3.261 \(\int (c+d x)^2 \sin (a+b x) \tan ^2(a+b x) \, dx\)

Optimal. Leaf size=145 \[ -\frac {2 i d^2 \text {Li}_2\left (-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \text {Li}_2\left (i e^{i (a+b x)}\right )}{b^3}-\frac {2 d^2 \cos (a+b x)}{b^3}-\frac {2 d (c+d x) \sin (a+b x)}{b^2}+\frac {4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}+\frac {(c+d x)^2 \cos (a+b x)}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b} \]

[Out]

4*I*d*(d*x+c)*arctan(exp(I*(b*x+a)))/b^2-2*d^2*cos(b*x+a)/b^3+(d*x+c)^2*cos(b*x+a)/b-2*I*d^2*polylog(2,-I*exp(
I*(b*x+a)))/b^3+2*I*d^2*polylog(2,I*exp(I*(b*x+a)))/b^3+(d*x+c)^2*sec(b*x+a)/b-2*d*(d*x+c)*sin(b*x+a)/b^2

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Rubi [A]  time = 0.13, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4407, 3296, 2638, 4409, 4181, 2279, 2391} \[ -\frac {2 i d^2 \text {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}+\frac {2 i d^2 \text {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {2 d (c+d x) \sin (a+b x)}{b^2}+\frac {4 i d (c+d x) \tan ^{-1}\left (e^{i (a+b x)}\right )}{b^2}-\frac {2 d^2 \cos (a+b x)}{b^3}+\frac {(c+d x)^2 \cos (a+b x)}{b}+\frac {(c+d x)^2 \sec (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*I)*d*(c + d*x)*ArcTan[E^(I*(a + b*x))])/b^2 - (2*d^2*Cos[a + b*x])/b^3 + ((c + d*x)^2*Cos[a + b*x])/b - ((
2*I)*d^2*PolyLog[2, (-I)*E^(I*(a + b*x))])/b^3 + ((2*I)*d^2*PolyLog[2, I*E^(I*(a + b*x))])/b^3 + ((c + d*x)^2*
Sec[a + b*x])/b - (2*d*(c + d*x)*Sin[a + b*x])/b^2

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 2638

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

Rule 3296

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

Rule 4181

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

Rule 4407

Int[((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> -Int[
(c + d*x)^m*Sin[a + b*x]^n*Tan[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Sin[a + b*x]^(n - 2)*Tan[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 4409

Int[((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.)*Tan[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Simp[
((c + d*x)^m*Sec[a + b*x]^n)/(b*n), x] - Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Sec[a + b*x]^n, x], x] /; Fre
eQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rubi steps

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

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Mathematica [B]  time = 3.12, size = 362, normalized size = 2.50 \[ \frac {\cos (b x) \left (\cos (a) \left (b^2 (c+d x)^2-2 d^2\right )-2 b d \sin (a) (c+d x)\right )-\sin (b x) \left (\sin (a) \left (b^2 (c+d x)^2-2 d^2\right )+2 b d \cos (a) (c+d x)\right )+b^2 \sec (a) (c+d x)^2+\frac {b^2 \sin \left (\frac {b x}{2}\right ) (c+d x)^2}{\left (\cos \left (\frac {a}{2}\right )-\sin \left (\frac {a}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (a+b x)\right )-\sin \left (\frac {1}{2} (a+b x)\right )\right )}-\frac {b^2 \sin \left (\frac {b x}{2}\right ) (c+d x)^2}{\left (\sin \left (\frac {a}{2}\right )+\cos \left (\frac {a}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (a+b x)\right )+\cos \left (\frac {1}{2} (a+b x)\right )\right )}-4 b c d \tanh ^{-1}\left (\cos (a) \tan \left (\frac {b x}{2}\right )+\sin (a)\right )+\frac {2 d^2 \csc (a) \left (i \text {Li}_2\left (-e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-i \text {Li}_2\left (e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+\left (b x-\tan ^{-1}(\cot (a))\right ) \left (\log \left (1-e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-\log \left (1+e^{i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )\right )\right )}{\sqrt {\csc ^2(a)}}-4 d^2 \tan ^{-1}(\cot (a)) \tanh ^{-1}\left (\cos (a) \tan \left (\frac {b x}{2}\right )+\sin (a)\right )}{b^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*b*c*d*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] - 4*d^2*ArcTan[Cot[a]]*ArcTanh[Sin[a] + Cos[a]*Tan[(b*x)/2]] +
(2*d^2*Csc[a]*((b*x - ArcTan[Cot[a]])*(Log[1 - E^(I*(b*x - ArcTan[Cot[a]]))] - Log[1 + E^(I*(b*x - ArcTan[Cot
[a]]))]) + I*PolyLog[2, -E^(I*(b*x - ArcTan[Cot[a]]))] - I*PolyLog[2, E^(I*(b*x - ArcTan[Cot[a]]))]))/Sqrt[Csc
[a]^2] + b^2*(c + d*x)^2*Sec[a] + Cos[b*x]*((-2*d^2 + b^2*(c + d*x)^2)*Cos[a] - 2*b*d*(c + d*x)*Sin[a]) - (2*b
*d*(c + d*x)*Cos[a] + (-2*d^2 + b^2*(c + d*x)^2)*Sin[a])*Sin[b*x] + (b^2*(c + d*x)^2*Sin[(b*x)/2])/((Cos[a/2]
- Sin[a/2])*(Cos[(a + b*x)/2] - Sin[(a + b*x)/2])) - (b^2*(c + d*x)^2*Sin[(b*x)/2])/((Cos[a/2] + Sin[a/2])*(Co
s[(a + b*x)/2] + Sin[(a + b*x)/2])))/b^3

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fricas [B]  time = 0.53, size = 511, normalized size = 3.52 \[ \frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) + i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right )\right ) - i \, d^{2} \cos \left (b x + a\right ) {\rm Li}_2\left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right ) + {\left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} - 2 \, d^{2}\right )} \cos \left (b x + a\right )^{2} - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) + \sin \left (b x + a\right ) + 1\right ) + {\left (b d^{2} x + a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-i \, \cos \left (b x + a\right ) - \sin \left (b x + a\right ) + 1\right ) - {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + i\right ) + {\left (b c d - a d^{2}\right )} \cos \left (b x + a\right ) \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + i\right ) - 2 \, {\left (b d^{2} x + b c d\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right )}{b^{3} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 + I*d^2*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a)) + I*d^2*cos(b*x
+ a)*dilog(I*cos(b*x + a) - sin(b*x + a)) - I*d^2*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - I*d^2*
cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) + (b^2*d^2*x^2 + 2*b^2*c*d*x + b^2*c^2 - 2*d^2)*cos(b*x + a
)^2 - (b*c*d - a*d^2)*cos(b*x + a)*log(cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*log(c
os(b*x + a) - I*sin(b*x + a) + I) - (b*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + (b
*d^2*x + a*d^2)*cos(b*x + a)*log(I*cos(b*x + a) - sin(b*x + a) + 1) - (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*co
s(b*x + a) + sin(b*x + a) + 1) + (b*d^2*x + a*d^2)*cos(b*x + a)*log(-I*cos(b*x + a) - sin(b*x + a) + 1) - (b*c
*d - a*d^2)*cos(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + (b*c*d - a*d^2)*cos(b*x + a)*log(-cos(b*x +
a) - I*sin(b*x + a) + I) - 2*(b*d^2*x + b*c*d)*cos(b*x + a)*sin(b*x + a))/(b^3*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 345, normalized size = 2.38 \[ \frac {\left (d^{2} x^{2} b^{2}+2 b^{2} c d x +2 i b \,d^{2} x +b^{2} c^{2}+2 i b c d -2 d^{2}\right ) {\mathrm e}^{i \left (b x +a \right )}}{2 b^{3}}+\frac {\left (d^{2} x^{2} b^{2}+2 b^{2} c d x -2 i b \,d^{2} x +b^{2} c^{2}-2 i b c d -2 d^{2}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{2 b^{3}}+\frac {2 \,{\mathrm e}^{i \left (b x +a \right )} \left (d^{2} x^{2}+2 c d x +c^{2}\right )}{b \left (1+{\mathrm e}^{2 i \left (b x +a \right )}\right )}+\frac {4 i d c \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {2 d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 d^{2} \ln \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {2 d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {2 i d^{2} \dilog \left (1+i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {2 i d^{2} \dilog \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {4 i d^{2} a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(d^2*x^2*b^2+2*b^2*c*d*x+b^2*c^2+2*I*b*d^2*x-2*d^2+2*I*b*c*d)/b^3*exp(I*(b*x+a))+1/2*(d^2*x^2*b^2+2*b^2*c*
d*x+b^2*c^2-2*I*b*d^2*x-2*d^2-2*I*b*c*d)/b^3*exp(-I*(b*x+a))+2*exp(I*(b*x+a))*(d^2*x^2+2*c*d*x+c^2)/b/(1+exp(2
*I*(b*x+a)))+4*I*d/b^2*c*arctan(exp(I*(b*x+a)))+2/b^2*d^2*ln(1+I*exp(I*(b*x+a)))*x+2/b^3*d^2*ln(1+I*exp(I*(b*x
+a)))*a-2/b^2*d^2*ln(1-I*exp(I*(b*x+a)))*x-2/b^3*d^2*ln(1-I*exp(I*(b*x+a)))*a-2*I/b^3*d^2*dilog(1+I*exp(I*(b*x
+a)))+2*I/b^3*d^2*dilog(1-I*exp(I*(b*x+a)))-4*I*d^2/b^3*a*arctan(exp(I*(b*x+a)))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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