3.27 \(\int \csc ^2(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=42 \[ -\frac {a^2 \cot (c+d x)}{d}+\frac {2 a b \log (\tan (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3516, 43} \[ -\frac {a^2 \cot (c+d x)}{d}+\frac {2 a b \log (\tan (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \csc ^2(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^2}{x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac {b \operatorname {Subst}\left (\int \left (1+\frac {a^2}{x^2}+\frac {2 a}{x}\right ) \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac {a^2 \cot (c+d x)}{d}+\frac {2 a b \log (\tan (c+d x))}{d}+\frac {b^2 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 0.58, size = 91, normalized size = 2.17 \[ -\frac {\cos (c+d x) (a+b \tan (c+d x))^2 \left (a \cos (c+d x) (a \cot (c+d x)+2 b (\log (\cos (c+d x))-\log (\sin (c+d x))))-b^2 \sin (c+d x)\right )}{d (a \cos (c+d x)+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]

[Out]

-((Cos[c + d*x]*(a*Cos[c + d*x]*(a*Cot[c + d*x] + 2*b*(Log[Cos[c + d*x]] - Log[Sin[c + d*x]])) - b^2*Sin[c + d
*x])*(a + b*Tan[c + d*x])^2)/(d*(a*Cos[c + d*x] + b*Sin[c + d*x])^2))

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fricas [B]  time = 0.46, size = 96, normalized size = 2.29 \[ -\frac {a b \cos \left (d x + c\right ) \log \left (\cos \left (d x + c\right )^{2}\right ) \sin \left (d x + c\right ) - a b \cos \left (d x + c\right ) \log \left (-\frac {1}{4} \, \cos \left (d x + c\right )^{2} + \frac {1}{4}\right ) \sin \left (d x + c\right ) + {\left (a^{2} + b^{2}\right )} \cos \left (d x + c\right )^{2} - b^{2}}{d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.87, size = 51, normalized size = 1.21 \[ \frac {2 \, a b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + b^{2} \tan \left (d x + c\right ) - \frac {2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

(2*a*b*log(abs(tan(d*x + c))) + b^2*tan(d*x + c) - (2*a*b*tan(d*x + c) + a^2)/tan(d*x + c))/d

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maple [A]  time = 0.47, size = 43, normalized size = 1.02 \[ -\frac {a^{2} \cot \left (d x +c \right )}{d}+\frac {2 a b \ln \left (\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2*(a+b*tan(d*x+c))^2,x)

[Out]

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

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maxima [A]  time = 0.47, size = 39, normalized size = 0.93 \[ \frac {2 \, a b \log \left (\tan \left (d x + c\right )\right ) + b^{2} \tan \left (d x + c\right ) - \frac {a^{2}}{\tan \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

(2*a*b*log(tan(d*x + c)) + b^2*tan(d*x + c) - a^2/tan(d*x + c))/d

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mupad [B]  time = 3.64, size = 44, normalized size = 1.05 \[ \frac {b^2\,\mathrm {tan}\left (c+d\,x\right )}{d}-\frac {a^2}{d\,\mathrm {tan}\left (c+d\,x\right )}+\frac {2\,a\,b\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^2/sin(c + d*x)^2,x)

[Out]

(b^2*tan(c + d*x))/d - a^2/(d*tan(c + d*x)) + (2*a*b*log(tan(c + d*x)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**2*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*csc(c + d*x)**2, x)

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