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

Optimal. Leaf size=103 \[ \frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (-6 a^2 b^2+a^4+b^4\right )+\frac{b (a+b \tan (c+d x))^3}{3 d}+\frac{a b (a+b \tan (c+d x))^2}{d} \]

[Out]

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*
b*(a + b*Tan[c + d*x])^2)/d + (b*(a + b*Tan[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.156476, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3086, 3482, 3528, 3525, 3475} \[ \frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+x \left (-6 a^2 b^2+a^4+b^4\right )+\frac{b (a+b \tan (c+d x))^3}{3 d}+\frac{a b (a+b \tan (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4 - 6*a^2*b^2 + b^4)*x - (4*a*b*(a^2 - b^2)*Log[Cos[c + d*x]])/d + (b^2*(3*a^2 - b^2)*Tan[c + d*x])/d + (a*
b*(a + b*Tan[c + d*x])^2)/d + (b*(a + b*Tan[c + d*x])^3)/(3*d)

Rule 3086

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

Rule 3482

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Int[(a^2 - b^2 + 2*a*b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n - 2), x] /; FreeQ[{a, b, c, d}, x] && NeQ
[a^2 + b^2, 0] && GtQ[n, 1]

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3525

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[(b*d*Tan[e + f*x])/f, x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a \cos (c+d x)+b \sin (c+d x))^4 \, dx &=\int (a+b \tan (c+d x))^4 \, dx\\ &=\frac{b (a+b \tan (c+d x))^3}{3 d}+\int (a+b \tan (c+d x))^2 \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}+\int (a+b \tan (c+d x)) \left (a \left (a^2-3 b^2\right )+b \left (3 a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x+\frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}+\left (4 a b \left (a^2-b^2\right )\right ) \int \tan (c+d x) \, dx\\ &=\left (a^4-6 a^2 b^2+b^4\right ) x-\frac{4 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac{b^2 \left (3 a^2-b^2\right ) \tan (c+d x)}{d}+\frac{a b (a+b \tan (c+d x))^2}{d}+\frac{b (a+b \tan (c+d x))^3}{3 d}\\ \end{align*}

Mathematica [C]  time = 0.39153, size = 105, normalized size = 1.02 \[ \frac{-6 b^2 \left (b^2-6 a^2\right ) \tan (c+d x)+12 a b^3 \tan ^2(c+d x)+3 i (a-i b)^4 \log (\tan (c+d x)+i)-3 i (a+i b)^4 \log (-\tan (c+d x)+i)+2 b^4 \tan ^3(c+d x)}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*(a + I*b)^4*Log[I - Tan[c + d*x]] + (3*I)*(a - I*b)^4*Log[I + Tan[c + d*x]] - 6*b^2*(-6*a^2 + b^2)*Tan
[c + d*x] + 12*a*b^3*Tan[c + d*x]^2 + 2*b^4*Tan[c + d*x]^3)/(6*d)

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Maple [A]  time = 0.14, size = 145, normalized size = 1.4 \begin{align*}{a}^{4}x+{\frac{{a}^{4}c}{d}}-4\,{\frac{{a}^{3}b\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-6\,{a}^{2}{b}^{2}x+6\,{\frac{\tan \left ( dx+c \right ){a}^{2}{b}^{2}}{d}}-6\,{\frac{{a}^{2}{b}^{2}c}{d}}+2\,{\frac{a{b}^{3} \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{d}}+4\,{\frac{a{b}^{3}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{4} \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{b}^{4}\tan \left ( dx+c \right ) }{d}}+{b}^{4}x+{\frac{{b}^{4}c}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^4*x+1/d*a^4*c-4/d*a^3*b*ln(cos(d*x+c))-6*a^2*b^2*x+6/d*tan(d*x+c)*a^2*b^2-6/d*a^2*b^2*c+2/d*a*b^3*tan(d*x+c)
^2+4/d*a*b^3*ln(cos(d*x+c))+1/3/d*b^4*tan(d*x+c)^3-b^4*tan(d*x+c)/d+b^4*x+1/d*b^4*c

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Maxima [A]  time = 1.83705, size = 157, normalized size = 1.52 \begin{align*} \frac{3 \,{\left (d x + c\right )} a^{4} - 18 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} b^{2} +{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} b^{4} - 6 \, a b^{3}{\left (\frac{1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )} - 6 \, a^{3} b \log \left (-\sin \left (d x + c\right )^{2} + 1\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(d*x + c)*a^4 - 18*(d*x + c - tan(d*x + c))*a^2*b^2 + (tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*b
^4 - 6*a*b^3*(1/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)^2 - 1)) - 6*a^3*b*log(-sin(d*x + c)^2 + 1))/d

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Fricas [A]  time = 0.514935, size = 282, normalized size = 2.74 \begin{align*} \frac{3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right )^{3} + 6 \, a b^{3} \cos \left (d x + c\right ) - 12 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) +{\left (b^{4} + 2 \,{\left (9 \, a^{2} b^{2} - 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(a^4 - 6*a^2*b^2 + b^4)*d*x*cos(d*x + c)^3 + 6*a*b^3*cos(d*x + c) - 12*(a^3*b - a*b^3)*cos(d*x + c)^3*l
og(-cos(d*x + c)) + (b^4 + 2*(9*a^2*b^2 - 2*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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

[Out]

Timed out

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Giac [A]  time = 1.19885, size = 140, normalized size = 1.36 \begin{align*} \frac{b^{4} \tan \left (d x + c\right )^{3} + 6 \, a b^{3} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b^{2} \tan \left (d x + c\right ) - 3 \, b^{4} \tan \left (d x + c\right ) + 3 \,{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )} + 6 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b^4*tan(d*x + c)^3 + 6*a*b^3*tan(d*x + c)^2 + 18*a^2*b^2*tan(d*x + c) - 3*b^4*tan(d*x + c) + 3*(a^4 - 6*a
^2*b^2 + b^4)*(d*x + c) + 6*(a^3*b - a*b^3)*log(tan(d*x + c)^2 + 1))/d