\(\int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx\) [220]

Optimal result

Integrand size = 20, antiderivative size = 20 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\frac {\cos (a+b x)}{d (c+d x)}+\frac {b \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{d^2}+\frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^2}+\text {Int}\left (\frac {\sec (a+b x)}{(c+d x)^2},x\right ) \]

[Out]

cos(b*x+a)/d/(d*x+c)+b*cos(a-b*c/d)*Si(b*c/d+b*x)/d^2+b*Ci(b*c/d+b*x)*sin(a-b*c/d)/d^2+Unintegrable(sec(b*x+a)
/(d*x+c)^2,x)

Rubi [N/A]

Not integrable

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

[In]

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

[Out]

Cos[a + b*x]/(d*(c + d*x)) + (b*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/d^2 + (b*Cos[a - (b*c)/d]*SinInte
gral[(b*c)/d + b*x])/d^2 + Defer[Int][Sec[a + b*x]/(c + d*x)^2, x]

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (a+b x)}{(c+d x)^2} \, dx+\int \frac {\sec (a+b x)}{(c+d x)^2} \, dx \\ & = \frac {\cos (a+b x)}{d (c+d x)}+\frac {b \int \frac {\sin (a+b x)}{c+d x} \, dx}{d}+\int \frac {\sec (a+b x)}{(c+d x)^2} \, dx \\ & = \frac {\cos (a+b x)}{d (c+d x)}+\frac {\left (b \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{d}+\frac {\left (b \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{d}+\int \frac {\sec (a+b x)}{(c+d x)^2} \, dx \\ & = \frac {\cos (a+b x)}{d (c+d x)}+\frac {b \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{d^2}+\frac {b \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^2}+\int \frac {\sec (a+b x)}{(c+d x)^2} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 7.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Not integrable

Time = 0.90 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

\[\int \frac {\sec \left (x b +a \right ) \sin \left (x b +a \right )^{2}}{\left (d x +c \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

integral(-(cos(b*x + a)^2 - 1)*sec(b*x + a)/(d^2*x^2 + 2*c*d*x + c^2), x)

Sympy [N/A]

Not integrable

Time = 2.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin ^{2}{\left (a + b x \right )} \sec {\left (a + b x \right )}}{\left (c + d x\right )^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.87 (sec) , antiderivative size = 270, normalized size of antiderivative = 13.50 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

1/2*((exp_integral_e(2, (I*b*d*x + I*b*c)/d) + exp_integral_e(2, -(I*b*d*x + I*b*c)/d))*cos(-(b*c - a*d)/d) +
4*(d^2*x + c*d)*integrate((cos(2*b*x + 2*a)*cos(b*x + a) + sin(2*b*x + 2*a)*sin(b*x + a) + cos(b*x + a))/(d^2*
x^2 + 2*c*d*x + (d^2*x^2 + 2*c*d*x + c^2)*cos(2*b*x + 2*a)^2 + (d^2*x^2 + 2*c*d*x + c^2)*sin(2*b*x + 2*a)^2 +
c^2 + 2*(d^2*x^2 + 2*c*d*x + c^2)*cos(2*b*x + 2*a)), x) + (-I*exp_integral_e(2, (I*b*d*x + I*b*c)/d) + I*exp_i
ntegral_e(2, -(I*b*d*x + I*b*c)/d))*sin(-(b*c - a*d)/d))/(d^2*x + c*d)

Giac [N/A]

Not integrable

Time = 3.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{2}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 24.92 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {\sin (a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^2}{\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^2} \,d x \]

[In]

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

[Out]

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