3.1340 \(\int \frac {\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=133 \[ -\frac {b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2}+\frac {2 a^3 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2902, 2606, 8, 3473, 2735, 2660, 618, 204} \[ \frac {2 a^3 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac {b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac {a \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {a^2 x}{b \left (a^2-b^2\right )}+\frac {b x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*x)/(b*(a^2 - b^2))) + (b*x)/(a^2 - b^2) + (2*a^3*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(b*(
a^2 - b^2)^(3/2)*d) + (a*Sec[c + d*x])/((a^2 - b^2)*d) - (b*Tan[c + d*x])/((a^2 - b^2)*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rubi steps

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

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Mathematica [A]  time = 0.84, size = 152, normalized size = 1.14 \[ \frac {\frac {b (a-b \sin (c+d x))-\left (a^2-b^2\right ) (c+d x) \cos (c+d x)}{(a-b) (a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 a^3 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^3*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + (-((a^2 - b^2)*(c + d*x)*Cos[c +
d*x]) + b*(a - b*Sin[c + d*x]))/((a - b)*(a + b)*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])))/(b*d)

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fricas [A]  time = 0.75, size = 369, normalized size = 2.77 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} a^{3} \cos \left (d x + c\right ) \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b - 2 \, a b^{3} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) - 2 \, {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}, -\frac {\sqrt {a^{2} - b^{2}} a^{3} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{3} b + a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(-a^2 + b^2)*a^3*cos(d*x + c)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 -
2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) -
a^2 - b^2)) + 2*a^3*b - 2*a*b^3 - 2*(a^4 - 2*a^2*b^2 + b^4)*d*x*cos(d*x + c) - 2*(a^2*b^2 - b^4)*sin(d*x + c))
/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c)), -(sqrt(a^2 - b^2)*a^3*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^
2)*cos(d*x + c)))*cos(d*x + c) - a^3*b + a*b^3 + (a^4 - 2*a^2*b^2 + b^4)*d*x*cos(d*x + c) + (a^2*b^2 - b^4)*si
n(d*x + c))/((a^4*b - 2*a^2*b^3 + b^5)*d*cos(d*x + c))]

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giac [A]  time = 0.29, size = 131, normalized size = 0.98 \[ \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a^{3}}{{\left (a^{2} b - b^{3}\right )} \sqrt {a^{2} - b^{2}}} - \frac {d x + c}{b} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a^3/((a^2*
b - b^3)*sqrt(a^2 - b^2)) - (d*x + c)/b + 2*(b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2
- 1)))/d

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maple [A]  time = 0.40, size = 138, normalized size = 1.04 \[ -\frac {16}{d \left (16 a +16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b}+\frac {16}{d \left (16 a -16 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {2 a^{3} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \left (a -b \right ) \left (a +b \right ) b \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

-16/d/(16*a+16*b)/(tan(1/2*d*x+1/2*c)-1)-2/d/b*arctan(tan(1/2*d*x+1/2*c))+16/d/(16*a-16*b)/(tan(1/2*d*x+1/2*c)
+1)+2/d/(a-b)/(a+b)*a^3/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 14.05, size = 1538, normalized size = 11.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)^3/(cos(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

(a^5*cos(c + d*x) + a^5)/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (2*a^6*atan(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2)))/(b*d*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b*(a^4*sin(c + d*x) - 6*a^4*cos(c + d*x)*at
an(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^4*(a + a
*cos(c + d*x)))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^2*(2*a^3*cos(c + d*x) + 2*a^3))/(d*c
os(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) + (b^3*(2*a^2*sin(c + d*x) - 6*a^2*cos(c + d*x)*atan(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2)) - (b^5*(sin(c + d*x) - 2
*cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/(d*cos(c + d*x)*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2
)) - (a^3*atan(((2*b^14*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^14*sin(c/2 + (d*x)/
2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 2*a^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3
/2) - 6*a^3*b^11*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 15*a^5*b^9*cos(c/2 + (d*x)/2)*
(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 20*a^7*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^
(1/2) + 15*a^9*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 6*a^11*b^3*cos(c/2 + (d*x)/2
)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 3*a^6*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)
^(3/2) - 13*a^2*b^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 36*a^4*b^10*sin(c/2 + (d*x
)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 56*a^6*b^8*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*
b^2)^(1/2) + 54*a^8*b^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) - 33*a^10*b^4*sin(c/2 + (
d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + 12*a^12*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*
a^4*b^2)^(1/2) + a^7*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2) + a*b^13*cos(c/2 + (d*x)/2
)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2) + a^13*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1
/2))*1i)/((a^4*b + b^5 - 2*a^2*b^3)*(2*b^12*sin(c/2 + (d*x)/2) + a*b^11*cos(c/2 + (d*x)/2) - 3*a^11*b*cos(c/2
+ (d*x)/2) - 6*a^3*b^9*cos(c/2 + (d*x)/2) + 15*a^5*b^7*cos(c/2 + (d*x)/2) - 19*a^7*b^5*cos(c/2 + (d*x)/2) + 12
*a^9*b^3*cos(c/2 + (d*x)/2) - 12*a^2*b^10*sin(c/2 + (d*x)/2) + 30*a^4*b^8*sin(c/2 + (d*x)/2) - 38*a^6*b^6*sin(
c/2 + (d*x)/2) + 24*a^8*b^4*sin(c/2 + (d*x)/2) - 6*a^10*b^2*sin(c/2 + (d*x)/2))))*(-(a + b)^3*(a - b)^3)^(1/2)
*2i)/(b*d*(a^2 - b^2)*(a^4 + b^4 - 2*a^2*b^2))

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sympy [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(sec(d*x+c)**2*sin(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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