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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 143, 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, 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 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]))

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 ^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*}

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Mathematica [A]  time = 0.55, 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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fricas [A]  time = 0.49, size = 142, normalized size = 0.99 \[ \frac {70 \, a b^{3} \cos \left (d x + c\right ) + 105 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} + {\left (2 \, {\left (35 \, a^{4} - 42 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{6} + {\left (35 \, a^{4} - 42 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, b^{4} + 6 \, {\left (21 \, a^{2} b^{2} - 4 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{105 \, d \cos \left (d x + c\right )^{7}} \]

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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giac [A]  time = 0.43, size = 144, normalized size = 1.01 \[ \frac {15 \, b^{4} \tan \left (d x + c\right )^{7} + 70 \, a b^{3} \tan \left (d x + c\right )^{6} + 126 \, a^{2} b^{2} \tan \left (d x + c\right )^{5} + 21 \, b^{4} \tan \left (d x + c\right )^{5} + 105 \, a^{3} b \tan \left (d x + c\right )^{4} + 105 \, a b^{3} \tan \left (d x + c\right )^{4} + 35 \, a^{4} \tan \left (d x + c\right )^{3} + 210 \, a^{2} b^{2} \tan \left (d x + c\right )^{3} + 210 \, a^{3} b \tan \left (d x + c\right )^{2} + 105 \, a^{4} \tan \left (d x + c\right )}{105 \, d} \]

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

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maple [A]  time = 52.37, size = 171, normalized size = 1.20 \[ \frac {-a^{4} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {a^{3} b}{\cos \left (d x +c \right )^{4}}+6 a^{2} b^{2} \left (\frac {\sin ^{3}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}\right )+4 a \,b^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{6 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{12 \cos \left (d x +c \right )^{4}}\right )+b^{4} \left (\frac {\sin ^{5}\left (d x +c \right )}{7 \cos \left (d x +c \right )^{7}}+\frac {2 \left (\sin ^{5}\left (d x +c \right )\right )}{35 \cos \left (d x +c \right )^{5}}\right )}{d} \]

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 = 0.33, size = 151, normalized size = 1.06 \[ \frac {35 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{4} + 42 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 5 \, \tan \left (d x + c\right )^{3}\right )} a^{2} b^{2} + 3 \, {\left (5 \, \tan \left (d x + c\right )^{7} + 7 \, \tan \left (d x + c\right )^{5}\right )} b^{4} - \frac {35 \, {\left (3 \, \sin \left (d x + c\right )^{2} - 1\right )} a b^{3}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} + \frac {105 \, a^{3} b}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{105 \, d} \]

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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mupad [B]  time = 1.10, size = 186, normalized size = 1.30 \[ \frac {\frac {b^4\,\sin \left (c+d\,x\right )}{7}-{\cos \left (c+d\,x\right )}^3\,\left (a\,b^3-a^3\,b\right )-{\cos \left (c+d\,x\right )}^2\,\left (\frac {8\,b^4\,\sin \left (c+d\,x\right )}{35}-\frac {6\,a^2\,b^2\,\sin \left (c+d\,x\right )}{5}\right )+{\cos \left (c+d\,x\right )}^4\,\left (\frac {\sin \left (c+d\,x\right )\,a^4}{3}-\frac {2\,\sin \left (c+d\,x\right )\,a^2\,b^2}{5}+\frac {\sin \left (c+d\,x\right )\,b^4}{35}\right )+{\cos \left (c+d\,x\right )}^6\,\left (\frac {2\,\sin \left (c+d\,x\right )\,a^4}{3}-\frac {4\,\sin \left (c+d\,x\right )\,a^2\,b^2}{5}+\frac {2\,\sin \left (c+d\,x\right )\,b^4}{35}\right )+\frac {2\,a\,b^3\,\cos \left (c+d\,x\right )}{3}}{d\,{\cos \left (c+d\,x\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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