\(\int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx\) [226]

Optimal result

Integrand size = 22, antiderivative size = 22 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=-\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {\sin (2 a+2 b x)}{2 d (c+d x)}+\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\text {Int}\left (\frac {\tan (a+b x)}{(c+d x)^2},x\right ) \]

[Out]

-b*Ci(2*b*c/d+2*b*x)*cos(2*a-2*b*c/d)/d^2+b*Si(2*b*c/d+2*b*x)*sin(2*a-2*b*c/d)/d^2+1/2*sin(2*b*x+2*a)/d/(d*x+c
)+Unintegrable(tan(b*x+a)/(d*x+c)^2,x)

Rubi [N/A]

Not integrable

Time = 0.20 (sec) , antiderivative size = 22, 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 = {} \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (a+b x) \sin (a+b x)}{(c+d x)^2} \, dx+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ & = -\int \frac {\sin (2 a+2 b x)}{2 (c+d x)^2} \, dx+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ & = -\left (\frac {1}{2} \int \frac {\sin (2 a+2 b x)}{(c+d x)^2} \, dx\right )+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ & = \frac {\sin (2 a+2 b x)}{2 d (c+d x)}-\frac {b \int \frac {\cos (2 a+2 b x)}{c+d x} \, dx}{d}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ & = \frac {\sin (2 a+2 b x)}{2 d (c+d x)}-\frac {\left (b \cos \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\cos \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}+\frac {\left (b \sin \left (2 a-\frac {2 b c}{d}\right )\right ) \int \frac {\sin \left (\frac {2 b c}{d}+2 b x\right )}{c+d x} \, dx}{d}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ & = -\frac {b \cos \left (2 a-\frac {2 b c}{d}\right ) \operatorname {CosIntegral}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\frac {\sin (2 a+2 b x)}{2 d (c+d x)}+\frac {b \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^2}+\int \frac {\tan (a+b x)}{(c+d x)^2} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 2.96 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Not integrable

Time = 1.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00

\[\int \frac {\sec \left (x b +a \right ) \sin \left (x b +a \right )^{3}}{\left (d x +c \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [N/A]

Not integrable

Time = 0.77 (sec) , antiderivative size = 242, normalized size of antiderivative = 11.00 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

1/4*((-I*exp_integral_e(2, 2*(-I*b*d*x - I*b*c)/d) + I*exp_integral_e(2, -2*(-I*b*d*x - I*b*c)/d))*cos(-2*(b*c
- a*d)/d) + 8*(d^2*x + c*d)*integrate(sin(2*b*x + 2*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) + (exp_integral_e(2, 2*(-I*b*d*x - I*b*c)/d) + exp_integral_e(2, -2*(-I*b*d*x - I*b*c)/d))*sin(-2*(b*c
- a*d)/d))/(d^2*x + c*d)

Giac [N/A]

Not integrable

Time = 2.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int { \frac {\sec \left (b x + a\right ) \sin \left (b x + a\right )^{3}}{{\left (d x + c\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 25.57 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {\sin ^2(a+b x) \tan (a+b x)}{(c+d x)^2} \, dx=\int \frac {{\sin \left (a+b\,x\right )}^3}{\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^2} \,d x \]

[In]

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

[Out]

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