3.388 \(\int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx\)

Optimal. Leaf size=129 \[ -\text {Int}\left (\frac {\tan (a+b x)}{(c+d x)^3},x\right )-\frac {4 b^2 \sin \left (2 a-\frac {2 b c}{d}\right ) \text {Ci}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {4 b^2 \cos \left (2 a-\frac {2 b c}{d}\right ) \text {Si}\left (\frac {2 b c}{d}+2 b x\right )}{d^3}-\frac {2 b \cos (2 a+2 b x)}{d^2 (c+d x)}-\frac {\sin (2 a+2 b x)}{d (c+d x)^2} \]

[Out]

-2*b*cos(2*b*x+2*a)/d^2/(d*x+c)-4*b^2*cos(2*a-2*b*c/d)*Si(2*b*c/d+2*b*x)/d^3-4*b^2*Ci(2*b*c/d+2*b*x)*sin(2*a-2
*b*c/d)/d^3-sin(2*b*x+2*a)/d/(d*x+c)^2-Unintegrable(tan(b*x+a)/(d*x+c)^3,x)

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Rubi [A]  time = 0.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

(-2*b*Cos[2*a + 2*b*x])/(d^2*(c + d*x)) - (4*b^2*CosIntegral[(2*b*c)/d + 2*b*x]*Sin[2*a - (2*b*c)/d])/d^3 - Si
n[2*a + 2*b*x]/(d*(c + d*x)^2) - (4*b^2*Cos[2*a - (2*b*c)/d]*SinIntegral[(2*b*c)/d + 2*b*x])/d^3 - Defer[Int][
Tan[a + b*x]/(c + d*x)^3, x]

Rubi steps

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

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Mathematica [A]  time = 6.02, size = 0, normalized size = 0.00 \[ \int \frac {\sec (a+b x) \sin (3 a+3 b x)}{(c+d x)^3} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right )}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.64, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (b x +a \right ) \sin \left (3 b x +3 a \right )}{\left (d x +c \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (i \, E_{3}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) - i \, E_{3}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \cos \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right ) + 2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )} \int \frac {\sin \left (2 \, b x + 2 \, a\right )}{{\left (d x + c\right )}^{3} {\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )}}\,{d x} + {\left (E_{3}\left (\frac {2 i \, b d x + 2 i \, b c}{d}\right ) + E_{3}\left (-\frac {2 i \, b d x + 2 i \, b c}{d}\right )\right )} \sin \left (-\frac {2 \, {\left (b c - a d\right )}}{d}\right )}{d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((I*exp_integral_e(3, (2*I*b*d*x + 2*I*b*c)/d) - I*exp_integral_e(3, -(2*I*b*d*x + 2*I*b*c)/d))*cos(-2*(b*c -
a*d)/d) + 2*(d^3*x^2 + 2*c*d^2*x + c^2*d)*integrate(sin(2*b*x + 2*a)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3
+ (d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*cos(2*b*x + 2*a)^2 + (d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*si
n(2*b*x + 2*a)^2 + 2*(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*cos(2*b*x + 2*a)), x) + (exp_integral_e(3, (2*I
*b*d*x + 2*I*b*c)/d) + exp_integral_e(3, -(2*I*b*d*x + 2*I*b*c)/d))*sin(-2*(b*c - a*d)/d))/(d^3*x^2 + 2*c*d^2*
x + c^2*d)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sin \left (3\,a+3\,b\,x\right )}{\cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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