3.2.51 \(\int (a+b \sin (c+d x))^2 \tan (c+d x) \, dx\) [151]

Optimal. Leaf size=78 \[ -\frac {(a+b)^2 \log (1-\sin (c+d x))}{2 d}-\frac {(a-b)^2 \log (1+\sin (c+d x))}{2 d}-\frac {2 a b \sin (c+d x)}{d}-\frac {b^2 \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2800, 815, 647, 31} \begin {gather*} -\frac {2 a b \sin (c+d x)}{d}-\frac {(a-b)^2 \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^2 \log (1-\sin (c+d x))}{2 d}-\frac {b^2 \sin ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((a + b)^2*Log[1 - Sin[c + d*x]])/d - ((a - b)^2*Log[1 + Sin[c + d*x]])/(2*d) - (2*a*b*Sin[c + d*x])/d -
(b^2*Sin[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 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

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Mathematica [A]
time = 0.09, size = 64, normalized size = 0.82 \begin {gather*} -\frac {(a+b)^2 \log (1-\sin (c+d x))+(a-b)^2 \log (1+\sin (c+d x))+4 a b \sin (c+d x)+b^2 \sin ^2(c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.19, size = 69, normalized size = 0.88

method	result	size
derivativedivides	\(\frac {-a^{2} \ln \left (\cos \left (d x +c \right )\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(69\)
default	\(\frac {-a^{2} \ln \left (\cos \left (d x +c \right )\right )+2 a b \left (-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+b^{2} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(69\)
risch	\(i a^{2} x +i b^{2} x +\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}-\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {2 i a^{2} c}{d}+\frac {2 i b^{2} c}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}+\frac {b^{2} \cos \left (2 d x +2 c \right )}{4 d}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 70, normalized size = 0.90 \begin {gather*} -\frac {b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.37, size = 74, normalized size = 0.95 \begin {gather*} \frac {b^{2} \cos \left (d x + c\right )^{2} - 4 \, a b \sin \left (d x + c\right ) - {\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{2} \tan {\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))**2*tan(d*x+c),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 7855 vs. \(2 (72) = 144\).
time = 36.44, size = 7855, normalized size = 100.71 \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))^2*tan(d*x+c),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 6.37, size = 150, normalized size = 1.92 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+b^2\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (a+b\right )}^2}{d}-\frac {2\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,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 )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left (a-b\right )}^2}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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