*b - b^3)*cos(d*x + c)^2 + 8*(8*(7*a^3 - 3*a*b^2)*cos(d*x + c)^7 + 4*(7*a^3 - 3*a*b
^2)*cos(d*x + c)^5 + 45*a*b^2*cos(d*x + c) + 3*(7*a^3 - 3*a*b^2)*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)
^8)

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giac [A]  time = 0.43, size = 166, normalized size = 0.95 \[ \frac {105 \, b^{3} \tan \left (d x + c\right )^{8} + 360 \, a b^{2} \tan \left (d x + c\right )^{7} + 420 \, a^{2} b \tan \left (d x + c\right )^{6} + 280 \, b^{3} \tan \left (d x + c\right )^{6} + 168 \, a^{3} \tan \left (d x + c\right )^{5} + 1008 \, a b^{2} \tan \left (d x + c\right )^{5} + 1260 \, a^{2} b \tan \left (d x + c\right )^{4} + 210 \, b^{3} \tan \left (d x + c\right )^{4} + 560 \, a^{3} \tan \left (d x + c\right )^{3} + 840 \, a b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b \tan \left (d x + c\right )^{2} + 840 \, a^{3} \tan \left (d x + c\right )}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/840*(105*b^3*tan(d*x + c)^8 + 360*a*b^2*tan(d*x + c)^7 + 420*a^2*b*tan(d*x + c)^6 + 280*b^3*tan(d*x + c)^6 +
168*a^3*tan(d*x + c)^5 + 1008*a*b^2*tan(d*x + c)^5 + 1260*a^2*b*tan(d*x + c)^4 + 210*b^3*tan(d*x + c)^4 + 560
*a^3*tan(d*x + c)^3 + 840*a*b^2*tan(d*x + c)^3 + 1260*a^2*b*tan(d*x + c)^2 + 840*a^3*tan(d*x + c))/d

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maple [A]  time = 9.66, size = 173, normalized size = 0.99 \[ \frac {-a^{3} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )+\frac {a^{2} b}{2 \cos \left (d x +c \right )^{6}}+3 b^{2} a \left (\frac {\sin ^{3}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {4 \left (\sin ^{3}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{8 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{24 \cos \left (d x +c \right )^{4}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^3,x)

[Out]

1/d*(-a^3*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+1/2*a^2*b/cos(d*x+c)^6+3*b^2*a*(1/7*sin(d*x+c)
^3/cos(d*x+c)^7+4/35*sin(d*x+c)^3/cos(d*x+c)^5+8/105*sin(d*x+c)^3/cos(d*x+c)^3)+b^3*(1/8*sin(d*x+c)^4/cos(d*x+
c)^8+1/12*sin(d*x+c)^4/cos(d*x+c)^6+1/24*sin(d*x+c)^4/cos(d*x+c)^4))

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maxima [A]  time = 0.33, size = 154, normalized size = 0.89 \[ \frac {56 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 24 \, {\left (15 \, \tan \left (d x + c\right )^{7} + 42 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3}\right )} a b^{2} + \frac {35 \, {\left (4 \, \sin \left (d x + c\right )^{2} - 1\right )} b^{3}}{\sin \left (d x + c\right )^{8} - 4 \, \sin \left (d x + c\right )^{6} + 6 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{2} + 1} - \frac {420 \, a^{2} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^9*(a*cos(d*x+c)+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/840*(56*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 24*(15*tan(d*x + c)^7 + 42*tan(d*x +
c)^5 + 35*tan(d*x + c)^3)*a*b^2 + 35*(4*sin(d*x + c)^2 - 1)*b^3/(sin(d*x + c)^8 - 4*sin(d*x + c)^6 + 6*sin(d*x
+ c)^4 - 4*sin(d*x + c)^2 + 1) - 420*a^2*b/(sin(d*x + c)^2 - 1)^3)/d

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mupad [B]  time = 1.15, size = 156, normalized size = 0.90 \[ \frac {{\cos \left (c+d\,x\right )}^3\,\left (\frac {a^3\,\sin \left (c+d\,x\right )}{5}-\frac {3\,a\,b^2\,\sin \left (c+d\,x\right )}{35}\right )+{\cos \left (c+d\,x\right )}^5\,\left (\frac {4\,a^3\,\sin \left (c+d\,x\right )}{15}-\frac {4\,a\,b^2\,\sin \left (c+d\,x\right )}{35}\right )+{\cos \left (c+d\,x\right )}^7\,\left (\frac {8\,a^3\,\sin \left (c+d\,x\right )}{15}-\frac {8\,a\,b^2\,\sin \left (c+d\,x\right )}{35}\right )+{\cos \left (c+d\,x\right )}^2\,\left (\frac {a^2\,b}{2}-\frac {b^3}{6}\right )+\frac {b^3}{8}+\frac {3\,a\,b^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{7}}{d\,{\cos \left (c+d\,x\right )}^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))^3/cos(c + d*x)^9,x)

[Out]

(cos(c + d*x)^3*((a^3*sin(c + d*x))/5 - (3*a*b^2*sin(c + d*x))/35) + cos(c + d*x)^5*((4*a^3*sin(c + d*x))/15 -
(4*a*b^2*sin(c + d*x))/35) + cos(c + d*x)^7*((8*a^3*sin(c + d*x))/15 - (8*a*b^2*sin(c + d*x))/35) + cos(c + d
*x)^2*((a^2*b)/2 - b^3/6) + b^3/8 + (3*a*b^2*cos(c + d*x)*sin(c + d*x))/7)/(d*cos(c + d*x)^8)

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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)**9*(a*cos(d*x+c)+b*sin(d*x+c))**3,x)

[Out]

Timed out

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