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

Optimal. Leaf size=138 \[ \frac{\left (a^2+b^2\right )^2}{3 a^3 b^2 d (a \cot (c+d x)+b)^3}+\frac{\frac{1}{a^3}+\frac{3 a}{b^4}}{d (a \cot (c+d x)+b)}+\frac{\frac{a}{b^3}-\frac{b}{a^3}}{d (a \cot (c+d x)+b)^2}-\frac{4 a \log (\tan (c+d x))}{b^5 d}-\frac{4 a \log (a \cot (c+d x)+b)}{b^5 d}+\frac{\tan (c+d x)}{b^4 d} \]

[Out]

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

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Rubi [A]  time = 0.16002, antiderivative size = 138, 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} \[ \frac{\left (a^2+b^2\right )^2}{3 a^3 b^2 d (a \cot (c+d x)+b)^3}+\frac{\frac{1}{a^3}+\frac{3 a}{b^4}}{d (a \cot (c+d x)+b)}+\frac{\frac{a}{b^3}-\frac{b}{a^3}}{d (a \cot (c+d x)+b)^2}-\frac{4 a \log (\tan (c+d x))}{b^5 d}-\frac{4 a \log (a \cot (c+d x)+b)}{b^5 d}+\frac{\tan (c+d x)}{b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.20459, size = 133, normalized size = 0.96 \[ \frac{-4 \left (a^2+b^2\right ) \left (a^2+3 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)+b^2\right )+6 a (a+b \tan (c+d x)) \left (a^2-4 a (a+b \tan (c+d x))-2 (a+b \tan (c+d x))^2 \log (a+b \tan (c+d x))+b^2\right )+3 b^4 \sec ^4(c+d x)}{3 b^5 d (a+b \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.285, size = 188, normalized size = 1.4 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{{b}^{4}d}}-4\,{\frac{a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{{b}^{5}d}}+2\,{\frac{{a}^{3}}{{b}^{5}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}+2\,{\frac{a}{d{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{2}}}-6\,{\frac{{a}^{2}}{{b}^{5}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}-2\,{\frac{1}{d{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{{a}^{4}}{3\,{b}^{5}d \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{2\,{a}^{2}}{3\,d{b}^{3} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}}-{\frac{1}{3\,db \left ( a+b\tan \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.27017, size = 194, normalized size = 1.41 \begin{align*} -\frac{\frac{13 \, a^{4} + 2 \, a^{2} b^{2} + b^{4} + 6 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )^{2} + 6 \,{\left (5 \, a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )}{b^{8} \tan \left (d x + c\right )^{3} + 3 \, a b^{7} \tan \left (d x + c\right )^{2} + 3 \, a^{2} b^{6} \tan \left (d x + c\right ) + a^{3} b^{5}} + \frac{12 \, a \log \left (b \tan \left (d x + c\right ) + a\right )}{b^{5}} - \frac{3 \, \tan \left (d x + c\right )}{b^{4}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/3*((13*a^4 + 2*a^2*b^2 + b^4 + 6*(3*a^2*b^2 + b^4)*tan(d*x + c)^2 + 6*(5*a^3*b + a*b^3)*tan(d*x + c))/(b^8*
tan(d*x + c)^3 + 3*a*b^7*tan(d*x + c)^2 + 3*a^2*b^6*tan(d*x + c) + a^3*b^5) + 12*a*log(b*tan(d*x + c) + a)/b^5
- 3*tan(d*x + c)/b^4)/d

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Fricas [B]  time = 0.702726, size = 1191, normalized size = 8.63 \begin{align*} \frac{3 \, a^{2} b^{4} + 3 \, b^{6} - 4 \,{\left (9 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - 2 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 6 \,{\left (5 \, a^{4} b^{2} + a^{2} b^{4} - 2 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left ({\left (a^{6} - 2 \, a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left ({\left (3 \, a^{5} b + 2 \, a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \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^{6} - 2 \, a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} + 3 \,{\left (a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} +{\left ({\left (3 \, a^{5} b + 2 \, a^{3} b^{3} - a b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (a^{3} b^{3} + a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2}\right ) + 2 \,{\left (2 \,{\left (3 \, a^{5} b - 7 \, a^{3} b^{3} - 6 \, a b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (11 \, a^{3} b^{3} + 9 \, a b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left ({\left (a^{5} b^{5} - 2 \, a^{3} b^{7} - 3 \, a b^{9}\right )} d \cos \left (d x + c\right )^{4} + 3 \,{\left (a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right )^{2} +{\left ({\left (3 \, a^{4} b^{6} + 2 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{3} +{\left (a^{2} b^{8} + b^{10}\right )} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/3*(3*a^2*b^4 + 3*b^6 - 4*(9*a^4*b^2 + 3*a^2*b^4 - 2*b^6)*cos(d*x + c)^4 + 6*(5*a^4*b^2 + a^2*b^4 - 2*b^6)*co
s(d*x + c)^2 - 6*((a^6 - 2*a^4*b^2 - 3*a^2*b^4)*cos(d*x + c)^4 + 3*(a^4*b^2 + a^2*b^4)*cos(d*x + c)^2 + ((3*a^
5*b + 2*a^3*b^3 - a*b^5)*cos(d*x + c)^3 + (a^3*b^3 + a*b^5)*cos(d*x + c))*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^6 - 2*a^4*b^2 - 3*a^2*b^4)*cos(d*x + c)^4 + 3*(a^4*b
^2 + a^2*b^4)*cos(d*x + c)^2 + ((3*a^5*b + 2*a^3*b^3 - a*b^5)*cos(d*x + c)^3 + (a^3*b^3 + a*b^5)*cos(d*x + c))
*sin(d*x + c))*log(cos(d*x + c)^2) + 2*(2*(3*a^5*b - 7*a^3*b^3 - 6*a*b^5)*cos(d*x + c)^3 + (11*a^3*b^3 + 9*a*b
^5)*cos(d*x + c))*sin(d*x + c))/((a^5*b^5 - 2*a^3*b^7 - 3*a*b^9)*d*cos(d*x + c)^4 + 3*(a^3*b^7 + a*b^9)*d*cos(
d*x + c)^2 + ((3*a^4*b^6 + 2*a^2*b^8 - b^10)*d*cos(d*x + c)^3 + (a^2*b^8 + b^10)*d*cos(d*x + c))*sin(d*x + c))

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

[Out]

Timed out

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Giac [A]  time = 1.19932, size = 186, normalized size = 1.35 \begin{align*} -\frac{\frac{12 \, a \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{b^{5}} - \frac{3 \, \tan \left (d x + c\right )}{b^{4}} - \frac{22 \, a b^{3} \tan \left (d x + c\right )^{3} + 48 \, a^{2} b^{2} \tan \left (d x + c\right )^{2} - 6 \, b^{4} \tan \left (d x + c\right )^{2} + 36 \, a^{3} b \tan \left (d x + c\right ) - 6 \, a b^{3} \tan \left (d x + c\right ) + 9 \, a^{4} - 2 \, a^{2} b^{2} - b^{4}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3} b^{5}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2/(a*cos(d*x+c)+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/3*(12*a*log(abs(b*tan(d*x + c) + a))/b^5 - 3*tan(d*x + c)/b^4 - (22*a*b^3*tan(d*x + c)^3 + 48*a^2*b^2*tan(d
*x + c)^2 - 6*b^4*tan(d*x + c)^2 + 36*a^3*b*tan(d*x + c) - 6*a*b^3*tan(d*x + c) + 9*a^4 - 2*a^2*b^2 - b^4)/((b
*tan(d*x + c) + a)^3*b^5))/d