Optimal. Leaf size=359 \[ \frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (-3 a^3 B+7 a^2 A b-4 a b^2 B+8 A b^3\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (-3 a^5 B+8 a^4 A b-17 a^3 b^2 B+30 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \sqrt {\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {(A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \]
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Rubi [A] time = 1.64, antiderivative size = 359, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3609, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 b \left (30 a^2 A b^3+8 a^4 A b-17 a^3 b^2 B-3 a^5 B-8 a b^4 B+16 A b^5\right ) \sqrt {\tan (c+d x)}}{3 a^4 d \left (a^2+b^2\right )^2 \sqrt {a+b \tan (c+d x)}}+\frac {2 b \left (7 a^2 A b-3 a^3 B-4 a b^2 B+8 A b^3\right ) \sqrt {\tan (c+d x)}}{3 a^3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {(A+i B) \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
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Rule 93
Rule 203
Rule 206
Rule 3609
Rule 3615
Rule 3616
Rule 3649
Rubi steps
\begin {align*} \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {3}{2} (2 A b-a B)+\frac {3}{2} a A \tan (c+d x)+3 A b \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx}{3 a}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \int \frac {-\frac {3}{4} \left (a^2 A-8 A b^2+4 a b B\right )-\frac {3}{4} a^2 B \tan (c+d x)+3 b (2 A b-a B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}} \, dx}{3 a^2}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {8 \int \frac {-\frac {3}{8} \left (3 a^4 A-14 a^2 A b^2-16 A b^4+9 a^3 b B+8 a b^3 B\right )+\frac {9}{8} a^3 (A b-a B) \tan (c+d x)+\frac {3}{4} b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{9 a^3 \left (a^2+b^2\right )}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}+\frac {16 \int \frac {-\frac {9}{16} a^4 \left (a^2 A-A b^2+2 a b B\right )+\frac {9}{16} a^4 \left (2 a A b-a^2 B+b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{9 a^4 \left (a^2+b^2\right )^2}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a-i b)^2}-\frac {(A+i B) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 (a+i b)^2}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a-i b)^2 d}-\frac {(A+i B) \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 (a+i b)^2 d}\\ &=-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}-\frac {(A-i B) \operatorname {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a-i b)^2 d}-\frac {(A+i B) \operatorname {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^2 d}\\ &=\frac {(A+i B) \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 A}{3 a d \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}+\frac {2 (2 A b-a B)}{a^2 d \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (7 a^2 A b+8 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{3 a^3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 b \left (8 a^4 A b+30 a^2 A b^3+16 A b^5-3 a^5 B-17 a^3 b^2 B-8 a b^4 B\right ) \sqrt {\tan (c+d x)}}{3 a^4 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 3.53, size = 383, normalized size = 1.07 \[ \frac {\frac {6 b \left (-3 a^3 B+7 a^2 A b-4 a b^2 B+8 A b^3\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {\frac {6 b \left (-3 a^5 B+8 a^4 A b-17 a^3 b^2 B+30 a^2 A b^3-8 a b^4 B+16 A b^5\right ) \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+9 (-1)^{3/4} a^4 \left (\frac {(a-i b)^2 (A+i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {(a+i b)^2 (A-i B) \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}\right )}{a^3 \left (a^2+b^2\right )^2}+\frac {6 (6 A b-3 a B)}{a \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {6 A}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}}{9 a d} \]
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 3.26, size = 2981039, normalized size = 8303.73 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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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.
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