3.274 \(\int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx\)

Optimal. Leaf size=88 \[ -\frac {i d^2 \text {Li}_2\left (e^{4 i (a+b x)}\right )}{2 b^3}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac {2 i (c+d x)^2}{b} \]

[Out]

-2*I*(d*x+c)^2/b-2*(d*x+c)^2*cot(2*b*x+2*a)/b+2*d*(d*x+c)*ln(1-exp(4*I*(b*x+a)))/b^2-1/2*I*d^2*polylog(2,exp(4
*I*(b*x+a)))/b^3

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Rubi [A]  time = 0.19, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4419, 4184, 3717, 2190, 2279, 2391} \[ -\frac {i d^2 \text {PolyLog}\left (2,e^{4 i (a+b x)}\right )}{2 b^3}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac {2 i (c+d x)^2}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*(c + d*x)^2)/b - (2*(c + d*x)^2*Cot[2*a + 2*b*x])/b + (2*d*(c + d*x)*Log[1 - E^((4*I)*(a + b*x))])/b^2
- ((I/2)*d^2*PolyLog[2, E^((4*I)*(a + b*x))])/b^3

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[m]

Rubi steps

\begin {align*} \int (c+d x)^2 \csc ^2(a+b x) \sec ^2(a+b x) \, dx &=4 \int (c+d x)^2 \csc ^2(2 a+2 b x) \, dx\\ &=-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac {(4 d) \int (c+d x) \cot (2 a+2 b x) \, dx}{b}\\ &=-\frac {2 i (c+d x)^2}{b}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}-\frac {(8 i d) \int \frac {e^{2 i (2 a+2 b x)} (c+d x)}{1-e^{2 i (2 a+2 b x)}} \, dx}{b}\\ &=-\frac {2 i (c+d x)^2}{b}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac {\left (2 d^2\right ) \int \log \left (1-e^{2 i (2 a+2 b x)}\right ) \, dx}{b^2}\\ &=-\frac {2 i (c+d x)^2}{b}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}+\frac {\left (i d^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i (2 a+2 b x)}\right )}{2 b^3}\\ &=-\frac {2 i (c+d x)^2}{b}-\frac {2 (c+d x)^2 \cot (2 a+2 b x)}{b}+\frac {2 d (c+d x) \log \left (1-e^{4 i (a+b x)}\right )}{b^2}-\frac {i d^2 \text {Li}_2\left (e^{4 i (a+b x)}\right )}{2 b^3}\\ \end {align*}

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Mathematica [B]  time = 1.72, size = 277, normalized size = 3.15 \[ \frac {2 b^2 \csc (2 a) \sin (2 b x) (c+d x)^2 \csc (2 (a+b x))-\frac {i e^{4 i a} \left (4 e^{-4 i a} b^2 (c+d x)^2+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1-e^{-i (a+b x)}\right )+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1+e^{-i (a+b x)}\right )+2 i \left (1-e^{-4 i a}\right ) b d (c+d x) \log \left (1+e^{-2 i (a+b x)}\right )-2 \left (1-e^{-4 i a}\right ) d^2 \text {Li}_2\left (-e^{-i (a+b x)}\right )-2 \left (1-e^{-4 i a}\right ) d^2 \text {Li}_2\left (e^{-i (a+b x)}\right )-\left (1-e^{-4 i a}\right ) d^2 \text {Li}_2\left (-e^{-2 i (a+b x)}\right )\right )}{-1+e^{4 i a}}}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*E^((4*I)*a)*((4*b^2*(c + d*x)^2)/E^((4*I)*a) + (2*I)*b*d*(1 - E^((-4*I)*a))*(c + d*x)*Log[1 - E^((-I)*(
a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)*a))*(c + d*x)*Log[1 + E^((-I)*(a + b*x))] + (2*I)*b*d*(1 - E^((-4*I)*a))
*(c + d*x)*Log[1 + E^((-2*I)*(a + b*x))] - 2*d^2*(1 - E^((-4*I)*a))*PolyLog[2, -E^((-I)*(a + b*x))] - 2*d^2*(1
- E^((-4*I)*a))*PolyLog[2, E^((-I)*(a + b*x))] - d^2*(1 - E^((-4*I)*a))*PolyLog[2, -E^((-2*I)*(a + b*x))]))/(
-1 + E^((4*I)*a)) + 2*b^2*(c + d*x)^2*Csc[2*a]*Csc[2*(a + b*x)]*Sin[2*b*x])/b^3

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fricas [B]  time = 0.56, size = 950, normalized size = 10.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d^2*x^2 + 2*b^2*c*d*x - I*d^2*cos(b*x + a)*dilog(cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) + I*d^2*cos(
b*x + a)*dilog(cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x
+ a))*sin(b*x + a) - I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + a))*sin(b*x + a) - I*d^2*cos(b*x + a
)*dilog(-I*cos(b*x + a) + sin(b*x + a))*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)
)*sin(b*x + a) + I*d^2*cos(b*x + a)*dilog(-cos(b*x + a) + I*sin(b*x + a))*sin(b*x + a) - I*d^2*cos(b*x + a)*di
log(-cos(b*x + a) - I*sin(b*x + a))*sin(b*x + a) + b^2*c^2 + (b*d^2*x + b*c*d)*cos(b*x + a)*log(cos(b*x + a) +
I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*
x + a) + (b*d^2*x + b*c*d)*cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*
cos(b*x + a)*log(cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*
x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1
)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) + sin(b*x + a) + 1)*sin(b*x + a) + (b*d^2*
x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*lo
g(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-1/2*cos(b*x +
a) - 1/2*I*sin(b*x + a) + 1/2)*sin(b*x + a) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x +
a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I)*sin(b*x + a) + (b*
d^2*x + a*d^2)*cos(b*x + a)*log(-cos(b*x + a) - I*sin(b*x + a) + 1)*sin(b*x + a) + (b*c*d - a*d^2)*cos(b*x + a
)*log(-cos(b*x + a) - I*sin(b*x + a) + I)*sin(b*x + a) - 2*(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2)*cos(b*x + a)^
2)/(b^3*cos(b*x + a)*sin(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 351, normalized size = 3.99 \[ -\frac {4 i \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{b \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 d c \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {8 d c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {4 i d^{2} a^{2}}{b^{3}}+\frac {2 d^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}-\frac {4 i d^{2} x^{2}}{b}-\frac {i d^{2} \polylog \left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 d^{2} \ln \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i d^{2} \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {8 i d^{2} a x}{b^{2}}-\frac {2 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {8 d^{2} a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*I/b^3*d^2*polylog(2,-exp(I*(b*x+a)))+2/b^2*d*c*ln(exp(I*(b*x+a))-1)+2*d/b^2*c*ln(1+exp(2*I*(b*x+a)))+2/b^2*
d*c*ln(exp(I*(b*x+a))+1)-8/b^2*d*c*ln(exp(I*(b*x+a)))-2*I*d^2*polylog(2,exp(I*(b*x+a)))/b^3+2/b^2*d^2*ln(exp(I
*(b*x+a))+1)*x-4*I*(d^2*x^2+2*c*d*x+c^2)/b/(1+exp(2*I*(b*x+a)))/(exp(2*I*(b*x+a))-1)-4*I*d^2/b^3*a^2+2*d^2/b^2
*ln(1+exp(2*I*(b*x+a)))*x+2/b^2*d^2*ln(1-exp(I*(b*x+a)))*x+2/b^3*d^2*ln(1-exp(I*(b*x+a)))*a-I*d^2*polylog(2,-e
xp(2*I*(b*x+a)))/b^3-4*I*d^2/b*x^2-8*I*d^2/b^2*a*x-2/b^3*d^2*a*ln(exp(I*(b*x+a))-1)+8/b^3*d^2*a*ln(exp(I*(b*x+
a)))

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maxima [B]  time = 0.69, size = 777, normalized size = 8.83 \[ -\frac {4 \, b^{2} c^{2} + {\left (2 \, b d^{2} x + 2 \, b c d - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + {\left (2 \, b d^{2} x + 2 \, b c d - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) - {\left (2 i \, b d^{2} x + 2 i \, b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - {\left (2 \, b c d \cos \left (4 \, b x + 4 \, a\right ) + 2 i \, b c d \sin \left (4 \, b x + 4 \, a\right ) - 2 \, b c d\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + {\left (2 \, b d^{2} x \cos \left (4 \, b x + 4 \, a\right ) + 2 i \, b d^{2} x \sin \left (4 \, b x + 4 \, a\right ) - 2 \, b d^{2} x\right )} \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 4 \, {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 2 \, {\left (d^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, d^{2} \sin \left (4 \, b x + 4 \, a\right ) - d^{2}\right )} {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (i \, b d^{2} x + i \, b c d + {\left (-i \, b d^{2} x - i \, b c d\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (b d^{2} x + b c d\right )} \sin \left (4 \, b x + 4 \, a\right )\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (-4 i \, b^{2} d^{2} x^{2} - 8 i \, b^{2} c d x\right )} \sin \left (4 \, b x + 4 \, a\right )}{-i \, b^{3} \cos \left (4 \, b x + 4 \, a\right ) + b^{3} \sin \left (4 \, b x + 4 \, a\right ) + i \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*b^2*c^2 + (2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(4*b*x + 4*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(4*b*
x + 4*a))*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) + (2*b*d^2*x + 2*b*c*d - 2*(b*d^2*x + b*c*d)*cos(4*b
*x + 4*a) - (2*I*b*d^2*x + 2*I*b*c*d)*sin(4*b*x + 4*a))*arctan2(sin(b*x + a), cos(b*x + a) + 1) - (2*b*c*d*cos
(4*b*x + 4*a) + 2*I*b*c*d*sin(4*b*x + 4*a) - 2*b*c*d)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (2*b*d^2*x*cos
(4*b*x + 4*a) + 2*I*b*d^2*x*sin(4*b*x + 4*a) - 2*b*d^2*x)*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 4*(b^2*d^
2*x^2 + 2*b^2*c*d*x)*cos(4*b*x + 4*a) + (d^2*cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(2*I*b*
x + 2*I*a)) + 2*(d^2*cos(4*b*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(-e^(I*b*x + I*a)) + 2*(d^2*cos(4*b
*x + 4*a) + I*d^2*sin(4*b*x + 4*a) - d^2)*dilog(e^(I*b*x + I*a)) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*
d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(
2*b*x + 2*a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b
*x + 4*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (I*b*d^2*x + I*b*c*d + (-I*b*d^2*x - I*
b*c*d)*cos(4*b*x + 4*a) + (b*d^2*x + b*c*d)*sin(4*b*x + 4*a))*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x
+ a) + 1) - (-4*I*b^2*d^2*x^2 - 8*I*b^2*c*d*x)*sin(4*b*x + 4*a))/(-I*b^3*cos(4*b*x + 4*a) + b^3*sin(4*b*x + 4*
a) + I*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,x\right )}^2}{{\cos \left (a+b\,x\right )}^2\,{\sin \left (a+b\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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