3.1348 \(\int \frac{\sec ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=117 \[ -\frac{a b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}+\frac{\sec ^2(c+d x) (a-b \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac{b \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{b \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.165122, antiderivative size = 117, normalized size of antiderivative = 1., 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, 823, 801} \[ -\frac{a b^2 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}+\frac{\sec ^2(c+d x) (a-b \sin (c+d x))}{2 d \left (a^2-b^2\right )}-\frac{b \log (1-\sin (c+d x))}{4 d (a+b)^2}+\frac{b \log (\sin (c+d x)+1)}{4 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 12

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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]

Rubi steps

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

Mathematica [A]  time = 0.374064, size = 162, normalized size = 1.38 \[ \frac{-\frac{4 a b^2 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^2}+\frac{1}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{1}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{(a+b)^2}+\frac{2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{(a-b)^2}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.067, size = 123, normalized size = 1.1 \begin{align*} -{\frac{a{b}^{2}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{1}{d \left ( 4\,a+4\,b \right ) \left ( \sin \left ( dx+c \right ) -1 \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) b}{4\,d \left ( a+b \right ) ^{2}}}+{\frac{1}{d \left ( 4\,a-4\,b \right ) \left ( 1+\sin \left ( dx+c \right ) \right ) }}+{\frac{b\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{4\, \left ( a-b \right ) ^{2}d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.993699, size = 178, normalized size = 1.52 \begin{align*} -\frac{\frac{4 \, a b^{2} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{b \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{b \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (b \sin \left (d x + c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2} + b^{2}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.84681, size = 370, normalized size = 3.16 \begin{align*} -\frac{4 \, a b^{2} \cos \left (d x + c\right )^{2} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, a^{3} + 2 \, a b^{2} + 2 \,{\left (a^{2} b - b^{3}\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21551, size = 230, normalized size = 1.97 \begin{align*} -\frac{\frac{4 \, a b^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac{b \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{b \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{2 \,{\left (a b^{2} \sin \left (d x + c\right )^{2} - a^{2} b \sin \left (d x + c\right ) + b^{3} \sin \left (d x + c\right ) + a^{3} - 2 \, a b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (\sin \left (d x + c\right )^{2} - 1\right )}}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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