3.1.46 \(\int (c+d x) (a+b \tan (e+f x))^2 \, dx\) [46]

Optimal. Leaf size=136 \[ -b^2 c x-\frac {1}{2} b^2 d x^2+\frac {a^2 (c+d x)^2}{2 d}+\frac {i a b (c+d x)^2}{d}-\frac {2 a b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {b^2 d \log (\cos (e+f x))}{f^2}+\frac {i a b d \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f} \]

[Out]

-b^2*c*x-1/2*b^2*d*x^2+1/2*a^2*(d*x+c)^2/d+I*a*b*(d*x+c)^2/d-2*a*b*(d*x+c)*ln(1+exp(2*I*(f*x+e)))/f+b^2*d*ln(c
os(f*x+e))/f^2+I*a*b*d*polylog(2,-exp(2*I*(f*x+e)))/f^2+b^2*(d*x+c)*tan(f*x+e)/f

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Rubi [A]
time = 0.12, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3803, 3800, 2221, 2317, 2438, 3801, 3556} \begin {gather*} \frac {a^2 (c+d x)^2}{2 d}-\frac {2 a b (c+d x) \log \left (1+e^{2 i (e+f x)}\right )}{f}+\frac {i a b (c+d x)^2}{d}+\frac {i a b d \text {Li}_2\left (-e^{2 i (e+f x)}\right )}{f^2}+\frac {b^2 (c+d x) \tan (e+f x)}{f}-b^2 c x+\frac {b^2 d \log (\cos (e+f x))}{f^2}-\frac {1}{2} b^2 d x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*c*x) - (b^2*d*x^2)/2 + (a^2*(c + d*x)^2)/(2*d) + (I*a*b*(c + d*x)^2)/d - (2*a*b*(c + d*x)*Log[1 + E^((2*
I)*(e + f*x))])/f + (b^2*d*Log[Cos[e + f*x]])/f^2 + (I*a*b*d*PolyLog[2, -E^((2*I)*(e + f*x))])/f^2 + (b^2*(c +
d*x)*Tan[e + f*x])/f

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3556

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

Rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 3801

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

Rule 3803

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 2.21, size = 200, normalized size = 1.47 \begin {gather*} \frac {\cos (e+f x) \left (\cos (e+f x) \left (-\left ((e+f x) \left (-2 i a b d (e+f x)+a^2 (d e-2 c f-d f x)+b^2 (-d e+2 c f+d f x)\right )\right )-4 a b d (e+f x) \log \left (1+e^{2 i (e+f x)}\right )+2 b (b d+2 a d e-2 a c f) \log (\cos (e+f x))\right )+2 i a b d \cos (e+f x) \text {PolyLog}\left (2,-e^{2 i (e+f x)}\right )+2 b^2 f (c+d x) \sin (e+f x)\right ) (a+b \tan (e+f x))^2}{2 f^2 (a \cos (e+f x)+b \sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(Cos[e + f*x]*(-((e + f*x)*((-2*I)*a*b*d*(e + f*x) + a^2*(d*e - 2*c*f - d*f*x) + b^2*(-(d*e) + 2
*c*f + d*f*x))) - 4*a*b*d*(e + f*x)*Log[1 + E^((2*I)*(e + f*x))] + 2*b*(b*d + 2*a*d*e - 2*a*c*f)*Log[Cos[e + f
*x]]) + (2*I)*a*b*d*Cos[e + f*x]*PolyLog[2, -E^((2*I)*(e + f*x))] + 2*b^2*f*(c + d*x)*Sin[e + f*x])*(a + b*Tan
[e + f*x])^2)/(2*f^2*(a*Cos[e + f*x] + b*Sin[e + f*x])^2)

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Maple [A]
time = 0.23, size = 238, normalized size = 1.75

method	result	size
risch	\(-\frac {b^{2} d \,x^{2}}{2}+\frac {2 i b^{2} \left (d x +c \right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {a^{2} d \,x^{2}}{2}-b^{2} c x +i a b d \,x^{2}+a^{2} c x +\frac {4 i b a d e x}{f}+\frac {b^{2} d \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f^{2}}-\frac {2 b^{2} d \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b a c \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {4 b a c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}-\frac {4 b a d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-2 i a b c x +\frac {2 i b a d \,e^{2}}{f^{2}}+\frac {i a b d \polylog \left (2, -{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a d x}{f}\)	\(238\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (127) = 254\).
time = 0.58, size = 559, normalized size = 4.11 \begin {gather*} \frac {2 \, {\left (f x + e\right )} a^{2} c + \frac {{\left (f x + e\right )}^{2} a^{2} d}{f} - \frac {2 \, {\left (f x + e\right )} a^{2} d e}{f} + 4 \, a b c \log \left (\sec \left (f x + e\right )\right ) - \frac {4 \, a b d e \log \left (\sec \left (f x + e\right )\right )}{f} + \frac {2 \, {\left ({\left (2 \, a b + i \, b^{2}\right )} {\left (f x + e\right )}^{2} d + 4 \, b^{2} c f - 4 \, b^{2} d e + 2 \, {\left (i \, b^{2} c f - i \, b^{2} d e\right )} {\left (f x + e\right )} - 2 \, {\left (2 \, {\left (f x + e\right )} a b d - b^{2} d + {\left (2 \, {\left (f x + e\right )} a b d - b^{2} d\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left (-2 i \, {\left (f x + e\right )} a b d + i \, b^{2} d\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) + {\left ({\left (2 \, a b + i \, b^{2}\right )} {\left (f x + e\right )}^{2} d + 2 \, {\left (i \, b^{2} c f + b^{2} d {\left (-i \, e - 2\right )}\right )} {\left (f x + e\right )}\right )} \cos \left (2 \, f x + 2 \, e\right ) + 2 \, {\left (a b d \cos \left (2 \, f x + 2 \, e\right ) + i \, a b d \sin \left (2 \, f x + 2 \, e\right ) + a b d\right )} {\rm Li}_2\left (-e^{\left (2 i \, f x + 2 i \, e\right )}\right ) - {\left (-2 i \, {\left (f x + e\right )} a b d + i \, b^{2} d + {\left (-2 i \, {\left (f x + e\right )} a b d + i \, b^{2} d\right )} \cos \left (2 \, f x + 2 \, e\right ) + {\left (2 \, {\left (f x + e\right )} a b d - b^{2} d\right )} \sin \left (2 \, f x + 2 \, e\right )\right )} \log \left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - {\left ({\left (-2 i \, a b + b^{2}\right )} {\left (f x + e\right )}^{2} d + 2 \, {\left (b^{2} c f - b^{2} d {\left (e - 2 i\right )}\right )} {\left (f x + e\right )}\right )} \sin \left (2 \, f x + 2 \, e\right )\right )}}{-2 i \, f \cos \left (2 \, f x + 2 \, e\right ) + 2 \, f \sin \left (2 \, f x + 2 \, e\right ) - 2 i \, f}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(f*x + e)*a^2*c + (f*x + e)^2*a^2*d/f - 2*(f*x + e)*a^2*d*e/f + 4*a*b*c*log(sec(f*x + e)) - 4*a*b*d*e*l
og(sec(f*x + e))/f + 2*((2*a*b + I*b^2)*(f*x + e)^2*d + 4*b^2*c*f - 4*b^2*d*e + 2*(I*b^2*c*f - I*b^2*d*e)*(f*x
+ e) - 2*(2*(f*x + e)*a*b*d - b^2*d + (2*(f*x + e)*a*b*d - b^2*d)*cos(2*f*x + 2*e) - (-2*I*(f*x + e)*a*b*d +
I*b^2*d)*sin(2*f*x + 2*e))*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) + ((2*a*b + I*b^2)*(f*x + e)^2*d +
2*(I*b^2*c*f + b^2*d*(-I*e - 2))*(f*x + e))*cos(2*f*x + 2*e) + 2*(a*b*d*cos(2*f*x + 2*e) + I*a*b*d*sin(2*f*x +
2*e) + a*b*d)*dilog(-e^(2*I*f*x + 2*I*e)) - (-2*I*(f*x + e)*a*b*d + I*b^2*d + (-2*I*(f*x + e)*a*b*d + I*b^2*d
)*cos(2*f*x + 2*e) + (2*(f*x + e)*a*b*d - b^2*d)*sin(2*f*x + 2*e))*log(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2
+ 2*cos(2*f*x + 2*e) + 1) - ((-2*I*a*b + b^2)*(f*x + e)^2*d + 2*(b^2*c*f - b^2*d*(e - 2*I))*(f*x + e))*sin(2*
f*x + 2*e))/(-2*I*f*cos(2*f*x + 2*e) + 2*f*sin(2*f*x + 2*e) - 2*I*f))/f

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Fricas [A]
time = 0.37, size = 228, normalized size = 1.68 \begin {gather*} \frac {{\left (a^{2} - b^{2}\right )} d f^{2} x^{2} + 2 \, {\left (a^{2} - b^{2}\right )} c f^{2} x - i \, a b d {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) + i \, a b d {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1} + 1\right ) - {\left (2 \, a b d f x + 2 \, a b c f - b^{2} d\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (2 \, a b d f x + 2 \, a b c f - b^{2} d\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (f x + e\right ) - 1\right )}}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} d f x + b^{2} c f\right )} \tan \left (f x + e\right )}{2 \, f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^2 - b^2)*d*f^2*x^2 + 2*(a^2 - b^2)*c*f^2*x - I*a*b*d*dilog(2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)
+ 1) + I*a*b*d*dilog(2*(-I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1) + 1) - (2*a*b*d*f*x + 2*a*b*c*f - b^2*d)*lo
g(-2*(I*tan(f*x + e) - 1)/(tan(f*x + e)^2 + 1)) - (2*a*b*d*f*x + 2*a*b*c*f - b^2*d)*log(-2*(-I*tan(f*x + e) -
1)/(tan(f*x + e)^2 + 1)) + 2*(b^2*d*f*x + b^2*c*f)*tan(f*x + e))/f^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tan {\left (e + f x \right )}\right )^{2} \left (c + d x\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2\,\left (c+d\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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