3.1347 \(\int \frac {\sec (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=116 \[ \frac {a^2 b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {a \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac {a \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2837, 12, 1647, 801} \[ \frac {a^2 b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac {\sec ^2(c+d x) (b-a \sin (c+d x))}{2 d \left (a^2-b^2\right )}+\frac {a \log (1-\sin (c+d x))}{4 d (a+b)^2}-\frac {a \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 801

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 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 2837

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

Rubi steps

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

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Mathematica [A]  time = 0.47, size = 108, normalized size = 0.93 \[ -\frac {-\frac {4 a^2 b \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+\frac {1}{(a+b) (\sin (c+d x)-1)}+\frac {1}{(a-b) (\sin (c+d x)+1)}-\frac {a \log (1-\sin (c+d x))}{(a+b)^2}+\frac {a \log (\sin (c+d x)+1)}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 154, normalized size = 1.33 \[ \frac {4 \, a^{2} b \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{2} b + 2 \, b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 168, normalized size = 1.45 \[ \frac {\frac {4 \, a^{2} b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac {a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right )^{2} - a^{3} \sin \left (d x + c\right ) + a b^{2} \sin \left (d x + c\right ) - b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*a^2*b^2*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) - a*log(abs(sin(d*x + c) + 1))/(a^2 - 2*
a*b + b^2) + a*log(abs(sin(d*x + c) - 1))/(a^2 + 2*a*b + b^2) + 2*(a^2*b*sin(d*x + c)^2 - a^3*sin(d*x + c) + a
*b^2*sin(d*x + c) - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(sin(d*x + c)^2 - 1)))/d

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maple [A]  time = 0.44, size = 123, normalized size = 1.06 \[ -\frac {1}{d \left (4 a +4 b \right ) \left (\sin \left (d x +c \right )-1\right )}+\frac {\ln \left (\sin \left (d x +c \right )-1\right ) a}{4 d \left (a +b \right )^{2}}+\frac {a^{2} b \ln \left (a +b \sin \left (d x +c \right )\right )}{d \left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {1}{d \left (4 a -4 b \right ) \left (1+\sin \left (d x +c \right )\right )}-\frac {a \ln \left (1+\sin \left (d x +c \right )\right )}{4 \left (a -b \right )^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 132, normalized size = 1.14 \[ \frac {\frac {4 \, a^{2} b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {a \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac {a \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {2 \, {\left (a \sin \left (d x + c\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.22, size = 206, normalized size = 1.78 \[ \frac {\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^2-b^2}+\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a^2-b^2}-\frac {2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2-b^2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{2\,d\,{\left (a+b\right )}^2}+\frac {a^2\,b\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^4-2\,a^2\,b^2+b^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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