3.89 \(\int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=201 \[ \frac{2 b^2 \left (3 a^2+b^2\right ) \tan ^7(c+d x)}{7 d}+\frac{2 a b \left (a^2+2 b^2\right ) \tan ^6(c+d x)}{3 d}+\frac{\left (12 a^2 b^2+a^4+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (2 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{2 a^2 \left (a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{a b^3 \tan ^8(c+d x)}{2 d}+\frac{b^4 \tan ^9(c+d x)}{9 d} \]

[Out]

(a^4*Tan[c + d*x])/d + (2*a^3*b*Tan[c + d*x]^2)/d + (2*a^2*(a^2 + 3*b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*(2*a^2 +
b^2)*Tan[c + d*x]^4)/d + ((a^4 + 12*a^2*b^2 + b^4)*Tan[c + d*x]^5)/(5*d) + (2*a*b*(a^2 + 2*b^2)*Tan[c + d*x]^
6)/(3*d) + (2*b^2*(3*a^2 + b^2)*Tan[c + d*x]^7)/(7*d) + (a*b^3*Tan[c + d*x]^8)/(2*d) + (b^4*Tan[c + d*x]^9)/(9
*d)

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Rubi [A]  time = 0.171375, antiderivative size = 201, normalized size of antiderivative = 1., 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} \[ \frac{2 b^2 \left (3 a^2+b^2\right ) \tan ^7(c+d x)}{7 d}+\frac{2 a b \left (a^2+2 b^2\right ) \tan ^6(c+d x)}{3 d}+\frac{\left (12 a^2 b^2+a^4+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (2 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{2 a^2 \left (a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{a b^3 \tan ^8(c+d x)}{2 d}+\frac{b^4 \tan ^9(c+d x)}{9 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(a^4*Tan[c + d*x])/d + (2*a^3*b*Tan[c + d*x]^2)/d + (2*a^2*(a^2 + 3*b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*(2*a^2 +
b^2)*Tan[c + d*x]^4)/d + ((a^4 + 12*a^2*b^2 + b^4)*Tan[c + d*x]^5)/(5*d) + (2*a*b*(a^2 + 2*b^2)*Tan[c + d*x]^
6)/(3*d) + (2*b^2*(3*a^2 + b^2)*Tan[c + d*x]^7)/(7*d) + (a*b^3*Tan[c + d*x]^8)/(2*d) + (b^4*Tan[c + d*x]^9)/(9
*d)

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])

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]))

Rubi steps

\begin{align*} \int \sec ^{10}(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^4 \left (1+x^2\right )^2}{x^{10}} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^4}{x^{10}}+\frac{4 a b^3}{x^9}+\frac{2 \left (3 a^2 b^2+b^4\right )}{x^8}+\frac{4 a b \left (a^2+2 b^2\right )}{x^7}+\frac{a^4+12 a^2 b^2+b^4}{x^6}+\frac{4 a b \left (2 a^2+b^2\right )}{x^5}+\frac{2 \left (a^4+3 a^2 b^2\right )}{x^4}+\frac{4 a^3 b}{x^3}+\frac{a^4}{x^2}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac{a^4 \tan (c+d x)}{d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{2 a^2 \left (a^2+3 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{a b \left (2 a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{\left (a^4+12 a^2 b^2+b^4\right ) \tan ^5(c+d x)}{5 d}+\frac{2 a b \left (a^2+2 b^2\right ) \tan ^6(c+d x)}{3 d}+\frac{2 b^2 \left (3 a^2+b^2\right ) \tan ^7(c+d x)}{7 d}+\frac{a b^3 \tan ^8(c+d x)}{2 d}+\frac{b^4 \tan ^9(c+d x)}{9 d}\\ \end{align*}

Mathematica [A]  time = 0.847717, size = 115, normalized size = 0.57 \[ \frac{\frac{2}{7} \left (3 a^2+b^2\right ) (a+b \tan (c+d x))^7-\frac{2}{3} a \left (a^2+b^2\right ) (a+b \tan (c+d x))^6+\frac{1}{5} \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^5+\frac{1}{9} (a+b \tan (c+d x))^9-\frac{1}{2} a (a+b \tan (c+d x))^8}{b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

(((a^2 + b^2)^2*(a + b*Tan[c + d*x])^5)/5 - (2*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^6)/3 + (2*(3*a^2 + b^2)*(a +
b*Tan[c + d*x])^7)/7 - (a*(a + b*Tan[c + d*x])^8)/2 + (a + b*Tan[c + d*x])^9/9)/(b^5*d)

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Maple [A]  time = 0.147, size = 236, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{a}^{4} \left ( -{\frac{8}{15}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \tan \left ( dx+c \right ) +{\frac{2\,{a}^{3}b}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+6\,{a}^{2}{b}^{2} \left ( 1/7\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{35\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{105\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +4\,a{b}^{3} \left ( 1/8\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{8}}}+1/12\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+1/24\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \right ) +{b}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{9\, \left ( \cos \left ( dx+c \right ) \right ) ^{9}}}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{63\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{315\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

1/d*(-a^4*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+2/3*a^3*b/cos(d*x+c)^6+6*a^2*b^2*(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)+4*a*b^3*(1/8*sin(d*x+c)^4/co
s(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)+b^4*(1/9*sin(d*x+c)^5/cos(d*x+c)^9+4
/63*sin(d*x+c)^5/cos(d*x+c)^7+8/315*sin(d*x+c)^5/cos(d*x+c)^5))

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Maxima [A]  time = 1.24377, size = 261, normalized size = 1.3 \begin{align*} \frac{42 \,{\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^{4} + 36 \,{\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^{2} b^{2} + 2 \,{\left (35 \, \tan \left (d x + c\right )^{9} + 90 \, \tan \left (d x + c\right )^{7} + 63 \, \tan \left (d x + c\right )^{5}\right )} b^{4} + \frac{105 \,{\left (4 \, \sin \left (d x + c\right )^{2} - 1\right )} a 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^{3} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{3}}}{630 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/630*(42*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^4 + 36*(15*tan(d*x + c)^7 + 42*tan(d*x +
c)^5 + 35*tan(d*x + c)^3)*a^2*b^2 + 2*(35*tan(d*x + c)^9 + 90*tan(d*x + c)^7 + 63*tan(d*x + c)^5)*b^4 + 105*(4
*sin(d*x + c)^2 - 1)*a*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^3*b/(sin(d*x + c)^2 - 1)^3)/d

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Fricas [A]  time = 0.540231, size = 400, normalized size = 1.99 \begin{align*} \frac{315 \, a b^{3} \cos \left (d x + c\right ) + 420 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \,{\left (8 \,{\left (21 \, a^{4} - 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{8} + 4 \,{\left (21 \, a^{4} - 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{6} + 3 \,{\left (21 \, a^{4} - 18 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 10 \,{\left (27 \, a^{2} b^{2} - 5 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{630 \, d \cos \left (d x + c\right )^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/630*(315*a*b^3*cos(d*x + c) + 420*(a^3*b - a*b^3)*cos(d*x + c)^3 + 2*(8*(21*a^4 - 18*a^2*b^2 + b^4)*cos(d*x
+ c)^8 + 4*(21*a^4 - 18*a^2*b^2 + b^4)*cos(d*x + c)^6 + 3*(21*a^4 - 18*a^2*b^2 + b^4)*cos(d*x + c)^4 + 35*b^4
+ 10*(27*a^2*b^2 - 5*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^9)

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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)**10*(a*cos(d*x+c)+b*sin(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.23587, size = 289, normalized size = 1.44 \begin{align*} \frac{70 \, b^{4} \tan \left (d x + c\right )^{9} + 315 \, a b^{3} \tan \left (d x + c\right )^{8} + 540 \, a^{2} b^{2} \tan \left (d x + c\right )^{7} + 180 \, b^{4} \tan \left (d x + c\right )^{7} + 420 \, a^{3} b \tan \left (d x + c\right )^{6} + 840 \, a b^{3} \tan \left (d x + c\right )^{6} + 126 \, a^{4} \tan \left (d x + c\right )^{5} + 1512 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 126 \, b^{4} \tan \left (d x + c\right )^{5} + 1260 \, a^{3} b \tan \left (d x + c\right )^{4} + 630 \, a b^{3} \tan \left (d x + c\right )^{4} + 420 \, a^{4} \tan \left (d x + c\right )^{3} + 1260 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 1260 \, a^{3} b \tan \left (d x + c\right )^{2} + 630 \, a^{4} \tan \left (d x + c\right )}{630 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/630*(70*b^4*tan(d*x + c)^9 + 315*a*b^3*tan(d*x + c)^8 + 540*a^2*b^2*tan(d*x + c)^7 + 180*b^4*tan(d*x + c)^7
+ 420*a^3*b*tan(d*x + c)^6 + 840*a*b^3*tan(d*x + c)^6 + 126*a^4*tan(d*x + c)^5 + 1512*a^2*b^2*tan(d*x + c)^5 +
126*b^4*tan(d*x + c)^5 + 1260*a^3*b*tan(d*x + c)^4 + 630*a*b^3*tan(d*x + c)^4 + 420*a^4*tan(d*x + c)^3 + 1260
*a^2*b^2*tan(d*x + c)^3 + 1260*a^3*b*tan(d*x + c)^2 + 630*a^4*tan(d*x + c))/d