3.130 \(\int \frac {\sec ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=141 \[ -\frac {4 a \left (a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}-\frac {4 a \left (a^2+b^2\right ) \log (a \cot (c+d x)+b)}{b^5 d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2+b^2\right )^2}{a b^4 d (a \cot (c+d x)+b)}-\frac {a \tan ^2(c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d} \]

[Out]

(a^2+b^2)^2/a/b^4/d/(b+a*cot(d*x+c))-4*a*(a^2+b^2)*ln(b+a*cot(d*x+c))/b^5/d-4*a*(a^2+b^2)*ln(tan(d*x+c))/b^5/d
+(3*a^2+2*b^2)*tan(d*x+c)/b^4/d-a*tan(d*x+c)^2/b^3/d+1/3*tan(d*x+c)^3/b^2/d

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Rubi [A]  time = 0.15, antiderivative size = 141, 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} \[ \frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}+\frac {\left (a^2+b^2\right )^2}{a b^4 d (a \cot (c+d x)+b)}-\frac {4 a \left (a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}-\frac {4 a \left (a^2+b^2\right ) \log (a \cot (c+d x)+b)}{b^5 d}-\frac {a \tan ^2(c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(a^2 + b^2)^2/(a*b^4*d*(b + a*Cot[c + d*x])) - (4*a*(a^2 + b^2)*Log[b + a*Cot[c + d*x]])/(b^5*d) - (4*a*(a^2 +
b^2)*Log[Tan[c + d*x]])/(b^5*d) + ((3*a^2 + 2*b^2)*Tan[c + d*x])/(b^4*d) - (a*Tan[c + d*x]^2)/(b^3*d) + Tan[c
+ d*x]^3/(3*b^2*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 \frac {\sec ^4(c+d x)}{(a \cos (c+d x)+b \sin (c+d x))^2} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^2}{x^4 (b+a x)^2} \, dx,x,\cot (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {1}{b^2 x^4}-\frac {2 a}{b^3 x^3}+\frac {3 a^2+2 b^2}{b^4 x^2}-\frac {4 a \left (a^2+b^2\right )}{b^5 x}+\frac {\left (a^2+b^2\right )^2}{b^4 (b+a x)^2}+\frac {4 a^2 \left (a^2+b^2\right )}{b^5 (b+a x)}\right ) \, dx,x,\cot (c+d x)\right )}{d}\\ &=\frac {\left (a^2+b^2\right )^2}{a b^4 d (b+a \cot (c+d x))}-\frac {4 a \left (a^2+b^2\right ) \log (b+a \cot (c+d x))}{b^5 d}-\frac {4 a \left (a^2+b^2\right ) \log (\tan (c+d x))}{b^5 d}+\frac {\left (3 a^2+2 b^2\right ) \tan (c+d x)}{b^4 d}-\frac {a \tan ^2(c+d x)}{b^3 d}+\frac {\tan ^3(c+d x)}{3 b^2 d}\\ \end {align*}

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Mathematica [A]  time = 2.78, size = 122, normalized size = 0.87 \[ \frac {4 b \left (2 a^2+b^2\right ) \tan (c+d x)+\frac {b^4 \sec ^4(c+d x)-4 \left (a^2+b^2\right ) \left (3 a^2 \log (a+b \tan (c+d x))+a^2+3 a b \tan (c+d x) \log (a+b \tan (c+d x))+b^2\right )}{a+b \tan (c+d x)}-2 a b^2 \tan ^2(c+d x)}{3 b^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a*Cos[c + d*x] + b*Sin[c + d*x])^2,x]

[Out]

(4*b*(2*a^2 + b^2)*Tan[c + d*x] - 2*a*b^2*Tan[c + d*x]^2 + (b^4*Sec[c + d*x]^4 - 4*(a^2 + b^2)*(a^2 + b^2 + 3*
a^2*Log[a + b*Tan[c + d*x]] + 3*a*b*Log[a + b*Tan[c + d*x]]*Tan[c + d*x]))/(a + b*Tan[c + d*x]))/(3*b^5*d)

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fricas [B]  time = 0.46, size = 281, normalized size = 1.99 \[ -\frac {4 \, {\left (3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{4} - b^{4} - 2 \, {\left (3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) - 6 \, {\left ({\left (a^{4} + a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + {\left (a^{3} b + a b^{3}\right )} \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) + 2 \, {\left (a b^{3} \cos \left (d x + c\right ) - 2 \, {\left (3 \, a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{3 \, {\left (a b^{5} d \cos \left (d x + c\right )^{4} + b^{6} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right )\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-1/3*(4*(3*a^2*b^2 + 2*b^4)*cos(d*x + c)^4 - b^4 - 2*(3*a^2*b^2 + 2*b^4)*cos(d*x + c)^2 + 6*((a^4 + a^2*b^2)*c
os(d*x + c)^4 + (a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a^2 - b^2)
*cos(d*x + c)^2 + b^2) - 6*((a^4 + a^2*b^2)*cos(d*x + c)^4 + (a^3*b + a*b^3)*cos(d*x + c)^3*sin(d*x + c))*log(
cos(d*x + c)^2) + 2*(a*b^3*cos(d*x + c) - 2*(3*a^3*b + 2*a*b^3)*cos(d*x + c)^3)*sin(d*x + c))/(a*b^5*d*cos(d*x
+ c)^4 + b^6*d*cos(d*x + c)^3*sin(d*x + c))

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giac [A]  time = 0.25, size = 149, normalized size = 1.06 \[ -\frac {\frac {12 \, {\left (a^{3} + a b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5}} - \frac {b^{4} \tan \left (d x + c\right )^{3} - 3 \, a b^{3} \tan \left (d x + c\right )^{2} + 9 \, a^{2} b^{2} \tan \left (d x + c\right ) + 6 \, b^{4} \tan \left (d x + c\right )}{b^{6}} - \frac {3 \, {\left (4 \, a^{3} b \tan \left (d x + c\right ) + 4 \, a b^{3} \tan \left (d x + c\right ) + 3 \, a^{4} + 2 \, a^{2} b^{2} - b^{4}\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )} b^{5}}}{3 \, d} \]

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))^2,x, algorithm="giac")

[Out]

-1/3*(12*(a^3 + a*b^2)*log(abs(b*tan(d*x + c) + a))/b^5 - (b^4*tan(d*x + c)^3 - 3*a*b^3*tan(d*x + c)^2 + 9*a^2
*b^2*tan(d*x + c) + 6*b^4*tan(d*x + c))/b^6 - 3*(4*a^3*b*tan(d*x + c) + 4*a*b^3*tan(d*x + c) + 3*a^4 + 2*a^2*b
^2 - b^4)/((b*tan(d*x + c) + a)*b^5))/d

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maple [A]  time = 0.34, size = 174, normalized size = 1.23 \[ \frac {\tan ^{3}\left (d x +c \right )}{3 b^{2} d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{b^{3} d}+\frac {3 a^{2} \tan \left (d x +c \right )}{d \,b^{4}}+\frac {2 \tan \left (d x +c \right )}{b^{2} d}-\frac {4 a^{3} \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,b^{5}}-\frac {4 a \ln \left (a +b \tan \left (d x +c \right )\right )}{d \,b^{3}}-\frac {a^{4}}{d \,b^{5} \left (a +b \tan \left (d x +c \right )\right )}-\frac {2 a^{2}}{d \,b^{3} \left (a +b \tan \left (d x +c \right )\right )}-\frac {1}{d b \left (a +b \tan \left (d x +c \right )\right )} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.33, size = 115, normalized size = 0.82 \[ -\frac {\frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}}{b^{6} \tan \left (d x + c\right ) + a b^{5}} - \frac {b^{2} \tan \left (d x + c\right )^{3} - 3 \, a b \tan \left (d x + c\right )^{2} + 3 \, {\left (3 \, a^{2} + 2 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{4}} + \frac {12 \, {\left (a^{3} + a b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}}}{3 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

-1/3*(3*(a^4 + 2*a^2*b^2 + b^4)/(b^6*tan(d*x + c) + a*b^5) - (b^2*tan(d*x + c)^3 - 3*a*b*tan(d*x + c)^2 + 3*(3
*a^2 + 2*b^2)*tan(d*x + c))/b^4 + 12*(a^3 + a*b^2)*log(b*tan(d*x + c) + a)/b^5)/d

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mupad [B]  time = 4.20, size = 1132, normalized size = 8.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a*cos(c + d*x) + b*sin(c + d*x))^2),x)

[Out]

((8*tan(c/2 + (d*x)/2)^2*(a^2 + b^2))/b^3 + (8*tan(c/2 + (d*x)/2)^6*(a^2 + b^2))/b^3 - (16*tan(c/2 + (d*x)/2)^
4*(3*a^2 + 2*b^2))/(3*b^3) - (2*tan(c/2 + (d*x)/2)^7*(4*a^4 + b^4 + 4*a^2*b^2))/(a*b^4) - (2*tan(c/2 + (d*x)/2
)^3*(36*a^4 + 9*b^4 + 44*a^2*b^2))/(3*a*b^4) + (2*tan(c/2 + (d*x)/2)^5*(36*a^4 + 9*b^4 + 44*a^2*b^2))/(3*a*b^4
) + (2*tan(c/2 + (d*x)/2)*(4*a^4 + b^4 + 4*a^2*b^2))/(a*b^4))/(d*(a + 2*b*tan(c/2 + (d*x)/2) - 4*a*tan(c/2 + (
d*x)/2)^2 + 6*a*tan(c/2 + (d*x)/2)^4 - 4*a*tan(c/2 + (d*x)/2)^6 + a*tan(c/2 + (d*x)/2)^8 - 6*b*tan(c/2 + (d*x)
/2)^3 + 6*b*tan(c/2 + (d*x)/2)^5 - 2*b*tan(c/2 + (d*x)/2)^7)) + (a*atan(((a*(a^2 + b^2)*((16*tan(c/2 + (d*x)/2
)*(4*a^5 + 4*a^3*b^2))/b^4 - (4*(8*a^2*b^7 + 8*a^4*b^5))/b^8 + (4*tan(c/2 + (d*x)/2)^2*(8*a^2*b^7 + 8*a^4*b^5)
)/b^8 + (4*a*(a^2 + b^2)*((4*(a*b^10 + 4*a^3*b^8))/b^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 +
16*a^2*b*tan(c/2 + (d*x)/2)))/b^5)*4i)/b^5 - (a*(a^2 + b^2)*((4*(8*a^2*b^7 + 8*a^4*b^5))/b^8 - (16*tan(c/2 +
(d*x)/2)*(4*a^5 + 4*a^3*b^2))/b^4 - (4*tan(c/2 + (d*x)/2)^2*(8*a^2*b^7 + 8*a^4*b^5))/b^8 + (4*a*(a^2 + b^2)*((
4*(a*b^10 + 4*a^3*b^8))/b^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2
)))/b^5)*4i)/b^5)/((8*(16*a^7 + 16*a^3*b^4 + 32*a^5*b^2))/b^8 + (8*tan(c/2 + (d*x)/2)^2*(16*a^7 + 16*a^3*b^4 +
32*a^5*b^2))/b^8 + (4*a*(a^2 + b^2)*((16*tan(c/2 + (d*x)/2)*(4*a^5 + 4*a^3*b^2))/b^4 - (4*(8*a^2*b^7 + 8*a^4*
b^5))/b^8 + (4*tan(c/2 + (d*x)/2)^2*(8*a^2*b^7 + 8*a^4*b^5))/b^8 + (4*a*(a^2 + b^2)*((4*(a*b^10 + 4*a^3*b^8))/
b^8 - (4*tan(c/2 + (d*x)/2)^2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5))/b^5 + (4*a*(a^
2 + b^2)*((4*(8*a^2*b^7 + 8*a^4*b^5))/b^8 - (16*tan(c/2 + (d*x)/2)*(4*a^5 + 4*a^3*b^2))/b^4 - (4*tan(c/2 + (d*
x)/2)^2*(8*a^2*b^7 + 8*a^4*b^5))/b^8 + (4*a*(a^2 + b^2)*((4*(a*b^10 + 4*a^3*b^8))/b^8 - (4*tan(c/2 + (d*x)/2)^
2*(3*a*b^10 + 4*a^3*b^8))/b^8 + 16*a^2*b*tan(c/2 + (d*x)/2)))/b^5))/b^5))*(a^2 + b^2)*8i)/(b^5*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a \cos {\left (c + d x \right )} + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]

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))**2,x)

[Out]

Integral(sec(c + d*x)**4/(a*cos(c + d*x) + b*sin(c + d*x))**2, x)

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