3.2.40 \(\int (a+b \sin (c+d x)) \tan ^3(c+d x) \, dx\) [140]

Optimal. Leaf size=88 \[ \frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-3 b) \log (1+\sin (c+d x))}{4 d}+\frac {3 b \sin (c+d x)}{2 d}+\frac {(a+b \sin (c+d x)) \tan ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2800, 833, 788, 647, 31} \begin {gather*} \frac {(2 a+3 b) \log (1-\sin (c+d x))}{4 d}+\frac {(2 a-3 b) \log (\sin (c+d x)+1)}{4 d}+\frac {\tan ^2(c+d x) (a+b \sin (c+d x))}{2 d}+\frac {3 b \sin (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

((2*a + 3*b)*Log[1 - Sin[c + d*x]])/(4*d) + ((2*a - 3*b)*Log[1 + Sin[c + d*x]])/(4*d) + (3*b*Sin[c + d*x])/(2*
d) + ((a + b*Sin[c + d*x])*Tan[c + d*x]^2)/(2*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 788

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*g*(x/c), x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rule 2800

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 77, normalized size = 0.88 \begin {gather*} -\frac {b \sin (c+d x) \tan ^2(c+d x)}{d}-\frac {3 b \left (\tanh ^{-1}(\sin (c+d x))-\sec (c+d x) \tan (c+d x)\right )}{2 d}+\frac {a \left (2 \log (\cos (c+d x))+\tan ^2(c+d x)\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

-((b*Sin[c + d*x]*Tan[c + d*x]^2)/d) - (3*b*(ArcTanh[Sin[c + d*x]] - Sec[c + d*x]*Tan[c + d*x]))/(2*d) + (a*(2
*Log[Cos[c + d*x]] + Tan[c + d*x]^2))/(2*d)

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Maple [A]
time = 0.17, size = 81, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(81\)
default	\(\frac {a \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(81\)
risch	\(-i a x -\frac {i b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {2 i a c}{d}-\frac {i \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}+2 i a \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b}{2 d}\)	\(177\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))*tan(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a*(1/2*tan(d*x+c)^2+ln(cos(d*x+c)))+b*(1/2*sin(d*x+c)^5/cos(d*x+c)^2+1/2*sin(d*x+c)^3+3/2*sin(d*x+c)-3/2*
ln(sec(d*x+c)+tan(d*x+c))))

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Maxima [A]
time = 0.28, size = 73, normalized size = 0.83 \begin {gather*} \frac {{\left (2 \, a - 3 \, b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a + 3 \, b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 4 \, b \sin \left (d x + c\right ) - \frac {2 \, {\left (b \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

1/4*((2*a - 3*b)*log(sin(d*x + c) + 1) + (2*a + 3*b)*log(sin(d*x + c) - 1) + 4*b*sin(d*x + c) - 2*(b*sin(d*x +
c) + a)/(sin(d*x + c)^2 - 1))/d

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Fricas [A]
time = 0.38, size = 90, normalized size = 1.02 \begin {gather*} \frac {{\left (2 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, b \cos \left (d x + c\right )^{2} + b\right )} \sin \left (d x + c\right ) + 2 \, a}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*((2*a - 3*b)*cos(d*x + c)^2*log(sin(d*x + c) + 1) + (2*a + 3*b)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*
(2*b*cos(d*x + c)^2 + b)*sin(d*x + c) + 2*a)/(d*cos(d*x + c)^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*tan(d*x+c)**3,x)

[Out]

Integral((a + b*sin(c + d*x))*tan(c + d*x)**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 26228 vs. \(2 (80) = 160\).
time = 230.66, size = 26228, normalized size = 298.05 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))*tan(d*x+c)^3,x, algorithm="giac")

[Out]

1/4*(3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + ta
n(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)
^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c
)^6*tan(c)^2 - 3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2
*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*ta
n(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6
*tan(1/2*c)^6*tan(c)^2 + 2*a*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)
^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 + 2*a*tan(d*x)^2*t
an(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 - 6*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2
*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d
*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*ta
n(d*x)*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c) + 6*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*
c) - 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan
(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 +
1))*tan(d*x)*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c) - 4*a*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d
*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(
c) - 3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + ta
n(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)
^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c
)^4*tan(c)^2 + 3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2
*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*ta
n(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6
*tan(1/2*c)^4*tan(c)^2 - 2*a*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)
^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^4*tan(c)^2 - 24*b*log(2*(tan(
1/2*d*x)^4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan
(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 -
2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^5*tan(1/2*c)^5*tan(c)^2 + 24*b*
log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x
)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(
1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^5*tan(1/2*c)^5*tan(c
)^2 - 16*a*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan(c)^2 + tan(d*x)^2 - 2*tan(d*x)*ta
n(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^5*tan(1/2*c)^5*tan(c)^2 - 12*b*tan(d*x)^2*tan(1/2*d*x)^6*tan
(1/2*c)^5*tan(c)^2 - 3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2*tan(1/2*d*x)^3*t
an(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d*x)*tan(1/2*c)^2
+ 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^2*tan(1/2*
d*x)^4*tan(1/2*c)^6*tan(c)^2 + 3*b*log(2*(tan(1/2*d*x)^4*tan(1/2*c)^2 - 2*tan(1/2*d*x)^4*tan(1/2*c) - 2*tan(1/
2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)^2*tan(1/2*c)^2 + 2*tan(1/2*d*x)^3 - 2*tan(1/2*d*x)*tan
(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 + 2*tan(1/2*d*x) + 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*tan(d*x)^
2*tan(1/2*d*x)^4*tan(1/2*c)^6*tan(c)^2 - 2*a*log(4*(tan(d*x)^4*tan(c)^2 - 2*tan(d*x)^3*tan(c) + tan(d*x)^2*tan
(c)^2 + tan(d*x)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(c)^2 + 1))*tan(d*x)^2*tan(1/2*d*x)^4*tan(1/2*c)^6*tan(c)^2 -
12*b*tan(d*x)^2*tan(1/2*d*x)^5*tan(1/2*c)^6*tan(c)^2 + 2*a*tan(d*x)^2*tan(1/2*d*x)^6*tan(1/2*c)^6 - 2*a*tan(d*
x)^2*tan(1/2*d*x)^6*tan(1/2*c)^4*tan(c)^2 - 16*a*tan(d*x)^2*tan(1/2*d*x)^5*tan(1/2*c)^5*tan(c)^2 - 2*a*tan(d*x
)^2*tan(1/2*d*x)^4*tan(1/2*c)^6*tan(c)^2 + 2*a*tan(1/2*d*x)^6*tan(1/2*c)^6*tan(c)^2 + 3*b*log(2*(tan(1/2*d*x)^
4*tan(1/2*c)^2 + 2*tan(1/2*d*x)^4*tan(1/2*c) + 2*tan(1/2*d*x)^3*tan(1/2*c)^2 + tan(1/2*d*x)^4 + 2*tan(1/2*d*x)
^2*tan(1/2*c)^2 - 2*tan(1/2*d*x)^3 + 2*tan(1/2*d*x)*tan(1/2*c)^2 + 2*tan(1/2*d*x)^2 + tan(1/2*c)^2 - 2*tan(1/2
*d*x) - 2*tan(1/2*c) + 1)/(tan(1/2*c)^2 + 1))*t...

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Mupad [B]
time = 6.73, size = 176, normalized size = 2.00 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (a+\frac {3\,b}{2}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-\frac {3\,b}{2}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + b*sin(c + d*x)),x)

[Out]

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

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