3.2.50 \(\int (a+b \sin (c+d x))^2 \tan ^3(c+d x) \, dx\) [150]

Optimal. Leaf size=111 \[ \frac {(a+b) (a+2 b) \log (1-\sin (c+d x))}{2 d}+\frac {(a-2 b) (a-b) \log (1+\sin (c+d x))}{2 d}+\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2800, 1659, 1643, 647, 31} \begin {gather*} \frac {2 a b \sin (c+d x)}{d}+\frac {(a+b) (a+2 b) \log (1-\sin (c+d x))}{2 d}+\frac {(a-2 b) (a-b) \log (\sin (c+d x)+1)}{2 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{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]^3,x]

[Out]

((a + b)*(a + 2*b)*Log[1 - Sin[c + d*x]])/(2*d) + ((a - 2*b)*(a - b)*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) + (Sec[c + d*x]^2*(a + b*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 1643

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1659

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*c*(p
+ 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - a*e*g*m + c*d*f*(2*p
+ 3) + c*e*f*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 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))^2 \tan ^3(c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {x^3 (a+x)^2}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac {\text {Subst}\left (\int \frac {(a+x) \left (-2 b^4-2 a b^2 x-2 b^2 x^2\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}+\frac {\text {Subst}\left (\int \left (4 a b^2+2 b^2 x-\frac {2 \left (3 a b^4+b^2 \left (a^2+2 b^2\right ) x\right )}{b^2-x^2}\right ) \, dx,x,b \sin (c+d x)\right )}{2 b^2 d}\\ &=\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {\text {Subst}\left (\int \frac {3 a b^4+b^2 \left (a^2+2 b^2\right ) x}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{b^2 d}\\ &=\frac {2 a b \sin (c+d x)}{d}+\frac {b^2 \sin ^2(c+d x)}{2 d}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}-\frac {((a-2 b) (a-b)) \text {Subst}\left (\int \frac {1}{-b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}-\frac {((a+b) (a+2 b)) \text {Subst}\left (\int \frac {1}{b-x} \, dx,x,b \sin (c+d x)\right )}{2 d}\\ &=\frac {(a+b) (a+2 b) \log (1-\sin (c+d x))}{2 d}+\frac {(a-2 b) (a-b) \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}+\frac {\sec ^2(c+d x) (a+b \sin (c+d x))^2}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.29, size = 108, normalized size = 0.97 \begin {gather*} \frac {2 (a+b) (a+2 b) \log (1-\sin (c+d x))+2 (a-2 b) (a-b) \log (1+\sin (c+d x))-\frac {(a+b)^2}{-1+\sin (c+d x)}+8 a b \sin (c+d x)+2 b^2 \sin ^2(c+d x)+\frac {(a-b)^2}{1+\sin (c+d x)}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

(2*(a + b)*(a + 2*b)*Log[1 - Sin[c + d*x]] + 2*(a - 2*b)*(a - b)*Log[1 + Sin[c + d*x]] - (a + b)^2/(-1 + Sin[c
+ d*x]) + 8*a*b*Sin[c + d*x] + 2*b^2*Sin[c + d*x]^2 + (a - b)^2/(1 + Sin[c + d*x]))/(4*d)

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Maple [A]
time = 0.19, size = 135, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(135\)
default	\(\frac {a^{2} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\ln \left (\cos \left (d x +c \right )\right )\right )+2 a b \left (\frac {\sin ^{5}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{2}+\frac {3 \sin \left (d x +c \right )}{2}-\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(135\)
risch	\(-\frac {4 i b^{2} c}{d}-\frac {2 i a^{2} c}{d}-\frac {b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-2 i b^{2} x -\frac {2 i \left (i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+i b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+a b \,{\mathrm e}^{3 i \left (d x +c \right )}-a \,{\mathrm e}^{i \left (d x +c \right )} b \right )}{d \left (1+{\mathrm e}^{2 i \left (d x +c \right )}\right )^{2}}-\frac {b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-i a^{2} x -\frac {i a b \,{\mathrm e}^{i \left (d x +c \right )}}{d}+\frac {i a b \,{\mathrm e}^{-i \left (d x +c \right )}}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{d}\)	\(301\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(d*x+c))^2*tan(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^2*(1/2*tan(d*x+c)^2+ln(cos(d*x+c)))+2*a*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)))+b^2*(1/2*sin(d*x+c)^6/cos(d*x+c)^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(cos(d*x
+c))))

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Maxima [A]
time = 0.31, size = 105, normalized size = 0.95 \begin {gather*} \frac {b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) + {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, a b \sin \left (d x + c\right ) + a^{2} + b^{2}}{\sin \left (d x + c\right )^{2} - 1}}{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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 140, normalized size = 1.26 \begin {gather*} -\frac {2 \, b^{2} \cos \left (d x + c\right )^{4} - b^{2} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} - 2 \, b^{2} - 4 \, {\left (2 \, a b \cos \left (d x + c\right )^{2} + a b\right )} \sin \left (d x + c\right )}{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))^2*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/4*(2*b^2*cos(d*x + c)^4 - b^2*cos(d*x + c)^2 - 2*(a^2 - 3*a*b + 2*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1)
- 2*(a^2 + 3*a*b + 2*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) - 2*a^2 - 2*b^2 - 4*(2*a*b*cos(d*x + c)^2 + a
*b)*sin(d*x + c))/(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 )^{2} \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))**2*tan(d*x+c)**3,x)

[Out]

Integral((a + b*sin(c + d*x))**2*tan(c + d*x)**3, x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)^3,x, algorithm="giac")

[Out]

Timed out

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Mupad [B]
time = 6.72, size = 232, normalized size = 2.09 \begin {gather*} \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a^2+4\,b^2\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+6\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+6\,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 )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+1\right )}-\frac {\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )\,\left (a^2+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 )\,\left (a+2\,b\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-b\right )\,\left (a-2\,b\right )}{d} \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))^2,x)

[Out]

(tan(c/2 + (d*x)/2)^2*(2*a^2 + 4*b^2) + tan(c/2 + (d*x)/2)^6*(2*a^2 + 4*b^2) + 4*a^2*tan(c/2 + (d*x)/2)^4 + 2*
a*b*tan(c/2 + (d*x)/2)^3 + 2*a*b*tan(c/2 + (d*x)/2)^5 + 6*a*b*tan(c/2 + (d*x)/2)^7 + 6*a*b*tan(c/2 + (d*x)/2))
/(d*(tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^4 + 1)) - (log(tan(c/2 + (d*x)/2)^2 + 1)*(a^2 + 2*b^2))/d + (
log(tan(c/2 + (d*x)/2) - 1)*(a + b)*(a + 2*b))/d + (log(tan(c/2 + (d*x)/2) + 1)*(a - b)*(a - 2*b))/d

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