3.87 \(\int \sec ^8(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=143 \[ \frac{b^2 \left (6 a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{a^2 \left (a^2+6 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{2 a b^3 \tan ^6(c+d x)}{3 d}+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

[Out]

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

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Rubi [A]  time = 0.123162, antiderivative size = 143, 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, 894} \[ \frac{b^2 \left (6 a^2+b^2\right ) \tan ^5(c+d x)}{5 d}+\frac{a b \left (a^2+b^2\right ) \tan ^4(c+d x)}{d}+\frac{a^2 \left (a^2+6 b^2\right ) \tan ^3(c+d x)}{3 d}+\frac{2 a^3 b \tan ^2(c+d x)}{d}+\frac{a^4 \tan (c+d x)}{d}+\frac{2 a b^3 \tan ^6(c+d x)}{3 d}+\frac{b^4 \tan ^7(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(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 + (a^2*(a^2 + 6*b^2)*Tan[c + d*x]^3)/(3*d) + (a*b*(a^2 + b^2
)*Tan[c + d*x]^4)/d + (b^2*(6*a^2 + b^2)*Tan[c + d*x]^5)/(5*d) + (2*a*b^3*Tan[c + d*x]^6)/(3*d) + (b^4*Tan[c +
d*x]^7)/(7*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 894

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] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.557767, size = 54, normalized size = 0.38 \[ \frac{(a+b \tan (c+d x))^5 \left (a^2-5 a b \tan (c+d x)+15 b^2 \tan ^2(c+d x)+21 b^2\right )}{105 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a*Cos[c + d*x] + b*Sin[c + d*x])^4,x]

[Out]

((a + b*Tan[c + d*x])^5*(a^2 + 21*b^2 - 5*a*b*Tan[c + d*x] + 15*b^2*Tan[c + d*x]^2))/(105*b^3*d)

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Maple [A]  time = 0.149, size = 171, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( -{a}^{4} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +{\frac{{a}^{3}b}{ \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+6\,{a}^{2}{b}^{2} \left ( 1/5\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+2/15\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) +4\,a{b}^{3} \left ( 1/6\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+1/12\,{\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}}{7\, \left ( \cos \left ( dx+c \right ) \right ) ^{7}}}+{\frac{2\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{35\, \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)^8*(a*cos(d*x+c)+b*sin(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.12172, size = 204, normalized size = 1.43 \begin{align*} \frac{35 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} + 42 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{2} b^{2} + 3 \,{\left (5 \, \tan \left (d x + c\right )^{7} + 7 \, \tan \left (d x + c\right )^{5}\right )} b^{4} - \frac{35 \,{\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + \frac{105 \, a^{3} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

1/105*(35*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^4 + 42*(3*tan(d*x + c)^5 + 5*tan(d*x + c)^3)*a^2*b^2 + 3*(5*tan(
d*x + c)^7 + 7*tan(d*x + c)^5)*b^4 - 35*(3*sin(d*x + c)^2 - 1)*a*b^3/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*si
n(d*x + c)^2 - 1) + 105*a^3*b/(sin(d*x + c)^2 - 1)^2)/d

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Fricas [A]  time = 0.511883, size = 333, normalized size = 2.33 \begin{align*} \frac{70 \, a b^{3} \cos \left (d x + c\right ) + 105 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} +{\left (2 \,{\left (35 \, a^{4} - 42 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{6} +{\left (35 \, a^{4} - 42 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, b^{4} + 6 \,{\left (21 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/105*(70*a*b^3*cos(d*x + c) + 105*(a^3*b - a*b^3)*cos(d*x + c)^3 + (2*(35*a^4 - 42*a^2*b^2 + 3*b^4)*cos(d*x +
c)^6 + (35*a^4 - 42*a^2*b^2 + 3*b^4)*cos(d*x + c)^4 + 15*b^4 + 6*(21*a^2*b^2 - 4*b^4)*cos(d*x + c)^2)*sin(d*x
+ c))/(d*cos(d*x + c)^7)

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

[Out]

Timed out

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Giac [A]  time = 1.17608, size = 194, normalized size = 1.36 \begin{align*} \frac{15 \, b^{4} \tan \left (d x + c\right )^{7} + 70 \, a b^{3} \tan \left (d x + c\right )^{6} + 126 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 21 \, b^{4} \tan \left (d x + c\right )^{5} + 105 \, a^{3} b \tan \left (d x + c\right )^{4} + 105 \, a b^{3} \tan \left (d x + c\right )^{4} + 35 \, a^{4} \tan \left (d x + c\right )^{3} + 210 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 210 \, a^{3} b \tan \left (d x + c\right )^{2} + 105 \, a^{4} \tan \left (d x + c\right )}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

1/105*(15*b^4*tan(d*x + c)^7 + 70*a*b^3*tan(d*x + c)^6 + 126*a^2*b^2*tan(d*x + c)^5 + 21*b^4*tan(d*x + c)^5 +
105*a^3*b*tan(d*x + c)^4 + 105*a*b^3*tan(d*x + c)^4 + 35*a^4*tan(d*x + c)^3 + 210*a^2*b^2*tan(d*x + c)^3 + 210
*a^3*b*tan(d*x + c)^2 + 105*a^4*tan(d*x + c))/d