\(\int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx\) [394]

Optimal result

Integrand size = 25, antiderivative size = 25 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=-\frac {2 b \cos (a+b x)}{d^2 (c+d x)}-\frac {2 b^2 \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{d^3}-\frac {2 \sin (a+b x)}{d (c+d x)^2}-\frac {2 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^3}-\text {Int}\left (\frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3},x\right ) \]

[Out]

-CannotIntegrate(sec(b*x+a)*tan(b*x+a)/(d*x+c)^3,x)-2*b*cos(b*x+a)/d^2/(d*x+c)-2*b^2*cos(a-b*c/d)*Si(b*c/d+b*x
)/d^3-2*b^2*Ci(b*c/d+b*x)*sin(a-b*c/d)/d^3-2*sin(b*x+a)/d/(d*x+c)^2

Rubi [N/A]

Not integrable

Time = 0.54 (sec) , antiderivative size = 25, 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 = {} \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]

[In]

Int[(Sec[a + b*x]^2*Sin[3*a + 3*b*x])/(c + d*x)^3,x]

[Out]

(-2*b*Cos[a + b*x])/(d^2*(c + d*x)) - (2*b^2*CosIntegral[(b*c)/d + b*x]*Sin[a - (b*c)/d])/d^3 - (2*Sin[a + b*x
])/(d*(c + d*x)^2) - (2*b^2*Cos[a - (b*c)/d]*SinIntegral[(b*c)/d + b*x])/d^3 - Defer[Int][(Sec[a + b*x]*Tan[a
+ b*x])/(c + d*x)^3, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3 \sin (a+b x)}{(c+d x)^3}-\frac {\sin (a+b x) \tan ^2(a+b x)}{(c+d x)^3}\right ) \, dx \\ & = 3 \int \frac {\sin (a+b x)}{(c+d x)^3} \, dx-\int \frac {\sin (a+b x) \tan ^2(a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {3 \sin (a+b x)}{2 d (c+d x)^2}+\frac {(3 b) \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{2 d}+\int \frac {\sin (a+b x)}{(c+d x)^3} \, dx-\int \frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {3 b \cos (a+b x)}{2 d^2 (c+d x)}-\frac {2 \sin (a+b x)}{d (c+d x)^2}-\frac {\left (3 b^2\right ) \int \frac {\sin (a+b x)}{c+d x} \, dx}{2 d^2}+\frac {b \int \frac {\cos (a+b x)}{(c+d x)^2} \, dx}{2 d}-\int \frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (a+b x)}{d^2 (c+d x)}-\frac {2 \sin (a+b x)}{d (c+d x)^2}-\frac {b^2 \int \frac {\sin (a+b x)}{c+d x} \, dx}{2 d^2}-\frac {\left (3 b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}-\frac {\left (3 b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}-\int \frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (a+b x)}{d^2 (c+d x)}-\frac {3 b^2 \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{2 d^3}-\frac {2 \sin (a+b x)}{d (c+d x)^2}-\frac {3 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{2 d^3}-\frac {\left (b^2 \cos \left (a-\frac {b c}{d}\right )\right ) \int \frac {\sin \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}-\frac {\left (b^2 \sin \left (a-\frac {b c}{d}\right )\right ) \int \frac {\cos \left (\frac {b c}{d}+b x\right )}{c+d x} \, dx}{2 d^2}-\int \frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ & = -\frac {2 b \cos (a+b x)}{d^2 (c+d x)}-\frac {2 b^2 \operatorname {CosIntegral}\left (\frac {b c}{d}+b x\right ) \sin \left (a-\frac {b c}{d}\right )}{d^3}-\frac {2 \sin (a+b x)}{d (c+d x)^2}-\frac {2 b^2 \cos \left (a-\frac {b c}{d}\right ) \text {Si}\left (\frac {b c}{d}+b x\right )}{d^3}-\int \frac {\sec (a+b x) \tan (a+b x)}{(c+d x)^3} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 17.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]

[In]

Integrate[(Sec[a + b*x]^2*Sin[3*a + 3*b*x])/(c + d*x)^3,x]

[Out]

Integrate[(Sec[a + b*x]^2*Sin[3*a + 3*b*x])/(c + d*x)^3, x]

Maple [N/A] (verified)

Not integrable

Time = 0.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00

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

[In]

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

[Out]

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

Fricas [N/A]

Not integrable

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.57 (sec) , antiderivative size = 1563, normalized size of antiderivative = 62.52 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

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

[Out]

2*(b*c*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b*c - a*d
)/d) - b*c*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*c - a*d)
/d) + (b*c*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b*c -
a*d)/d) - b*c*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*c -
a*d)/d) + (b*d*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b
*c - a*d)/d) - b*d*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*
c - a*d)/d))*x)*cos(2*b*x + 2*a)^2 + (b*c*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I
*b*d*x + I*b*c)/d))*cos(-(b*c - a*d)/d) - b*c*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*
b*d*x + I*b*c)/d))*sin(-(b*c - a*d)/d) + (b*d*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3,
-(I*b*d*x + I*b*c)/d))*cos(-(b*c - a*d)/d) - b*d*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3,
-(I*b*d*x + I*b*c)/d))*sin(-(b*c - a*d)/d))*x)*sin(2*b*x + 2*a)^2 - d*sin(2*b*x + 2*a)*sin(b*x + a) + (b*d*(-I
*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b*c - a*d)/d) - b*d
*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*c - a*d)/d))*x + (
2*b*c*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b*c - a*d)
/d) - 2*b*c*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*c - a*d
)/d) + 2*(b*d*(-I*exp_integral_e(3, (I*b*d*x + I*b*c)/d) + I*exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*cos(-(b*
c - a*d)/d) - b*d*(exp_integral_e(3, (I*b*d*x + I*b*c)/d) + exp_integral_e(3, -(I*b*d*x + I*b*c)/d))*sin(-(b*c
- a*d)/d))*x - d*cos(b*x + a))*cos(2*b*x + 2*a) - d*cos(b*x + a) - 3*(b*d^5*x^3 + 3*b*c*d^4*x^2 + 3*b*c^2*d^3
*x + b*c^3*d^2 + (b*d^5*x^3 + 3*b*c*d^4*x^2 + 3*b*c^2*d^3*x + b*c^3*d^2)*cos(2*b*x + 2*a)^2 + (b*d^5*x^3 + 3*b
*c*d^4*x^2 + 3*b*c^2*d^3*x + b*c^3*d^2)*sin(2*b*x + 2*a)^2 + 2*(b*d^5*x^3 + 3*b*c*d^4*x^2 + 3*b*c^2*d^3*x + b*
c^3*d^2)*cos(2*b*x + 2*a))*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))/(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4 + (b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^
2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*cos(2*b*x + 2*a)^2 + (b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d
*x + b*c^4)*sin(2*b*x + 2*a)^2 + 2*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + 4*b*c^3*d*x + b*c^4)*cos(2*b
*x + 2*a)), x))/(b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*d^2*x + b*c^3*d + (b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*d^
2*x + b*c^3*d)*cos(2*b*x + 2*a)^2 + (b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*d^2*x + b*c^3*d)*sin(2*b*x + 2*a)^2 +
2*(b*d^4*x^3 + 3*b*c*d^3*x^2 + 3*b*c^2*d^2*x + b*c^3*d)*cos(2*b*x + 2*a))

Giac [N/A]

Not integrable

Time = 7.78 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int { \frac {\sec \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right )}{{\left (d x + c\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 34.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {\sec ^2(a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )}{{\cos \left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^3} \,d x \]

[In]

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

[Out]

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