Optimal. Leaf size=311 \[ -\frac{\left (a^2 (m+1)-b^2\right ) (\sin (c+d x)+1)^3 \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1} \left (\frac{(a+b) (\sin (c+d x)+1)}{(a-b) (\sin (c+d x)-1)}\right )^{\frac{m-2}{2}} \, _2F_1\left (\frac{m}{2},m+1;m+2;-\frac{2 (a+b \sin (c+d x))}{(a-b) (\sin (c+d x)-1)}\right )}{d e^3 m (m+1) (a-b)^3}+\frac{(a (m+2)-2 b) (\sin (c+d x)-1) (\sin (c+d x)+1)^2 \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1}}{d e^3 m (m+2) (a-b)^2}+\frac{(\sin (c+d x)-1) (\sin (c+d x)+1) \sec ^4(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^{m+1}}{d e^3 (m+2) (a-b)} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.511881, antiderivative size = 420, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2700, 2698, 2920, 96, 132} \[ -\frac{b (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )^{m/2} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (m+1,\frac{m+2}{2};m+2;\frac{2 (a+b \sin (c+d x))}{(a+b) (\sin (c+d x)+1)}\right )}{d e (m+1) (m+2) \left (a^2-b^2\right )}+\frac{a (\sin (c+d x)+1) (e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) \left (a^2-b^2\right )}+\frac{a 2^{-m/2} (a m+a+b) (1-\sin (c+d x)) (e \cos (c+d x))^{-m-2} \left (\frac{(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{\frac{m+2}{2}} (a+b \sin (c+d x))^{m+1} \, _2F_1\left (-\frac{m}{2},\frac{m+2}{2};\frac{2-m}{2};\frac{(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right )}{d e m (m+2) (a-b) (a+b)^2}-\frac{(e \cos (c+d x))^{-m-2} (a+b \sin (c+d x))^{m+1}}{d e (m+2) (a-b)} \]
Warning: Unable to verify antiderivative.
[In]
[Out]
Rule 2700
Rule 2698
Rule 2920
Rule 96
Rule 132
Rubi steps
\begin{align*} \int (e \cos (c+d x))^{-3-m} (a+b \sin (c+d x))^m \, dx &=-\frac{(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}+\frac{a \int \frac{(e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m}{1-\sin (c+d x)} \, dx}{(a-b) e^2}-\frac{b \int (e \cos (c+d x))^{-1-m} (a+b \sin (c+d x))^m \, dx}{(a-b) e^2 (2+m)}\\ &=-\frac{(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac{b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac{2+m}{2};2+m;\frac{2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac{\left (a (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac{2+m}{2}} (1+\sin (c+d x))^{\frac{2+m}{2}}\right ) \operatorname{Subst}\left (\int (1-x)^{-1+\frac{1}{2} (-2-m)} (1+x)^{\frac{1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) d e}\\ &=-\frac{(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac{b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac{2+m}{2};2+m;\frac{2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac{a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac{\left (a (a+b+a m) (e \cos (c+d x))^{-2-m} (1-\sin (c+d x))^{\frac{2+m}{2}} (1+\sin (c+d x))^{\frac{2+m}{2}}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-2-m)} (1+x)^{\frac{1}{2} (-2-m)} (a+b x)^m \, dx,x,\sin (c+d x)\right )}{(a-b) (a+b) d e (2+m)}\\ &=-\frac{(e \cos (c+d x))^{-2-m} (a+b \sin (c+d x))^{1+m}}{(a-b) d e (2+m)}-\frac{b (e \cos (c+d x))^{-2-m} \, _2F_1\left (1+m,\frac{2+m}{2};2+m;\frac{2 (a+b \sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right ) (1-\sin (c+d x)) \left (-\frac{(a-b) (1-\sin (c+d x))}{(a+b) (1+\sin (c+d x))}\right )^{m/2} (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (1+m) (2+m)}+\frac{a (e \cos (c+d x))^{-2-m} (1+\sin (c+d x)) (a+b \sin (c+d x))^{1+m}}{\left (a^2-b^2\right ) d e (2+m)}+\frac{2^{-m/2} a (a+b+a m) (e \cos (c+d x))^{-2-m} \, _2F_1\left (-\frac{m}{2},\frac{2+m}{2};\frac{2-m}{2};\frac{(a-b) (1-\sin (c+d x))}{2 (a+b \sin (c+d x))}\right ) (1-\sin (c+d x)) \left (\frac{(a+b) (1+\sin (c+d x))}{a+b \sin (c+d x)}\right )^{\frac{2+m}{2}} (a+b \sin (c+d x))^{1+m}}{(a-b) (a+b)^2 d e m (2+m)}\\ \end{align*}
Mathematica [A] time = 5.05326, size = 319, normalized size = 1.03 \[ \frac{\sec ^2(c+d x) (e \cos (c+d x))^{-m} (a+b \sin (c+d x))^m \left (\frac{b (\sin (c+d x)+1) (a+b \sin (c+d x)) \left (\frac{(a+b) (\sin (c+d x)+1)}{(a-b) (\sin (c+d x)-1)}\right )^{m/2} \, _2F_1\left (m+1,\frac{m+2}{2};m+2;-\frac{2 (a+b \sin (c+d x))}{(a-b) (\sin (c+d x)-1)}\right )}{(m+1) (a-b)}+\frac{a (1-\sin (c+d x)) (\sin (c+d x)+1) \left (2^{-m/2} (a m+a+b) \left (\frac{(a+b) (\sin (c+d x)+1)}{a+b \sin (c+d x)}\right )^{m/2} \, _2F_1\left (-\frac{m}{2},\frac{m+2}{2};1-\frac{m}{2};-\frac{(a-b) (\sin (c+d x)-1)}{2 (a+b \sin (c+d x))}\right )-\frac{m (a+b \sin (c+d x))}{\sin (c+d x)-1}\right )}{m (a+b)}-a-b \sin (c+d x)\right )}{d e^3 (m+2) (a-b)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [F] time = 0.178, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{-3-m} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 3}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{-m - 3}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{-m - 3}{\left (b \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]